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the triangle, we see th at cos a = (d/6) ; (d/ 2) = 1/ 3; therefore, we have 7r  a ::;
=
d/(6cos(3)
=
= p = d/(6cos(7r 
¢))
= d/(6 cos¢).
Therefore, d/ (6 cos ¢) ::; p ::; d/2. So far , we have described the crosssection in one quadrant. The entire volume requires a revolution around the z axis, so its description is
__d_ < p < ~ 0 < 6 cos ¢  2' 
()  < 27r
and 7r  cos 1
(31 )
<_
A,
'i'
< _ 7r.
1.5: nDIMENSIONA L EUC LIDEAN SPACE GOALS 1. Be able to extend the ideas of the previous sections to
jR n.
2. Be able to multiply matrices.
STUDY HINTS 1. The space ]Rn . Most of this textbook deals with the Euclidean spaces that we can visualize, jR2 and jR3. Many of the same properties hold in jRn. Vector addition, scalar multiplication, vector lengths, the dot product and the triangle inequality are defined similarly.
2. No cross product analog. The cross product in
]R3
does not have an easy analog in
jRn,
n ~ 4.
3. Standard basis vectors. The analogs of i, j, k are defined e j . The vector ej is (0,0, ... ,1, ...0) with 1 in the zth position. The vectors ej and ej are orthogonal if i =f j.
13
THE GEOMETRY OF EUCLIDEAN SPACE
4. Matrices. A matrix is a rect angular array of num bers . Unlike a determinant, a matrix has no numerical value. You should remember th at an n x m matrix has n rows and m columns. The (i, j) entry is the number located in row i, column j.
5. Mat rix m ultiplication. You should practice until matrix multiplication becomes second nature to you. Let the components of A be a i j and let those of B be bkl, where A is an m x p matrix and B is a p x n matrix. Then the components of A B are p
(A B )mn
=L
amjbjn .
j =l
We can only multiply an m x p mat rix with a p x n matrix, i.e., [m x p][P x n]. Note that the number of columns of A and the number of rows of B must be equal (p in this case). The result is an m x n mat rix ("cancelling" the p).
6. Noncommutativity of m atrix multiplication. In general, AB # B A . In fact, AB may be defined when BA is undefined . However , matrix multiplication is associat ive; i.e., (AB)C = A(BC) if the product AB C is defined . 7. Matrices and mappings. An m x n matrix can represent a mapping from ]Rn to ]Rm. To see this, let A be the matrix and let x be a vector in ]R n, represented as an n x 1 matrix, and y be a vector in ]Rm, an m x 1 matrix. T hen the matrix A takes a point in ]R n to a point in ]Rm by the equation Ax = y.
SOLUTIONS TO SELE CTED EXERCI SES 2. (a) Use the properties of leng hs and dot products:
(x + y) . (x + y) + (x  y) . (x  y) x . x + 2x . y + y . y + x . x  2x . y + y . y 2x· x + 2y· y = 211xl12 + 211Y11 2.
12
The figure at the left depicts the equation geometrically. By the law of cosines, we have
and
We also note that a + j3 = 1r, so j3 = 1r  a . T herefore , cos j3 = cos( 1r  a) =  cos a. Adding the two equations from the law of cosines yields 211xl1 2 + 211Yl12 ;: Ilx  Yl12 + I'lx + Y112. 4. To verify the CauchySchwarz inequality, we compute
Ix· yl Ilxll Il,yll Thus, we indeed have compute
Ix .yl = v'I7v'I7 < v4fJ90 = II'xlillyll.
x + y = (4,8,6,7) Indeed, we have
+ (2)(4) + (6)(1)1 = 17. '1'1 + 0 + 4 + 36 = J41. J9 + 64 + 16 + 1 = V90. 1(1)(3) + (0)(8)
and
For the triangle inequality, we
IIx + yll = J16 + 64 + 36 + 49 = v'l65 < 13.
Ilx + yll < 13 < 15 = 6 + 9 < AT + v'9O = Ilxllllyll.
14
CHAPTER 1
8. We compute
AB~
r:
1 1 ~31 1 1
1
A+B~[:
and
Expanding by minors across the first row gives det A
3:1 02 1 1 I + 1111 ~
detE
11 1
0
+ B)
11.
(a) For n =
1 11
~ ~

41
~ ~
I
= 6
2 1
n
2= 4,
1 = 3,
1 1_ 1151 311 1 1
det(AE) det(A
~
I
0
~1
5 1 1 31 1 1 1=12
and
= 8.
2, det('xA) =
I
'xall 'xa21
For n = 3, 'xall Aa21 Aa31
det( AA)
Aal2 'xa13
'xa 22 'xa23 'x a32 'xa33
'xall det('xAd  'xa12 det('xA 2 ) ,X . ,X2(a11 det Al  al2 det A2 ,X3 det A,
+ 'xa13 det(AA3)
+ a1 3 det A3)
where AI , A2 and A3 are 2 x 2 matrices obtained by expanding across the first row . Assume that for n = k, det('xA) Ak det A. The for n k + 1, det(AA) can be found by a process analogous to the 3 x 3 case:
=
det(AA)
'xall det('xAd  'xa12 det('xA 2 ) ,Xk+l det A,
=
+ ... + (_l)k 'xal,k+l det('xAk+d
where AI, A 2 , ... , Ak+l are k x k matrices obtained by expanding across the first row . By induction, det('xA) An det A for an n x n matrix A.
=
14. Assume,
as in the book, that det(AE) = (det A)(det E). Then det(AEC) det(AE) det C = (det A)(det E)(det C).
17. Multiply the two matrices to get the identity matrix: a [ e
b]
d
1 ad  be
[de b] a
Similarly, we can show that
= det[(AB)C]
=
TH E GEOMETRY OF EUCLIDEAN SPACE
15
SOLU TIONS TO SELECT ED R EVIEW EXERCISES FOR CHAPT ER 1
= =
=
=
(3,4, 5) + (I , I , 1) (4, 3, 6) 4i + 3j + 6k; v + w is the di agonal of a parallelogram whose sides are v and w. 3v 3(3,4 ,5) (9 , 12, 15) 9i + 12j + 15k; 3v has the same direction as v with length 3 times the length of v. 6v + 8w 6{3, 4, 5) + 8(1, I , 1) (26, 16,38) 26i + 16j + 38k; 6v + 8w is the diagonal of a parallelogram whose sides are 6v and 8w. 6v has the same direction as v with length 6 times th at of v and simi larly for 8w.  2(3 , 4, 5) (6,  8, 10)  6i  8j  10k;  2v is a vector in the opposite  2v direction of v with length twice that of v.
1. v + w
=
=
=
=
=
=
=
=
y
y
x
=
v ·w (3)(1) + (4)(1) + (5)(1) angle between v and w .
= 4; v · w is the number II v llll w ll cos B, where B is the
v x w is perpendicular to both v and w . Its length is the area of the parallelogram spanned by v and w . 4. (a) Using the pointdirection form of the line, we get l(t ) = (0 , 1,0) + t (3, 0,1). (b) Using t he pointpoint for m of the line, we get
l(t ) = a + t(b  a)
= (0 , 1, 1) + t [(O, 1, 0 
0, 1, 1)]
= (0 , 1, 1) + t (O, 0, 1) .
(c) T he normal to the plane is (a, b, c) = ( 1,1, I) , so th equation of the plane is
a(x  xo) + b( y  Yo)
+ c(z 
i.e.,
 1( x  1) 5. (b) v · w = (1)(3) + (2)(1) i
~
j
+ l(y 
1)  1(z  1)
zo ) = 0,
= ° or
x  y+z
1
~ ~
= 1.
+ ( l ){O) = 5.
k
~1 = 1 ~ ~
1
~ ~
1k
=i 
3j  5k.
7. (b) We compute IIvll = VI + 4 + 1 = V6 and Il w ll = V9 + 1 + 0= of the dot product and the result of xercise 5(b) gives
v'lO.
T hen the definition
6. (b) v x w =
;
cosO
1j I
1j + 1
5 5 = Jlvvllll· ww il = Y6v'lO = 5V60 2.Jl5 ·
(HAPTER 1
16
8. (b) The area of the parallelogram is the length of v x w. Using the result of exercise 6(b), we get Ilv x wll VI + 9 + 25 v'35.,
=
=
12. We compute the following dot products using the fact that u, v and w are orthonormal:
a .u
a· v a·w
(au + f3v + , 'w) . u = a. (au + (3v + ,w) . v = ,B. (au + (3v + ,w) . w = ,.
/
I
v
Geometrically, a . u is the projection of a on u; similarly for the others.
15. From the definition of the dot product and the fact that cos ()
=
cos ()
U ·
v . a ~ lIall,b . a + IIblillall 2 II v II II all 11vlliiall v .b lIallllbW + IIblla . b IIvllllbll IIvllllbll
u = lIull 2 , we compute
_
b· a
+ IIblili a ll
IIvll lIalillbl,1+ b . a Ilvll
Since t he angle between a and v is the same as the angle between b and v, the vector v must bisect the angle between a and b.
°
18. (a) Given that a · b = a' . b, this implies that a· b  a' . b = or (a  a') . b = 0 for all b. Choose b = a  a' to get lIa  a'1I2 = 0, so a  a' = 0, which means that a = a'. (b) The answer is yes . Ifax b = a' x b for all b, we can conclude that (a  a') x b = 0 for all b. Choose b to be a unit vector orthogonal to a  a'. By definition of the cross product, this implies lI a  a'il = 0, so a = a'. 22. (e) Note that the x and y coordinates lie in the third quadrant
of the xy plane. The definitions from section 1.4 are used: For
cylindrical coordinates, we compute
r ()
=
v'12 + 4 = 4.
1 7r + tan (y/x) = 7r + tan 1(1/V3)
JX2
7r
+ y2
+ 7r/6 =
77r/6.
Thus, the cylindrical coordinates are (4,77r/6,3).
y x
For spherical coordinates, we compute p
=
J x 2 + y2 + z2 = V12 + 4 + 9 = 5. 4>
= cos1(z/p) = cos1(4/5).
Thus, the spherical coordinates are (5, 77r /6, cos 1(4/5)), 23 , (b) Using the formulas from section 1.4, we calculate
x
z
= r cos () = (3)( V3/2)
y
and
y = rsin() = (3)(1/2).
Thus, the corresponding cartesian coordinates (3V3/2, 3/2, ~4).
For the spherical coordinates, we compute
are
THE GEOMETRY OF EUCLIDEAN SPACE
17
p= Jx2 + y2 + Z2 = / 27 +~+ 16 = V4 4
1 cjJ = cos (z /p) = cos 1(4/ 5).
5.
Hence, the corresponding spherical coordinates are (5,1r/ 6,cos 1 ( 4/5)).
24. (b) Using th formu las from section 1.4, we compute:
y
pcosBsincjJ = 2(0)( 1/ 2) = 0, psinB sin
z
Pcos
x
z y
= 2(..;3/ 2) = v'3.
T hus, the corre ponding oordinates are (0 ,  1, ..;3).
x
For the cylindrical coordinates, we calculate: r =
J x 2 + y2
= JO+l = 1.
We note that x = 0 and y = 1 , so B = 311'/2. Therefo re , the corresponding cylindrical coordinates are (1, 3rr/2, V3). 28. Using the methods of sec ion 1.5, we get
AB = [ and
~
0 0 0
n[~ n[! n 1
[~ ;][l
Clearly, AB :j: B A.
=
0
0 1 0
BA =
0
0
0 0 0
0
0
nr: n 0 0
0
33. (a) r is t he vector 7i + 2j, so W = F · r = 70 co B + 20 sinB . (b) If F has a m agnitude of 6, then F = 6 cos Bi + 6sinBj. Since B = 11'/6, we have F 6( J3/2)i + 6(1/2)j , and so W = F . r = (21 V3 + 6) footlb.
=
36. Su btract the fir t row from the second and third rows , t hen expand by minors along the first column to get 1 X 1 Y 1 Z
x2
1
x
0 y x 0 z x
y2
Z2
x2 y2 _ x2 z2 _ x 2
2 y2 _ x =I y  X z2 _ x2 z  x
I
(y  x)(z2  x2)  (z  X)(y 2  x 2) = (y  x)(z  x )( z  y) :j: 0 if X , y, z are all differen t. The last step used the fact that (z2  x 2) = (z  x)(z + x ).
=
39. (a) We recogniz a· (b x c) as a t riple product . Let a (a1 ' a2 , a3 ) and use a similar notation for band c. T herefore, 1 a t a2 a3 V = 6" b1 b2 b3 Cl
C2
Cs
(b) Use t he formula from part (a) and subtract the first row from th second row;
1 1 1
1

1 1
1
6
1
0
T hus , the volume is 1/3.
1
1 6
1
1
1
~ ~2 ~
110  2 1 1 = 6" 1 1 = 3'
18
CHAPTER 1
44 . A vector which is normal to t.he first plane is v = 8i + j + k. A vector which is normai to the second plane is w = i  j  k. T he cross product v x w is orthogonal to both normal vectors, so it should be parallel to both planes. We compute v x w = 2i  7j  9k , so the unit vector in question is (2i  7j  9k)/v'f34. 47. We want
v = ai + .Bj + ,k to
we know that
be such that
v'3 v ·j = 2 = Ilvllllill'
o
cos 30
!lvll = 1. so v
From the definition of the dot product,
= 2v'3,1 + /3.'J +,k .
Since v makes equal angles wi h j and k , we must have that .B = 1 = 1/20. T herefore,
TEST F OR CHAPTER 1
.B = , .
Since
Ilvll ==
1, we determine
(The answers are at the back of this book .)
1. True or false . If false, explain why. (a) Given vectors u= i + j in IR 2, v = i + j +k in IR3 and w = 2i k in IR 3, we have u·v = 2 and v· w = 1. (b) Gi ven two m atrices A and B, det(AB) is defined if and only if det(BA ) is defined. (c) For nonzero vectors a and b, the set a , b , and a x b, in that order, always forms a righthanded system. (d) Suppose a , b, and c are nonzero vectors in IR3 and a· b multiple of c or b and c are parallel vectors.
= a · c = 0.
Then either b is a
(e) A cross product is 0 only if one of the vectors is O.
2. Find an equation for the plane passing through the point (0,0,1) and parallel to the line contain ing t he points (1, 0, 2) and (1,2, 1). 3. Let A = (2 , 1,3) , B = ( 3,2, 4) and C = (1 , 2, 2) be the vertices of a triangle. Let a be the vector from A to B and let b be t he vector from B to C. (a) Calculate a and b . (b) Describe the triangle parameters sand t.)
~ ABC
using vectors of the form sa + tb . (Find restrictions on the
(c) Calculate the area of ~ABC . 4. Let u = 3i + OJ + k and v = i  2j + k have a common tail . Let () be the acute angle between u and v. Calculate sin (). 5. A parallelpiped spanned by the vectors (3,  2,1), (0 ,2 ,1) and ( 1, 0, 1) is filled with sand. The sand will fill another parallel piped spanned by the vectors  i, 2j and zk. Compute all possible values of z. 6. (a) In
]R4,
fi nd an equation for the line passing through A
= (1,3,  2, 0) and B == (0,1, 1,1).
(b) Show that the line in part (a) is not orthogonal to the line l(r) = (1 , 3, 2 , 0) +r(2, 0, 3, 1). 7. Let
(a) Compute AB.
,.
"
THE GEOMETRY OF EUCLID EAN SPACE
19
(b) Compute det(AB) and interpret geometrically. (c) If det(BA ) exists, compute it and interpret it geometrically. 8. Convert the spherical coordinates ( 2, 11"/4, 27r/3) to (a) Cylindrical coordinates. (b) Rectangular coordinates. 9. Let u
= (2, 1, 0,1) and v = (1,  2, 1,2).
(a) Verify the CauchySchwarz inequality for the given vectors. (b) Calculate the projection of v onto u.
10. A unique house has an inclined floor where the owner keeps his dog . The dog 's play area is a fenced en closure with vertices at A = (2,4,1), B = (1, 2, 0) , C = (0,0, 1), D = (1, 1, 1) and E = (3,0,0), as shown in the diagram.
z A y
(a) All of the vertices lie on a plane. Find the equation
of the plane.
x (b) Find the area of the dog's play area. (c) An object is located at (1, 1, 3). W hat is the min
imum distance the dog must leap off the floor to
reach the object?
D
21
               
2
    
 

DIFFERENTIATIO N
2.1: THE GEOMETRY OF REALVALUED F UNCTIONS GOALS 1. Be able to define a graph, a level cur ve, a level set, and a section.
2. Be able to graph a function
f : ]R 2 + ]R .
ST UDY HINTS 1. Notation. f : A C ]Rn + ]Rm describes a function f The domain is A, which is a subset of ]Rn. Points in A a re mapped to points in the range, which is a subset of ]R m . The only restriction on nand m is that they have to be positive integers.
2. Realvalued functio n . In the not ation, f : A C ]R n + ]R m, we restrict m to be 1 in this section. A realvalued function assigns a unique real number to each point in A.
3. Graphs. T he graph of f : A C ]Rn + ]R is dr awn in the sp ace ]Rn+l. If x is a point of A, the the graph consists of all points in ]Rn+l wi th the form (x,f(x)), where f (x) is a real number. 4. Level set. This is the set of x such th at f(x) is a constant. In ]R2, such a set is called a level curve and in ]R3, it is called a level surface. Level sets are important for graphing. 5. Sections. These are intersections of graphs with a vertical plane. Usually, the most helpful sections for graphs in ]R3 are he intersections with the planes x constant and y consta.nt.
=
=
6. Graphing in JR3. Often, the best way to d raw a graph in JR3 is to draw level curves for z constant. Then lift or drop the curves to the appropri ate "height" for z constant. Analyzing the sections helps complete the graph. It is a good idea to review how to sketch the graph of an ellipse, a hyperbola, a circle and a parab ola from your calculus or precalculus text.
=
=
7. Sketching planes. Many of us are poor artists, and as a result, threedimensional geometry may be frustrating due to this problem rather than a lack of mathematical understanding. Planes are sometimes easily sketched by plotting three noncollinear points (usually on the coordinate axes) and then passing a plane through them. 8. Spheres. The equation (xa )2+(yb)2+(zc)2 = r2 represents a sphere ofradius reentered at (a, b, c) . 9. Cylin ders . A surface in ]R3 is called a cylinder if x, yor z is missing from the equation . A cylinder can be sketched by drawing the level curve in the plane where the missing variable is zero. Then move the curve along the axis of the missing varia ble. 10. Graphs in ]R4. Example 5 gives a fu nction whose graph cannot be drawn on paper. To see the "graph," one can make a movie which shows the concentric spheres expanding.
CHAPTE R 2
22 SOLUTIONS T O SELECTED EXERC ISES
1. (a) To determine the level curves, we look at t he equation c = x  y + 2, where c is a constant . T he equation is the same as y = x + (2  c), which is the equation of a line with slope 1 and y intercept 2  c. The graph of f is a plane with intercepts at ( 2, 0, 0), (0 ,2, 0) and (0 , 0,2).
=
=
2. (b) We look at c 1  x 2  y 2 for c constant . This rearran ges to x 2 + y 2 = 1  c, which is the equation for a circle of radius v'1='C, centered at t he origin if c < 1. If c = 1, then the level set is a point at the origin. Th re is no level curve if c > 1. 3. (3) Substitute x
Since z =
1'2
= rcos (} and y = rsinB to get
does not depend on () , the sh ape of the graph does not depend on () .
5. For constant c, the equation c = y'10 0  x 2  y2 is equivalent to c2 = 100  x 2  y2 or x 2 + y2 = 100  c2 . T h I vel curves are circles with radii VI00  c2 , centered at the origin. So, for c = 0, 2,4,6 8, 10 the radii are 10, v'96, j84, 8, 6 and 0, respectively. Dr awing the level curves and raising them to t he appropriate z values, we obtain the following graph. T he graph of f (x , y) is a hemisphere. T he level curves and the graph a re shown here.
y ,
x
10. The level curves have the equation c = x / y or y = x / c for onstant c. Th se level curves are lines thro ugh the origin with slope l /e. One restriction is that y f:. 0. Next we discuss the x/ yo Noti e t hat when x is held constant, the sect ion are the hyperbolas c yz , graph z and when y i held constant , we get the lines c = x / z . Putting all this information together, we get t. he "twisted plane" as t he graph.
=
=
23
DIFFERENTIATIOI\l
y
12. The level surfaces have the equation c == 4:z;2 + y2 + 9z 2. T here is no level surface if c < O. If c = 0, then t he level surface is the origin. If c > 0, then we should look at the level curves for constant k, i.e., analyze c  9k 2 = 4x 2 + y2 . We recognize that if 9k 2 < c, then the level curves are ellipses which get smaller as Ikl approaches v'C/ 3. Similarly, we see that level sections parallel to the yz plane have the equation c  4k 2 = y2 + 9z 2 , which are ellipses with decreasing "radii" as Ikl approaches .;c/2. Also , the level sect ions parallel to the xz plane have the equation c  k 2 = 4x 2 + 9z 2 , which are again ellipses which get smaller as Ikl approaches .;c. The level surfaces are ellipsoids if c is positive.
17. If c < 0, the level curve is empty. If c = 0, t he level curve is the xaxis. If c > 0, it is the pair of parallel lines Iyl = c; that is, the lines y = c and y = c. In the yz plane, we sketch the graph of z = Iyl. Since x does not appear in the equation , we get a "cy,)inder" and this sketch is shifted along the x axis to obtain the graph in 1R 3 .
y
x 22. The equation can be written as (x 2  2x + 1) + y2 = 1 or (x  1)2 + y'2 = 1. In the xy plane, this is a circle with r adius 1, centered at (1, 0). Since z is not in the equation, z may take on any value, so the circle is sh ifted up and down the z axis.
CH APTER 2
24
y  .:;,.<+
(1 ,0,0)
x ,
/
25 . First, we sketch the graph of z ::: x 2 in the xz plane. T hen shift the graph along the y axis to get a parabolic ylinder.
I
I
/
/ /
/
y
x
29 . An equi alent equation is 4x 2 + 2z2 ::: 3y2. When y ::: 0, t h level curve is the origin. When y :j:. 0, we have level sections which are ellipses centered around the y axis. T he major axes
are parallel to the z axis and the minor axes are parallel to the x axis. T he ellipses get larger as IYI increases. To complete the graph, we note that when x ::: 0, the section is the straight lines z ::: ±y'372y. T hus our graph is two cones.
y
32. Subs itute x ::: rcosB and y ::: rsinB. Therefore, if x 2 + y2 ::: r2 :j:. 0, then z ::: f (x, y) ::: 2x y/ (x 2 + y2 )::: 2(r cos 0)( l ' sin B) / r 2 ::: r2 (2 sin 0 cos 8)j r2 ::: sin 20. Thus, t he function reduces to z ::: sin 2B if r' :j:. O. If  1 ~ z ~ 1, the level curve is one or two straight lines through the or igin satisfying z ::: sinO (see sketch below at left). The level curves are raised to a height z ::: sin 2B to obtain the graph of a "wrinkled plane." (See graph below at right. The dot ted line is a portion of the xy plane.) If z > 1 or z <  1, t here is no level curve. Notice that the plane becomes flatter as I' gets larger.
25
DIF FERENTIATION
x
2.2: LIMITS AND CONTINU ITY
GOALS 1. Be able to defi ne the following: open disk, open set , neighborhood, boundary point , limit ,
continuous. 2. Be able to determine where a function is continuous. 3. Given a function, be able to compute a limi t or show that a li mit does not exist.
STUDY HINTS 1. Theoretical section . This section is not essential for computation al purposes. Your instructor may choose not to emphasize the material in this section . Use your lectures to determine how important the material is for your course. 2. Definitions , (a) An open disk around Xo is the se of points x such that Il x  xoll < 1' . It is denoted Dr (xo) . Note the strict inequality. (b) An open set U is a set such th at every xo has an open disk entirely within U. You will need to find an r when proving that a set is open. (c) A neighborhood of Xo is an open set containing Xo. (d) A boundary point of a set A has no neighborhood en irely inside or entirely outside of A.
3. Review. You should review the concepts of limit and continuity from your onevariable calculus textbook before continuing.
4. Limits. In the defin ition, be aware that Xo doesn't have to be in A ; Xo may be on the boundary. Also, f (xo) does not have to be defined. We are only interested in the points x n ar Xo . For proofs, we need to choose U, which is dependent upon N. 5. Properties of limits. Most of th se are what you would intuitively xpect. Note hat for multiplication and division, the m apping is into }R I.
6. Conti nuity. Analogous to the onevariable definition, a functi on is continuous at Xo if lim f(x ) = f( xo),
X~ X o
I.e. , the limit equ als the function value. T he limi t on the lefthand side is concerned about points near Xo. The righthand side, f(xo), is concerned about the point Xo itself.
CHAPTER 2
26
7. Nonexistent limits. Showing that the limit of f(x, y) does not exist is sometimes simple. To show a limit does not exist , one can, for example, look at the limit of f by first holding x constant, then holding y constant. If the two values differ, the limit does not exist.
SOLUTIONS TO SELECTED EXERCISES 2. Take
°<
r ~ y, th n for all points
(x , y), the open disk Dr (x,y) C B. Thus, B is open.
5. (c) Recall the defin ition of the derivative:
f ' (X o )  I'1m f(x o + h)h  f (xo) . h tO
By letting f(x ) = e"" and xo = 0, we get
e"  1 = lim eO+ h 
lim  hh t O
"to
d
eO
= _ eX dx
h
I x =o
= 1.
6. (c) From onevariable calcul us recall that lirn.,t 0 (sin :z: / x ) = 1. This fact an al 0 be verified by using I'Hopital's rule . Using this fact and the properties of limits, we compute
.2
lim sm 2 x _ I'1m
xtO
x
(.)2 ( sm.)2
x _ 12 Sill X
x t O
X
_
}'

1m
xt O
X


1
.
g. (c) It is obvious that the limit of the numerator is 0, and the limit of the denominator is 2 # 0, so t he limit of he quotient is 0/ 2 = O.
10. (b) First, hold y = 0 constant and let x approach O. Then use I'Hopital's ru le:
. cosx  1  (:z:2 / 2) . sinx  x hm = 11m X4 ""to 4x 3 xtO
. = xt hm O
cosx  1 . 12:z:2
This last limit tends to 00 since the numerator tends to  2 and the denominator tends to 0, Thus, the limit does not exist.
11, (a) Let t
= xy and use continuity of compositions (theorem 5) to get · I1m
sin xy xy
}'
sin t = 1,
  = 1m  
(x ,y ) t( O, O)
t tO
t
13. (c) Use the fact that the limit of a vector is the limit of each component (theorem 3(v)). So we get limx t l (x 2 ,e X ) = (l,e). 16. (a) We will use theorem 5 (continuity of compositions). Note that f (:z: ) = (1  x)8 + cos (1 + x3 ) is the sum of two functions . The first is u 8 with 'U = 1  x. Since tJ is continuous, theorem 5 says that (1  x)8 is continuous. The second function is cos v with v = 1 + x 3, Again, since v is continuous, theorem 5 says that cos (l + x 3 ) is continuous. T he sum of continuous functions is continuous, so f(x ) is continuous. 17. (a) We can make a function continuous by equating f (xo ) and limxt xo f(x) . As in exercise 11(a) , we can Ie t x + y , so
=
· 11m ( x ,y )t(o,O)
sin(:z: + y) _ l' sin t  1  Im  . x +Y tt O t
Thus, we let
sin (x + y)
~'''1 for x x +y
=
"
= y = 0,
27
DIFFERENTIATION
and we have a continuous function. (b) First , we note that if x = y , then xy
lim ') ( x . y ) ~ (o. O ) xOn the other hand , if x
.
+ y"
x'.l
1
= lim 2x., = 2 ' x~o
= y, then xy
lim
(x.y) ~ (o.O ) x 2
+ y2
.
_x 2
1
= xhm   =  ~o 2x 2 2
Since the Hmiting value depends on the direction of approach, the limit does not exist at the origin , so it is not possible to make the function continuous at the origin. 22. (b) We want to find a 6 for every N > 0 such that 0 < x < 6 implies that 1/1xl > N. Let 0 < 6 < l i N , then Ixl < liN implies that 1/1xl > N. T his is not true if t he absolute values are omitted , i.e., limx~o(llx) may be +ex> or ex> depending on which side of 0 we are approaching from (see the figure below.) y
x x
y=
l/lxl
y
= l/x
27. (a) By the triangle inequality, 1a. 3 + 3a 2 + al < la 2 1+ 31a 2 1+ lal, since lal < 1 (we assume it is small, since 6 ,is small), this is less than (or equal to) 514 Choose 6 < 1/500 , then for lal < 6, la 3 + 3a 2 + al ::; 1/100 (note that this is a very rough estimate; a bigger 6 would probably work if we work harder to improve the inequality.)
2.3: DIFFE RENTIATION GOALS 1. Be able to state the definition of partial derivat ives.
2. Be able to compute a partial derivative or a matrix of partial derivatives. 3. Be able to compute a gradient . 4. Be able to compute a tangent plane.
STUDY HIN TS 1. Notation. Class
en means that the nth derivative is continuous.
2. Partial derivatives . Know the definition
of = l In1 ' f( Xl, "" Xi + h, ... , x n)  OXi h~ O h
f(Xl' ... , Xi, ... , Xn)
=..:........:....:.'''''~~''=''''''''
To compute ofI OXi in examples, consider all variables except Xi to be constant and differentiate by onevariable methods . Differen iation is performed with respect to the variable Xi.
CHAPTER 2
28 3. No tation lor partial deriv atives. In many texts,
Ix
is used for 8f/8x. If we wish to evaluate
at a given point, we write
81
ax
1(xo,Yo)
'
fx l(xo ,yo) '
~: I(:Co ,Yo)
or
if
z
= f (x , y).
4. Tangent plane. The tangent plane to the graph of a function f (x , y) at (x o, Yo, f(xo, Yo) ) is given by
z= f(x o,yo)+ 8f ax l
81 1 (x xo) + {)
(Y Yo).
y (Xo ,Yo)
(Xo,Yo )
T his equation is also used to compute linear approxi m ation . Compare this equation to the equation of a tangent line and the linear approximation in the onevariable case. See section 2.6 for a gener lization.
5. Differ ntiability. Equation (2) in the text tells you that f : R 2 + is differentiable if t he tangent plane approaches f (xo,yo ) as (x,y) approaches (x o,yo ). Now, if f : U C ]Rn + ]Rffl, then · Il f(x )  I(xo )  T(x  x o) 11 0 I1m (2) X+Xo Il x  xoll  , where T : ]Rn + ]Rm is the derivative. You should be able to get equat ion (2) from this definition . T his definition of differe ntiab.ility is most important for theortical work .
6. Gradient. T he gradient is a vector whose components are the partial derivatives of f , with 8f/8x; in the ,t h position. Here, fis a realvalued function. Thi operation is denoted by the sy mbol 'V . Someti, es , it is denoted "grad." For a function I : ]R3 + JR. , you should remember the form ula
7. Derivative of vectorvalued fu.nctions. Con ider a fu nction I : R n + ]R m. T he derivative is an m x n matrix of partial derivativ s. The range consists of vectors with m components. Think of t he components as realvalued vector fu nctions; th n each row of the derivative matrix is a gradient. The derivative m atrix of f, evaluated at Xo, is denoted Df(xo) . 8. Important facts . Diffe rentiability implies continuity of the fun ct.ion, but continuity does not imply differen tiability. The existence of continuous partial derivatives im plies differentiability but the converse is not true. If a funct ion is differentiable, then its partial derivatives exist , but the converse is, again , not true.
SOLUTIONS TO SELECTED EXERCISES 1. (b) Holding y constant and differentiating with respect to x, we get 81/8x = ye xy . By symmetry, 8 f / 8y = xe XY • In this problem, all we did to compute fJ!I fJy was to switch x and y. T his is what we mean by "symmetry" ; it only works for functions that are unchanged when x and yare swapped.
2. (b) Hold y con taut and use the chain rule to differentiate wi th respect to x. We get f}z 
1
1
.
1
ox  J1+ xy 2y'1 + xy
At (1,2), 8z/8x
'y 
y
.
 2{1 + xy) '
. '1 1 ar y,
SImI
8z 8y
= 1/3 and 8z/ oy = 1/6. At (0,0), az/ax = az/ 8y = o.
3. (b) Holding y constant and using the quotient rule, we calculate:
8w _ 2x (x 2  y2 )  2x (x + 2 + y2) 8x (x 2 _ y2 )2
x
2(1
+ xy) .
DIFFERENTIATION
29
Holding x constant and differentiating with the quotient rule , we get:
ow _ 2y(x 2  y2)  (2y)(x ay (x 2  y2)2
+ 2 + y2)
_ 4x 2y 2  (x  y2)2 .
4. (b) We must show that the partials are continuous in the domain : of/ax = l/y  y/x 2 = (x 2  y2)/ x 2y, which is continuous for x f. 0 and y f. OJ a f/ ay = _X/y2 + l/x :;;:: (y2  x 2 )/xy2, which is continuous for x f. O. T hus, f(x) is C l since its partial derivatives are continuous.
6. (b) The equation of the tangent plane is given by Z
= Zo + [fx(xo, yo)](x 
xo)
+ [fy(x o, yo)](y 
Yo).
Using .the result of exercise l(b) , we compute:
afl  l' ax (0 ,1)  , Therefore, the tangent plane is z
afl =0', f( 0,1 ) = 1. oy (0,1)
= 1 + l(x 
0)  O(y  1) or z  1 + x.
7. (b) The first row contains the partial derivatives of xe Y + cosy. The second roW contains those of x, and the third row con tains those of x + eY • The fi rst column contains the partial derivatives with resp ect to x, and t he second colu mn contains those wi th respect to y. Thus, the matrix of partial derivatives is
eY xeY  sin y 1 0
[ 1
1.
eY
8. (b) The function f is a m apping from JR3 to JR 2, so the matrix of the partials is 2 x 3. Let It (x, y, z) x  y, the first component of f. Similarly, let Iz(x, y, z) y + z. Then
=
=
af l Df(x, y , z)
=[
ay
a;: aiz
alz
ax
ay
11. Using t.he result of exercise 1(b), x(of/ax )
%1 ] _ [1
olz

1 ()]
0
1
1
.
az
= xye xy
= y(af/ay) .
12. (b) Use the linear approximation form ula, which is the same as the equation of the tangent plane: z = Zo + [fx(xo, yo)](x  xo) + [f y(xo , yo)](y  Yo). Let z = f(x, y) = x 3 + y3  6xy, IO = 1, I = 0.99, Yo = 2 and y = 2. 01. We com pute:
8flax
= 3x 2 
6y,
a f /ay=3 y2_6x, Therefore, our linear approxim ation is z value is 2.8485 .
~
so
fx(1 ,2) = 9.
so
/ y(1, 2) =6.
 3 + (9)( 0.01)
+ 6(0.01) = 2.85.
13. (c) The gradient is defined as the vector (of/ox , of/oy , a f/az) . Thus,
14. (c) The tangent plane is defined by \If(xo) . (x  xo)
+ 2e (z 
Exercise 13 (c) gives us
= ei + 2ek,
1) = 0 or x + 2z = 3.
\l f (1, 0,1)
so the tangent plane is e(x  1)
= O.
The actual
CHAPTER 2
30 17. We compute 'V/ (0,0 , 1)
= (2:1:, 2y ,  2z )1(0 ,0,1) = (0 , 0,  2) =  2k.
20. We want to find T in equation (4). By linearity, / (x)  f( x o) by h , we want to find T so that ·
I hl:To
= f{ x 
xo ). Denoting x  Xo
°
II/ (h)  Thll Ilhll .
=
If we choose T f, the numerator vanishes for all h, so this T sati ties the condition; that is, t he deriv tive of a linear map is the map itself. For exam ple, in one variable, consider f (x) = ax . From onevariable calculus, T = f ' (x o) = a fo r all Xo.
2.4: INTROD U CTIO N T O P ATHS AND CURVES G OALS 1. Given a path, be abl to compute t he velocity vector. 2. Be able to find a tangent line for a given path.
ST U DY HIN TS 1. Paths . A path is a "formula" that describes a curve in space . The picture of the pat h, which we can draw on paper, is called the image of the path or the curve of the pa th . 2. Path images. Often, it is convenient to express a path in terms of x and y when you want to know the image of a path. This is done by elim inating the parameter. For example (x, y) = (t2,t4) means t so y = t 4 = (J x)4 = x 2. Caution: In this example, x = t 2, so x is always nonnegative.
=..;x,
3. Circular functions. If a path is parametr ized by an expression involving cos t nd sin t , the parameter can usually be eli minated by squaring and adding. Use the identity cos 2 t +sin 2 t = 1 and 0 her trigonometric identities. 4. Velocity. T he velocity vector 's components are first deri vat ives of the components of the path. The velocity vector is tangent to the path. 5. Tangent lines. It is easy to find a tangent line if you rem ember that a line can be described as x + tv. The vector x is chosen to be a point on the path at to and v is the velocity vector c' (t o). Thus, the tangent line to a path i
l(t ) = c(to ) + (t  to )c'(to). SOLUTIONS TO SELECT ED EXERC ISES 1. From y = 4 cost , we get y/4 = cost. Use the fact that
and substitution to obtain 1 Since 0 ~ t ::; 211", th curve is an ellipse with y intercepts at ± 4 and x intercepts at ± l .
4
x
31
DIF FE RENTIATION 3. As in example 1, c(t) has the for m (xo, Yo , zo) + tv,
where (xo, Yo , zo) = ( 1,2,0) and v = (2, 1,1). T hus,
c(t) is a line in R3. Specifically, it is the line t hrough ( 1, 2,0) with direction (2, 1,1 ).
z 2
(1,2,0)
x 7. A path's velocity vector is found by differentiating the individual components. In this case,
d 2 d 3 d ) ( dt (cos t ), dt (3t  t ), dt (t)
1" ( t)
( 2costsint,3  3t 2 , 1). 12. T he tangent vector to a curve c(t) tangent vector is (2t, 0).
= (x(t), y(t»
is c' (t )
= (x'(t),lI(t».
15. T he equation for the tangent line is l(t) = c(to) + (t  to) c'(t o). Here , to (3 cos 3t,  3 sin 3t, 5t 3 / 2 ). T herefore , the tangent line is
l(t) = c(l)
+ (t  l) c'(I)
(sin 3, co 3, 2)
In this case, the
= 1 and
c'(t) =
+ (t  1)(3 cos 3, 3 sin 3, 5) + t(3 cos 3,  3 sin 3, 5).
(sin 3  3 cos 3, cos 3 + 3 sin 3, 3)
17. F irst , we need to find the tangent lin at to. We compute c (2) = (4,0 ,0) and c'(t) (2t , 3t2  4,0) , so c' (2) (4,8, 0). T hus, the tangent line is l(t) (4,0,0) + (t  2)(4 ,8, 0). T he position of the particle a t t l = 3 is 1(3) = (4,0,0) + (3  2)(4 ,8, 0) = (8 , 8, 0).
=
=
2.5: PRO PERTIES OF THE DERIVATIVE
GOALS 1. Be able to state the ch ain rule. 2. Be able to compute a partial derivative by using the chain rule .
STUDY HINT S 1. Chain rule. Suppose
f is
a funct ion of Yl, Y2, ... , Yn and each Yi is a function of x. Then
df dY2 ' =of  .dYl  +of  . + ... +of . dYn
dx
aYI
dx
aY2
dx
aYn
dx '
ay,
Notice how each term appears to be dJl dx wit h a.nd dYi "cane ling." However, bewar that the "su m" on t he righthand side is df / dx, not n times df / dx. Also note the different d's: "a" is for a fun ction of many variables, while "d' is for a function of one variable. 2. Multivariable chain rule. T he multivar iable chain rule st ates th at
D(f 0 g)(xo ) =::; Df(yo) Dg (x o) , where Yo = g(xo) . T his is he product of two deriv tive m atrices , so any desired partial m y be obtained by multi plica ion .
32
CHAPTER 2
3. Chain rule, gradient relationship. Know that if fis r aivalued and h (t )
== f( c (t)) , then
dh
dt = \7f(c(t)) . c' (t ).
SOLUTIONS TO SELECTED EXERCISES
f is differentiable by the sum rule. Its derivative is
2. (b) The function
af af] [ az' ay
= [1 , 1] .
(f) The function is differentiable by the chain rule. We know that z2 and y2 are differentiable by he product rule and that 1 Z2  y2 is differentiable by the sum rule. Also, the square root function is differentiable (where its argument is positive), so th entire function is differentiable . Its derivative is
3. (b) This is a spe ial case of the first special case of the chain rule: dh af dx dx = ax· dz + af al az +
au .
5. (b) First, we compute I(c (t))
af du af dv au . dx + av . dx du af dv dx + av . dx·
= exp(3t 2 • t 3 ) = exp(3t 5 ), so f' (t)
= 15t4 exp(3t 5 ).
Next, by the chai n ruie, a 1 dx dy dl =  .  + aayf .dx ax dt dt
Substitute x
= ye '" Y . 6t + :te'" Y • 3t 2 .
= 3t 2 and y = t 3 to get dl j dt
= =
t 3 exp (3t 5 ) • 6t + 3t 2 exp(3t 5 ) · 3t 2 15t4 exp(3t6 ),
which is the same as we got from a direct computation. 6. (b) Take th derivatives of each component to get c'(t) = (6t,3t 2 ) . 9. Substitute u = e"' Y and v
=x 
y to get
By the chain rule, D(f 0 9)( X, y) = D/(u, v )D9(X, y). First, we calculat
Df(u, v) =
[ ~~~~~: ~~~~~~] = [ sec\uu
When (x, y) = (1 , I ), we have g(l , 1)
= (e 1  1 , 1 
D/(I, O) =
1)
1)
= (1, 0) . Hence,
[12 1° ].
=;:] .
DIFFERENTIATION
33
Next , we calculate
Dg(I, 1)
so Therefore
D(f 0 g)(I,
1) == [~
Alternatively, we may calcul ate D (f
0
~1] [~ =~]
=[
1 1
1 ] 1 .
[~ ~2]'
=
g)(x, y) directly from (f
g)(x, V) .
0
13. (a) By the chain rule,
dT = V'T(c(t)) . c'(t), dt
where c( t)
= (cos t, sin t).
Substituting x
Differentiate:
= cos t, Y = sin t gives
(2 cos te sin t
V'T(c(t)) c' (t)

sin 3 t, cos 2 te sin t

3 cos t sin 2 t),
(  sin t, cos t) .
Thus,
(b) Plug in x
' 2 ')
dT = 2sintcoste Sm. / +sin 4 t+cos 3 te smt 3cos tsin~t . dt = cos t, Y = sin t into the expression for T and get
T(t) = cos 2 te sin t

cost sin 3 t.
Using techniques from onevariable calculus, we have
. t cos team . t + cos 3 teS int + dT = 2 S1l1 ~
· 4
S1l1
' ? t  3 cos 2 t S1l1~ t,
which is the same as the answer obtained by the chain rule in part (a). 17. (a) If y(x) and G are differentiable, then by the chain rule, and the fact that G(x, y(x)) is constant, dC = aC . dx + aC . dy = aC + oG . dy = 0 dx ax dx ay dx ox oy dx . Solve for dy/dx: If oC/ay
¥ 0, then dy dx
oC/ox  oC/oy'
(b) As in part (a), we differentiate C I and G 2 by the ch ain r ule:
dCI dx
= aC I
+ aC I . ciYI + oC I . dY2 ox aYI dx 8Y2 dx
=0
and
dC 2 dx
= aC 2 + ox
aC 2 . dYI + oC 2 . dY2 = O.
OYI dx oY2 dx
Assuming YI (x), Y2 (x) and G are differentiable and
I
aCt/aYI oCdOYI
aC I /aY2 8C2/ aY2
11=0
for all x ,
then we can solve for dyt/dx and dY2/dx. Rewrite the two equations as
oC I dYI .  + oG _I .dY2 
oYI dx oG 2 dYI . OYI dx
oY2 dx oG 2 dY2 +_. aY2 dx
oC I
ox OC 2 ax
(1)
(2)
34 Mul tiply (1) by OG 2 / OYl and (2) by  8GI/OYl . Add t h two together to get :
8G 2 8G 2 8G l 8G 1 . + _ . dY2 8x 8Yl 8x 8Yl dx =  8" G=l ;<8~ G;:=2~;; 8;: G;2 ;8"G=:l" 8Y2 . 8Yl  8Y2 . 8Yl Sim ilarly,
8G 2 8G 2 . 8G l
 _8G l ._+
dYl
d;' 
8x 8Y2 8x OY2 8G l OG2 8G 2 8G l . ._  . 8Yl 8Y2 8Yl 8Y2
(x, y, z) = 0. L t x = I (y, z ), y = g( x , z ) and z = h(x , y); this means' F(f(y, z), y, z) = 0, F (x, g(x, z ), z ) = and F (x y, h(x , y)) = 0. Differentiating F(f(y, z) ,
18. Begin with
°
with respect to y and z , we get
and
8x
F~ 8z
= 0.
+ Fz
Similarly, differentiating F(x , g(x , z ),z) and F (x,y, h(x , y)) with resp ct to x and z and y z, respectively, we get
8y Fx + Fy 8x 8y Fy 8z + Fz
=
0,
0,
and
Fx
8z
+ Fz 8x =
8z Fy + Fz 8y
=
0, 0.
Solving (2) for 8x/ 8z, (3) for By/ ox and (6) for oz/ 8y gives
ox 8z assuming that none of the partials Fx , Fy and Fz are O. Multiply and get l. 20. (a) Use the definition of the partial derivative:
81 (0, 0) 8x
= lim 1(0 + h, 0) h t O
T he last step holds since O/ h =
°
1(0 , 0)
h
=
h(0) 2 _ 0 lim h 2 + 02 h
ht O
= lim Q= O. htO
h
for all h =F O. Similarly, O(h)2 _
°
81 (0,0) = lim 1(0 , + h)  1(0 ,0) = lim 02 + h2 8y h tO h h t O h
°
= lim Q= O. h tO
Therefore, 01/8x and 01/ 8y exist at (0 , 0) and equal O. (b) If g (t) = (at, btl, then
(at )(bt)2 (f 0 g )(t) = (at )2 + (bt )2
ab 2t 3
= (a2 + b2)t2 =
ab 2 a2 + b2t,
h
35
DIFFERENTIATION so
(J
0
, g) (t)
ab 2
= a 2 + b2 .
=
On the other hand, from part (a) , we have "V f(O, 0) (of /ax, of/o y)(O , 0) = (0 , 0). Aliso, we compute g'(t) = (a, b), so "V f (O, 0) . g'(O) = (0,0) . (a, b) O. This verifies, as required, that the chain rule does no t a pply to this function.
=
ow/ax on the lefthand side means the partial derivative of w(x , y,g(x , y)) with respect to x, holding y constant. T he term ow/ox on the righthand side means the partial derivative of w(x, y , z ) with respect to x holding yand z constant . These two terms are not equal because different independent variables are held constant. Thus, the reasoni ng is indeed flawed .
24. The t.erm
2.6: GRADIENTS A ND D IRE C TIO N A L D ERIVAT IVES GOALS 1. Be able to define a directional derivative.
2. Be able to compute a directional derivat ive. 3. Be able to explain the significance of the gradient. 4. Be able to understand the relationships among the direct ional derivatives, the gradient , the
tangent plane and level sets.
STUDY HINTS
=
1. Important example. Many ex m ples in this book use the fact that "V1' = r/1', where r (x , y , z ) and r = Ilrll = Jx 2 + y 2 + z2. T his is derived in example 1. Much time can be saved by remembering this result . 2. Definition. The directional derivative is defined to be
!
f (x + tv) It=o
or
. f(x I 1m
11+0
+ hv) h
f(x)
.
The directional derivative gives a rate of change of f in the direction of v at x .
3. Geometric interpretation. Suppose v = ai + bj is a unit vector , (xo , YO) is a given point and z := f(x , y) is a urface. T he d irectional derivative tells us the "slope" of a curve at (xo , Yo) in the direction of v . T he curve is formed by the intersection of the surface with the pl ane described by the set of points sv + tk. If v is not a unit vec tor , then the "slope" may be determined by norm alizing v to be a unit vector . 4. Relation to partial derivatives . T he partials of/ox , of/ oy and of/ OZ are t he directional derivatives of f in the directions i , j and k , respec ively. 5. Computing direction al derivatives. To compute t he direct ional derivative of fin the direction of v , the easiest formula to use is "V f (x)· v . T he directional derivative is a scalar, not a vector . The vector v is often chosen to be a unit vector . 6. Gradient pmperties. Recall that "V f = (af/ ox )i + (of/oy) j + (af/oz)k. You should know that "V f points in the d irection in which f is increasing fastest and  \l f points in the direction of fastest decrease. The gradient of f is always orthogonal to a level surface of f 7. Tangent plane. In terms of the grad ient , the equ ation of the tangent plane at Xo is "V f(xo) . (x  xo) = O. This generalizes the formul a given is section 2.3.
36
CHAPTEP
SOLUTIONS TO SELECTED EXERCISES 1. The directional derivative of / (x) at Xo in the direction v is "V / (xo) . v . In thi case,
V' / (1,1,2)· (1/ V5,2/ V5, 0)
=
(z2,3 y 2, 2xz )I(1 ,1,2)' (l/V5,2/ v'5,0)
(4, 3, 4)· (1/ vf:5,2/vI5,0)
= 1O/v15 = 2v'5. 2. (b) Given / (x,y)
= \n(lx2 + y2 ), we compute
"VI (x,y) =
= (1 , 0) and the directional derivative i V' /(1 ,0) · v = 2/ vI5. 3. (b) Given /(x, y, z ) = eX + yz, we calculate "V /(x, y, z) = eXi + zj + yk and thus V' / (1 ,1 , 1) = At (1 , 0) , V'/ (1 , 0)
ei + j + k. The unit vector parallel to (1, 1 , 1) is (1,  1, 1)/..)3. Thus, the directional derivati is V' /( 1, 1, 1) . (1,  1, 1)/..)3 = e/ ..j3. 4. (c) For a function of three variables, I (x, y , z), the tangent plane to the surface is
V' /(xo, Yo , zo) . (x  xo, y  Yo, z  zo) = 0. To use this formula, we need to describe t he surface f(x, y, z) = constant. In this Cal: f (x,y,z) = xyz = 1, so V'/ (x,y,z) = (yz,xz , xy) and at (1,1 1), V'f = (1, ] , 1) . Therefor the desired tangent plane is (1,1 , ] ) . (x  1, y  1, z  1) = 0, i.e., x + y + z 3.
= =  sinxeosy and, by symmetry, Zy =  sinycosx . .
5. (b) We have z = (cosx)(cosy), so Zx (0 , rr/ ,0), we compute zX = 0 and zy =  1. Therefore, the equation of the tangent plane i_
z
= Zo + [zx (xo, yo)]( x 
i.e., z + y
xo) + [Zy(xo . yo)]( y  yo ) = 0+ O(x  0)  l(y  rr/ 2) ,
= 7r/ 2.
6. (c) Given f (x, y, z) = 1/(x 2 + y2 + z2 ) = 1/1,2, we have Ix =  2x / (x 2 + y2 + z2)2 By symmetry, I y =  2Y/ I·4 and 10 =  2z/ r4 . Then "VI = ( 2/ r4 )( xi + yj + zk) where r = xi + yj + zk and r = l x2 + y2 + z2.
=  2x/r =  2r/r
7. ( ) The direction of fastest increase is along the gradient vector. U ing the result of exerci: 6(c), we get the direction of fastest increase as 1(1,1,1 ) = ( 2/9)(i + j + k ). 8. The gradient vector is normal to a surface. Here, we have I(x, y, z) = x 3 y'3 + y  z + 2 = _ so Ix = 3x2y'3, Iy = 3x 3 y2 + 1 and /z = 1. At the point (0 , 0,2), we compute Ix = 0, Iy = and f. = 1. Therefore, a normal vector is V' / (O, 0,2) = j  k. Normalize this to get the un normal: (l/.../2)(j  k) . 13. (b) By definition, "VI = (81/ 8x,8// 8y,8f/8z), so "VI = (yzexyz,xzexY"xye xyO). Giv g(t) = (6t,3t 2 ,t 3 ) , we differentiate each componen to get g' (t) = (6 ,6t,3t 2 ) _ From Seetio 2_5, we know that
(lo g)' (I ) =
=
V'/(g(I)) . g'( I) = exp(1 8t 6 )(3t 5 ,6t4 ,18t3 ) . (6 ,6t ,3t 2 )lt=1 e18 (18 + 36 + 54)
= 108e 1S _
17. By definition, "V I = (Ix , Iy ). Since f is independent of y, I y = 0 and given that I (x, y) = g( z we have Ix = 81/ 8x = g' (x ). Th refore, "V I(x, y) = (g'(x ), 0).
DIFF ERENTIATION
37
20. T he direction in which the alt it ud is increasing most rapidly at the poi nt (x, y) is \7z(x, y ) = (2ax ,  2by). At the point (1,1) , \7 z(l , 1) = 2(a, b) , so the desired direction is  (ai + bj )/Ja 2 + b2 • If a marble were released at (1, 1) , it will roll in t he di rection at which the altit ude i decreasing most rapidly, so the m arble will roll down in the direction \7z, i . . , (ai + bj )/Ja2 + b2 . 24 . (a) We want to maximize I (c(t)) = (cost )(sin t ). Set the first derivative equal to 0: df/dt :::: (sint)(sin t ) + (cost )(cos t) = 0, so we get cos2 t = si n 2 t or t :::: ±(7r/ 4 + n7r ), where n = 0, 1, 2, .. .. Since t .::; 211', we only want t = 7r/ 4,3 7r/4 , 511'/ 4, 711'/4. Evaluating at these points, we get l (c{7r/4)) l (c (57r/4)) 1/2 and l (c (37r/ 4)) :::: l(c(77r/ 4)) ::::  1/ 2. T herefore, the maximum value of lalong the curve c(t ) is 1/ 2 and the min imum value is  1/ 2.
°.: ;
=
=
SOLUTIONS T O SELECTED R EVIEW EXERCISES FOR CHAPTER 2 1. (a) Since 3x 2 and y2 are nonnegative, t here is no level curve if z < O. If z :::: 0, then t he I vel cu rve is the origin. If z > 0, then t he level curve is an ellipse wit h the m aj or axi parallel to the y axis and the minor axis parallel to the x axis. The ellipses get larger as z increase . Put all of these level curves together to get an elliptical paraboloid .
2. (c) First, consider the surface x yz :::: O. T he surface consists of the planes x :::: 0, y :::: 0 and z :::: O. Now consider x yz :::: 1. When z :::: k , a positi ve constant, th I vel curve is xy :::: l/ k . T hus we get a hyperbola in the fi rst and third quadrants with asy mptotes on the x and y axes. T he hyperbol as get closer to the origin as z gets larger. Thus, t he surface in the first octant has t he plane x :::: 0, y :::: 0 and z :::: 0 as asy mptotes. Similarly, there is a surface in t he octant where z > 0, x < 0 and y < O. Now , if z < 0, then there is a si milar surface in either of the octants where x < 0 and y > 0 or x > 0 and y < 0. T he level surfaces for xyz :::: c, where c is a positive constant , consist of four similar surfa.ces which move further from the origin as c gets larger. If c < 0, the level surface is positioned in the other fo ur 0 tants. 3. (b) The function takes a point from ~ 1 to ~ 2 , so the si ze of the derivative matrix is 2 x 1. The derivative is
D / (x) :::: [
~
].
5. We need to show the vector normal to the tangent plane of I (x , y) at the point (xo, Yo , I(x o, Yo) ) is parallel to the vector (xo, Yo , zo). T he partial derivatives of I are:
01 ox (xo, yo)
9 1 2x (1  x 2 ~
8I ( ) oy xo, Yo

Y
2
/!
)1 2
::::
(xa,ya)
1 2Y(I_X2_y2)1/2! 2
:::: (x a,Ya)
Xo 2
vI  Xo  Yo . /1
V

2 .
Yo 2 2' Xo  Yo
So the normal to the tangent plane is
 Xo
.
.
,1 +
VI  X6  Y6
Yo
.
J  k.
VI  X6 Y6
Multiply t he above through by ( 1  x 5  y6)1 / 2. We get (xo, Yo ,/(xo ,Yo)) = (xo, Yo, zo). Geomet ri cally, we are looking at t he sphere : x 2 + y 2 + z2 :::: 1. T he v ctors normal to the tangent planes are precisely the vectors r = (x, y , z). Those vectors that have the direction of e p (see exercise 7(a) in section 1.4) are perpendic ular to the sphere. 7. (c) T he equation of the tangent plane is
z :::: I (xo, Yo)
+ [(01/ fJ x )(xo , Yo) ]( x ~ xo) + [( 0I / oy)(xo, yo)](y  Yo).
In this case, we have 1(1, 1) :::: 1; ol / ox :::: y, (of/8x )(  1, 1) :::: 1; of/oy :::: x , (of/oy)(  I , 1 ) ::::  1. T herefor , the equation of t he tangent plane is z :::: 1  l (x + 1)
l(y+ 1), i.e., x +y+ z + 1:::: O.
. .
CHAPTER.
38
=
8. (b) If f (x , Y, z ) = constant, then the equation of t he tangent plane is "il f(x o) . (x  x o) where x = (x , y,z). In this case, f(x,y,z ) = x 3  2y3 + z3, so "ilf(x) = (3x 2 ,6y2,3; and "il f (l, 1, 1) = (3, 5 , 3). Therefore, the tangent plane is (3, 6, 3) . (x  1, Y  1, z  1) = 3x  5y + 3z = 0, i.e. , x  2y + z = O. 11. (b) The strategy here is to find a few "paths," compute the limit along those paths. L x = 2y; then the limi t as y goes to 0 is
lim (x,II)~(o,O)
f
X+YII y~ r Y1;1I v'3. yF ';=Y 3Y
On the other hand, take the path x = 4y. Then the limit as y goes to 0 is lim
(x, y )+ (O ,O )
JI +
x y x y
Y 1= y~O lim f5I = f§. =/: v'3. Y13"Y 1 V3
Since the limiting vain s depend upon the path taken, the limit does not exist .
++
+
12. (b) Hold y and z constant and use the hain rule to differentiate with respect to x. Th ofl ax = 10( x y z)9 . By symmetry, ofl ay = ofl az = 10(x y + z )9. T he gradient is t vector (of/ox, of/oy, of/oz ) = 10(x + y + z)9(i + j + k ). 16. We compute
~z (1,  2) = 2x l x
( 1, 2)
= 2; ~z (1,  2) = 2y l Y
=  4.
( 1,2 )
Then , the tangent plane is z = f(l,  2) + (8 zl ox )I(1, _2) (x  1) + (az l ay)I(1,_2) (y + 2), z = 5+ 2(x1)  4(y+ 2) or 2z 4y  z = 5. Geometrically, the gradient of f (x , y) = x 2 + at (1,  2) is perpendicular to the level urve z = 5. The tangent plane of the graph of f t he plane that contains the line perpendicular to the gradient of f at (1,  2) and lying in t horizontal plane z = 5, and the tangent plane has slope .../22 + 42 = 2v5 relative to the plane. 18. (b) The directional derivative in the direction ofv is "ilf(xo) , v /ll v il. We compute "ilf(x) = (y + z , x + z , x + y) , so "il f(l, 1,2) = (3, 3,2) . Also, we have
Ilvll =
) 10 2 + ( 1)2 + 22 =
.../105.
Thus, the directional derivative in the direction of v is (3, 3, 2) . (10 ,  1,2) I v105 = 31 / M 21. The bug should move in the direction of  "ilT(x,y ), since this is the direction of fast decrease. We compute  "ilT(x, y) =  4xi + 8yj and "ilT(1 , 2) = 4i + 16j, so the bug shou move in the direction 4i + 16j. 24. Let u = :z:  y, then
01
01 and 81
ox
ay
ou
Holding y constant, we get
8z =
ax
a!
du au . dy
of  AU .
(!:.) 81 = (!) 01 .
y
ax
y
au
Holding x constant and using the quotient rule , we calculate
az ay Therefore,
=~ [al . y _ 1(1)] = ~ [_ of . y  I]. y2 oy y2 au
39
DIFFERENT IATION
27. (ii) (a) The sum rule tells us that x 2 + y4 is differenti able. Also , x 2y2 is differentiable by the product rule. Finally, x 2 y2l(x 2 + y4) is differentiable by the quotient rule since (x , y) 1 (0 , 0), and so x 2 + y4 i O. Holding y constant, we get
of _ 2xy2(X 2 + y4)  2X(x 2 y2) _ 2x y6 2 (X2+y4)2 (x +y4)2 ·
ax
Holding x constant , we get
2x 2y(x 2 + y4) _ 4y.3 (X2 y2) (x 2 + y4)2
of oy
2x4y  2x 2 y 5 (x 2 + y4)2
At the origi n, we use the definition of the partial derivative:
o o f (0 0) ox'
lim f(O
::=
+ h , 0) 
f(O, 0)
lim h 2  0
::=
h
h+ Qo
= lim 0 = 0,
h
h+ O
h+ O
and similarly,
o
of (0 , 0)
By
= lim
f(O , O+ h)  f(O , 0)
= lim }l~ = lim 0 = o.
h
"+0
h
"+0
"+0
(b) The partials exist at (0,0) and
f( x, y)
lim (x ,y)+(O,o)
II(x, y) 
(0,0)11
= 0,
so by the definition of differentiability (equation (2), section 2.3), f is differentiable at (0,0). The function f is differentiable at all other points because the partials are continuous. Thus f is differentiable. However, as (x,y) approaches the origin, aflox and o f loy do not approach o (take, for example, the path x = y); t hus the partial derivatives are not continuous .
= (oflox, ofloy).
28. (b) The gradient vector is 'V f
If (x, y)
i
1)

2x y cos (X 2 +y2)2
1)

2x y2 cos (X 2 +y2)2
of = YSID. ( x 2 +y2 ox
2
(G, 0), then
(1)
x 2 +y2'
and by symmetry
of = oy
. (
X Sill
X2 +y2
(1)
1
x 2 +y2
·
Now, if (x, y) = (0 , 0), then the definition of the partial derivative gives us
of 0 
0X
_ .
(0 , 0)  hm
f(O
+ h, 0) h
"+ 0
Similarly, (ofl oy)(O , O)
'Vf(x,y)
=
f(O , 0) _.
 hm
h
h+O
h)~ + (0)2)
=
I.
0
1m 
h+O
h
= o.
= O. Therefore, if (x,y) i (0,0), then
[ . ( +1) YSID
x2
y2
2

+ and thus 'V f(O, 0) ~ Oi
(0 + h)(O) sin ((0 +
+ OJ.
2x y ') cos ( [ )] J• (x2 + y2 )~ x 2 + y2
[ . ( +1) XSIl1
x2
y2
(1)] .
2xy2  (x 2 + y2)2 cos . x 2 + y2
J,
CHAPTER _
40 29. (b) T he directional derivative is 'V J (x o) . v. Here,
aJ ax
=_
x
V x2
+ y2
sin(v
x2
+ y2 ),
2
+ y2 ) .
and by symm try,
~~ Therefore, 'V J (1, 1)
vx+ ZY
y2 sin( v x
= ((  1/ V2) sin (V2), ( 1/.;2) sin (.;2))
and the directional derivative is
(( 1/ v2) sin (v2), ( 1/ v2) sin (h )) . (1 / Y2, 1/ 0 ) =  sin (v2). 33. (b) Directly, we first compute g(u)
= J (h (u) ) = sin 2 31.1 + cos 8u . Then
!~ = 6(sin 3u)(cos3u) When u
8sinSu.
= 0, dg / du = O. By the chai n rule, ddgu
= =
Again, wh n u
] DJ(x,y) ·D h(u) = [2x, 1] [ 38cos ' 3U Su sm 6(sin 3u)(cos 3u )  8 sin Su.
. [ = [2sm3u, 1]
3 cos 31.1 ] ' 8u  S sm
= 0, dg / du = O.
35 . The normal to the surface J(x, y, z)
= x 2 + 2y2 + 3z 2 = 6 is
aJ aJ aJ ) ( ax' oy' oz
= (2x, 4y, 6z ) =
(x, 2y, 3z ).
At (1, 1, 1), the normal is (1,2,3), so he unit normal is (1,2, 3)/ Jf4, the direc ion offiight . T veloci y of the particle is the speed times the direction , or 10( 1,2, 3)/J14. The position of particle at any time can be found by finding the equation of t he line through (1 ,1, 1) with direction 10( 1,2, 3) / V14, and it is (1, 1, 1)+10t (l , 2, 3)/Jf4. This impli s tha.t x = 1+10t/ K y = 1 + 20t/ v1A and z = 1 + 30t/ ..;u. At som t ime T, the particle is on he sph x 2+ y2 + z2 = 103, which means that (1+ 10T ;./f4) 2 + (1 + 20T / ..;u)2 +( 1+ 30T/J14)2 = 1 Simplifying, we get 3 + (120/ v'14)T + 100T2  103 = O. Solving for T using the quadra formula., and taking th positive T only, we get T = (3 + V359 )/5V14. 3S. (a) By substitution, Z = (x + y)(x  y) (b) By the chain rule, we have
az ax
oz aU.
= x2 
y2, so az/ ax = 2x and az/oy
oz av
=  2y.
,
= au' ax + ov' ox = vl1) + u(l) = v+ u = 2x.
Also, we have
oz ay 41. Let u(t)
= J (t )g(t) dh
dh
az
ou
az ov . By
= ell.
By the chain rule, we have
= au . ay + av
and h(u) du
dt = du . dt
=e
U
[ dJ
= ve l ) + u(1) = v  u =  2y.
d9 ]
dtg (t) + J (t) dt
J 9 = [ddtg(t) + J(t) ddt ] exp [J(t)g(t)]..
46 . The velocity is d fined as the derivative of displacement. Therefore, we want to compu au/ot =  6cos(x  6t ) + 5cos(x + 5t ). When t = 1/3, x = 1 and the velocity is au./at = 6 cos (1) + 6 cos(3). Sine cos (  x) = cos x, the velocity is 6(cos (3)  cos( l )).
41
DIFFERENTIATION
49. (a) As given, P is a function of T and V. We can also write
so T is a function of P and V. Finally, we can write
p = RTV2  a (V  {3) V2(V  jJ)
.
Upon rearranging, we get pv 3

(PjJ  RT)V2
+ o:V 
0:{3
= O.
Since this is a cubic equation in V, it is theoretically possible to write Yin terms of P and T. Therefore, any two of V, P or T determines the third variable. (b) From the equation for T , we hold V constant and get (aT/a P )
= (V 
{3)/ R .
From the equation for P, we hold T constant and get (fJP/fJV)
= RT/(V 
jJ)2
+ 2Vo:/ V 4 = 
RT/(V  jJ)2
+ 20:/V 3.
Now, hold P constant and differentiate the equation for P by implici t differentiat.ion . We get aV _ R(V  jJ)  7jf( R T) o(V  {3)2
aV
+
or equivalently,
R
aV aT 
.
2V7jf0: _ ~ _ aV [ RT _ 20:] V4  V  {3 aT (V  jJp V3'
(RT 20:) . (V  jJ) (V _ {3)2  V3
(c) Using the results of part (b) , we get aT) (BP) (BV) ( aP BV fJT
(V  {3) (  R T
= ~
(V  jJ)2
20: ) [
+ V3
(V _ jJ)
R
((v~~p
_ ~~) 1=1.
50. (a) The question is asking for the directional deriva ive in the direction of a unit vector. Here, our unit vector is (1,1)/v'2. Also V h(x , y) = (0.00130x, 0.00048y) and Vh( 2, 4) == (0.00260,0.00196). Therefore, the directional derivative at (2, 4) in the direction of (1,1)/0 is 'Vh( 2, 4) . (1, 1)/v'2 = 0.00456//2. T his means th at the height increases (0 .00456//2) miles per horizontal mile traveled. (b) The direction of the steepest upward path is V h(  2, 4)
= (0.00260,0.00196) .
TEST FOR CHAP T ER 2 1. True or false. If false, explain why.
(a) If a 3variable function ! (x , y, z) has partial derivatives the function is differentiable at t he origin.
l x, Iy and Iz at the origin, then
(b) A grad ient vector for a function fin IR 2 is parallel to level curves of f
(c) A function f is continuous at a point Xo if
lim I(x}
X+Xo
= I(x o).
CHAPTER _
42 (d) In general , a ontinuous function is differentiable.
(e) Given a function !(w, x, y, z), th directional derivative in the direction of (0, 0,1,0) the same as 8f18y.
2. Let u = x 2 y + z and v = xyz. Also, let x = 2a + h, y = ab 2 and z = a + 2a 2 b. Fu r t her mor~ let a = d3 and b = c  d. Compute 8(1.£, v)18(c , d) at (c, d) = (1,1) . 3. Use the linear approximation to estimate th (8.01,3.9 1.04).
distance between the origin and the poi..
4. Let c(t ) be a mapping from ]R to ]R3 and let ! (x, y , z ) be a mapping fro m ]R3 to R. 10:: h(t ) = !(c(t )). When t = 1, we have (z, y, z ) = ( 1,2,4) , c'( I ) = (3,4, 2), dyldt = 1 ar dz l dt = 2. Is the rate of decrease fastest in the positive :c di r ction , the negative x directi the positive 11 direction r the negative y direction? 5. Let z = (z + y) 2  5x 3 + 2ye X be the equation of a surface in space. Find the equation f t tangent plane at x = 0, y = 3.
6. Let g(x, y) =
Y.., if x {
i= 0
x
o
. if x = 0
Compute 8g18z and 8g 18y at t he origin if they exist there. 7. Let u and v be functions of x and V, and let x and y be functions of s and t. Furthermore. is known t hat 8(u, v) z, y) = 0  1] and . [ 3 2 8(s, t) 0 1 8(z, y)
8(
[2 1]
Calcul ate 8u.18sand 8v 18t. 8. (a) Evaluate the following limit for !(w, z, y, z)
=w
z2y3z :
lim !(5, 2, 1, 1)  !(5, 2,  1 + h, 1). h
h + O
(b) Is it possible to define g(O, 0) so that g(x , y) 1R2? Explain why or why not .
= (x 2 + y2) /( x 2 + y) is continuous on all
9. Let u (x, y ) = 2x + y2 + eX. At (1,1) show that u increases faster in the direction parallel
the x axis t han in th direction parallel to the 11 axis.
10. A recent survey showed that patient satisfaction, s, at a pharmacy depended mainly up.. three factors  t he waiting time in minutes (t) , the patient's percei v d degree of illness (i) a scale from 0 to 10, and t he dollar cost of the prescription (c) . T h patient satisfaction ind is given by S (t, i, c) (1000  c) / it 2 .
=
(a) In what direction does patient satisfaction increase most rapidly at the point (1 0, 0.5, 10C (b) At the point (10, 0.5, 100) how fast and in what direction (positive or negative) d patient satisfaction change for each extra minute of waiting time? (c) The administrators do not want s to decrease by more than 1 per un it change of t, i c. Is this goal met with a price decrease of 3/ 5 dollars and an increase in waiting time 4/ 5 minutes? Explain .
..~~
43
3
HIGHERORDER DERIVATIVES; MAXIMA AND MINIMA
3.1: ITERATED PARTIAL DERIVATIVES GOA LS 1. Be able to comp ute iterated partial derivatives. 2. Be able to explain when mixed partials are equal.
STUDY H INT S 1. Iterated partial deri vatives. T hese are high rorder derivatives, such as second and th ird deriva tives. With several variables, higherorder deri vat ives may be taken with respect to different
:::y
variables. The notation
means
(:x ) (~~) ,
which is also denoted fy x .
2. EquaLity of m ixed partials. If the second partial deri vatives are cont inuous, t hen an iterated
partial derivative may be computed in any order. 3. W arning. Note that the theorem on equality of mixed partials requires continuous partial derivatives. If this requirement is viol a ed , different orders of differe nti ation may yield different
results. 4. EvaLu ating partials at a given point. Always remember to differentiate com pletely before sub
sti tuting given values. With mixed partials , you may substitute for a variable only after you have completed differentiating in that variable. 5. A pplications. T he heat equation, the wave equation and the potenti al (Laplace) equation are
famous examples of how higherorder derivatives occur in nature. T here is normally no need to memorize these equ ations in a vector calculus course.
SOLUTIONS TO SE LECTED E XERCISES 2. Note that when we differentiate wi t h re pect to either x or y, eZ is "constant ," so the first partial derivatives are of/ox 1/x 2 + e Y and of/oy  x e Y F inding the partial derivatives of the first partial derivatives gives us these second partial derivatives:
=
=
7. (b) Rewrite the func ion as z= provided x #O and y
# O.
oz ox
2X2
+ 7x 2y 3x y
2x
7x
fJ 2z ox 2
= ox
=+ 3y 3 '
W compu te: 2
7
= 3y + 3;
fJz fJy
2x
=  3 y2 ;
0 (Ozox ) = 0;
 ( HAPTE
44
a
2
/)2z a ( oZ ) 2 a (az ) z 8xay = 8x 8y =  3y2 = ay ax = ay8x;
2 a z 8 ( az ) 4x ay'.! = ay ay = 3y3 ·
The function is not continuous if either x = 0 or y = O. Hence, the funct ion is not differentia whenever x 0 or y O. 3 .. a f 8 (8 f )) af oa f a ( Oh ) 11. By defirutlOn, oxayaz = ax ay az . Let h = az ' so axayoz = Ox oy and 8 (8h) ah = ox a ( aozf ) = 8z 0 ( aaxf theore m 1, this is als equal to ay ax . Also, by theorem 1, ax
=
=
(0
and t herefore
(88z (88xf ) ) = 8yozax· oaf
o3f a 8xayaz = ay
15. (a) We are given f(x , y) = x arctan (x/ y), so we compute:
fx
arctan
(~) + ::' 1 y y
+
(\ /
~)
xy~
=
2 xy 2 x+ y
+ arctan ( ~) y
.
fy
=
fx x
=
_ x2 1 + (x 2 / y2 ) . :;J2 = x + y2 ·
1 y3 _ x 2y Y 2~
y(x 2 +y2)_ 2x 2y 1 (x 2 + y2)2 + 1 + (x 2 / y2 )  (x 2 + y2 )2 + x 2 + y2  (x 2 + y2) 2· 2x(x 2 + y2) + 2x 3 _ 2xy2 fyx = (x 2 + y2 )2 = (x2 + y2)2·
=
=~"7.
/:ey fyy
19. We have U x us 'U xx + 'U yy
x
x
2
Y
x2
2
(x2 + y2 )2
2y
= (x22x+ y2y )2 .
= 3x 2 
6xy , so U xx = 6x  6x = O. Also, u y = 3x 2, so U yy = o. Substitution gi = 0 + 0 = o. Thus, u (x, y) satisfies Laplace's equation and so u(x, y) is harmon
=
=
20. (b) For u = ;c2 + y2 , we get U x 2x, so U xx 2. Also, u y = 2y, so U yy = 2. Substitution in Laplace's equation gives us 'U x ., + U y y = 2 + 2 :f. 0, so x 2 + y2 is not harmonic. (d) For U = 11 + 3x 2y, we get u., = 6xy, so Uxx 6y. Also, U y 3y2 + 3x 2 , so U yy = Substitution into Laplace's equation gives us Uxx + U yy = 6y + 6y "I 0, so y3 + 3x 2 y is r harmonic.
=
23. Given V (x, y, z ) =  GmM/J x2 + y2
+ z2 =  GmM/ r,
=
we compute:
GmMx
24. (a) If (x, y)
af ax 8f ay
"I (0,0), we can compute the first partial derivatives in the usual way: = (y(x 2  y2) + 2x 2y)( x 2 + y2 ) _ 2x 2y(x 2 _ y2 ) _ x4 y _ yS + 4x 2y3 (x 2 + y2 )2 (x2 + y2 )2 2 2 5 2 (x(x  y2 ) _ 2xy2 )(x + y2 ) _ 2xy2(x _ y2) x _ 4x 3y2 _ xy4 (x 2 + y2 )2 (x2 + y2) 2
45
HIGHERORDER DERIVATIVES ; MAXIMA AND MINIMA
(b) To compute the part ial derivatives at (0,0), we need to use t he definition of t he partial derivati es. First, hold y constant at and differe ntiate with respect to x at Xo = 0. ate that f(O,O) is defined to be 0. Then
°
fj l (0 0) = lim 1 (0 fjx ' 11+0
+ h, 0) 
1 (0,0)
h
Similarly, we hold x constant at
=
h(0)( h 2  0 2 ) 2 lim ..:hc::.. .'+h,0,,2_ = lim h+O
°
h+ O
h
lim 0= 0. 11+0
and differentiate with respect to y at Yo = 0. Then
0(h)( 02  h2 ) {)I (OO)  I' I (O , O+ h ) I (O,O) _ l'1m 02 + h 2 ,  1m fjy 11 + 0 h h+O h (c) By definition,
Q=
:::y = :x( ~: ).
First, we use
~~
= "lim Q = 0. + 0 h
from part (a) and then perform
di ffe rentiation as in part (b) . By definition, {)f fj f {P I . {)y (O+h,O) {)y (O,O) {) {) (0,0) = hm h x Y h+O
h5
4h3(0)2  h(0)4 (h 2 + 02 ) 2 0 = hlim _ _'_ ..,!....J _ _ _ _ +O h 
.
h5/ h4
= 11hm   = l. + 0 h
Similarly, fj f 02 f  (0, (0,0) = lim {)% oyox
h+ O
of (0) 4h  h + 4(0 )2h °+ h)  (0,0) (0 + ° = lim '~3
5
2
h
fjx
h + O
h2 )2

h
(d) The mixed parti als are not equal, which is consistent wi th the fact that the second partials are not continuous at (0 , 0) .
3.2: TAYLOR'S THEOREM GOALS 1. Be able to write down t he first few terms of Taylor's formula for a given fun ction.
STUDY HINTS 3
1. Notation. The summation symbol
L:
means to sum all possible combinations of (i, j) with
i ,j = l
i and j ranging from 1 to 3, i.e., (i, j ) = (1,1), (1 ,2), (1 , 3), (2, 1), (2,2), (2, 3), (3, 1), (3, 2), and (3,3) . In general , if m indices are summed from 1 to n, there will be nm terms; in our case there are 32 = 9 terms.
2. Review. Before continuing , you may wish to review Taylor's for mula from your onevariable calculus text. Recall that the Taylor series can be used to approximate the values of functi ons. 3. Taylor 's fo rmula . You should know the pat tern for the general formula. As a reminder, Taylor's for m ula is
The second term , which involves the second partial derivatives , will become important in the com ing sections. The term involving the third partials sums up 3 n terms, so it may be unreasonable to ask you to com pute all of these terms unless n = 2.
CHAPTER
46
4. Computing Ta ylor 's f ormula. Remember that you will n ed to compute all of the partl derivatives of the same order. For example, when computing second partials, on must compu 82 f 82 f 82 f .. 82 f 82 f (j2 f 82' 82' ... , 2 as well as all of the mixed partials 8 8 '8 8 ' 8 ,etc. Note th a
aX n
X2
Xl
Xl
!
2
we do not need to compute
!.l
8
,i
U XiUX j
f.
X2
X2
Xl
Xl
aX3
j, twice because mixed partials are equal (assuminz
continuity) . 5. Taylor's formu la remainder. Recall that in onevariable calcul us, the remainder is determina. at some point between X o and X o + h. Now the r mainder is det rmin ed at some point on th line between the vectors Xo and Xo + h, where h = (hl ' h2 , ... , h n ).
SOLUTIONS TO SELEC TED EXERC ISES 1. Recall that Taylor's formula is a polynomial which approximates a functi on. If our function is
itself a polynomial, then this functio n must be its own Taylor series as well. Hence the second order Taylor formula for f is f (h 1 + h 2 ) = (hi + h2 )2 = h~ + 2h l h2 + h~ at X o = 0, Yo = C Alternatively, 8f/8x = a f / ay = 2{x + y) and 8 2 f/8x2 = 8 2 f / 8x8y = f}2f/8 ya x = z 8 // 8y2 = 2. At the point (0,0), f (x, y) and the first partials are all 0, and all of the second partials are 2. Thus, the Taylor approximation is (1/ 2)(2)(hI + 2h 1hz + h~ ) = (hi + h2)2 .
= = = = = =
=
5. Here, 1(0,0) 1, f x ycosxy  ysinxy, fy x cosxy  xsinxy, In == y2 sinxy  y 2 cosx y fxy fyx cos xy  x y sin x y  sin xy  xy cos x y and fy y _ X2 sin xy  x 2 cos xy. At (0 , 0) we have Ix fy f xx I yy 0 and fxy 1. Thus, the Taylor series is
=
f(h l , hz)
=
= 1 + Oh l
=
=
1
2
+ Ohz + "2(O h 1 + 2hl h2
+ Oh 2z ) = 1 +
h1h z + Rz(O, h).
7. (b) For x > 0 and x < 0, f (x) = exp( l / x) is infinitely d ifferentiable . For x = 0, we must use the definition of the derivative: 1'(0)
=
lim f( x )  f (O) x+o+ X
=
lim exp(  l / x ) . x+ o+ X
Let u = l/x. We must show that lim.. > oo u exp(u) = O. By I'H6pital's rule, lim ue  u u+ 00
= u+oo lim (.!!:..) = lim e tI + U
00
( .!..) U
e
= o.
In general,
= x+li mo+ /
f(n)(o)
f( nl)(o)
(n  1)(X) _
.
X
I"( x ) = .!..2 exp ( 1) , f(3 )( x ) = (32 + .!..) exp (  1) , x
X
X
:1;4
X
so if we can show t hat lim.. + oo une  u = 0 for all n, then we can conclude that fis Coo at x :::: 0 as well. Again, use I'Hopital's rule n times: .
lI m une u+oo
_
. un . n! " = u+ hm  = ... = lim  = O. oo etl u+ oo e"
Thus, f is C oo , with all derivatives equaling
f (O) + Hence,
f is not analytic.
f
I
°at x = 0. Now, 1 (0 + h ) = exp( l/h) > 0, but
f(k)(O) k (O)h + ... +  k ! h + .. . =
o.
HI GHERORDER DERIVATIVES ; MAXIMA AND MINIM A EXTREMA OF REALVALUED FUNCTIONS GOALS 1. Be able to fin d th critical points of a realvalued twovariable fu n tion. 2. Be able to use the Hessian to classify the critical points of a functi on as maxima, minima, or
saddles .
STUDY HINTS 1. Defini tions. (a) A local or relative extremum is a point Xo where ! (x o) is largest or smallest in a sm all neigh borhood of Xo . (b) An absolute extremu m is a point Xo where !(xo) is largest or smalles on the entire doma in under consideration . (c) Critica l points occu.r where all first partial derivatives are zero. (d) A saddle point. is a critical poin L which is not a local extremum . 2. Criti cal po in textrem um relatzonshlP. All extrem a occur a t critical points, but not all critical points are extrema. C ritical points m ay also be saddle points. 3. Realvalued f unctions. Note t hat we are comparing function values for re Ivalued fu nctions, not vectorval ued functions .
If a! / ox = 0 and {}!/ {}y = 0 have more than one solut ion , each com binat ion of (x, y) which satisfies t hese conditions must be considered. You must he complete in your analysis. See example 8.
4. F in ding extrema.
5. Hessian . T his is denoted by H! (x o)(h ) and it is equal to the second term of Taylor 's formu la, . . 1
wh ich
IS
"2
L n
; ,j= 1
[p! hi hj ~ ( xo). Xt
xJ
To determine definiteness in genet I, we need to know the determinants of the diagonal submatrices of the Hessian matrix. Starting from the upper lefthand corner , i.e., all , compute the deter minants . If they are all positive, t hen the Hessian is positive defini te . If a11 < 0 and t he signs alternate, then the Hessian is negative defi nite. ote that this test includes t heorem 6.
6. Determining definiteness .
7. Usefulne ss of H essian. At a critical point, the first partial derivatives are all zero , so Taylor 's fo rmul a reduces to f (xo + h ) ! (x o) + Hessian + remainder,
=
where the remainder is small compared to the Hessian. Thus , if H ! (xo) is positive definite , then !(Xo + h ) = ! (xo ) + Hessian + remainder > ! (xo), so ! (xo) is a rel ative minimum . Similarly, if H ! (x o) is negative definite, then ! (xo
+ h ) = !(xo) + Hessian + remainder < !(xo) ,
and f(xo) is a relative m aximum .
8. Classifyi ng a critical poin t Xo. (a) If H! (xo ) is positive definite, then Xo is a local m inimum. (b) If H ! (x o) is negative definite, then Xo is a local maximu m. (c) If Hf (x o) does not satisfy (a) or (b) and not aU of the submatrices are zero, t.hen Xo is a saddle point.
CHAPTE R 3
48
9. Second deriv ative test. T heorem 6 is a special case of how the Hessian is used and is often
used for problem solving. To use this test , one must compute the discriminant D
°
= ((J2 f) (()2f) _(fl) 2 2 ()X
()y2
() X()y
°
If ()2 f I ()X 2 > and D > at a critical point , then we have a local minimum. If ()2 /1 ()x 2 and D > 0 at a critical p oint, then we have a local m aximu m. And if ()2 /1 ()x 2 > and D at a crit.ical point, then we have a saddle point.
°
<0 <0
10. Minimizing distance. Example 8 shows that distances may be m inimized by analyzing d 2 .
This is justified by the chain rule. By differentiating d2 , the chain rule gives 2d(adl ox) and 2d(adlay), so we again are solving adlax = 0 and adlay = O. Since d ~ 0, the maximum (or minimum) of d 2 occurs at the same points as the maximum (or minimum) of d. 11. Guaranteeing absolute extrema. In lR! ', if a domain is closed and bounded, and tis continuou5 on the domain, then there exists an a bsolute minimum and an absolute maxi mum. All thre conditions are neces ary. Think about what happens if the domain is not closed, if t he domain is unbounded or if / is not continuous . Compare this to the extreme value theorem of one variable calculus.
12. Locating boundary extrema. In this section , one parametrizes the given bound ary of a region and then differentiates to determine the local minimum and maximum points. Another method is introduced in the next section.
SOLUTIONS T O SE LEC TED EXER CISES 3. We compute the partials aflax = 2x + 2y and a/lay = 2y + 2x . These partials vanish at points such t hat x = v, so these are the critical points. Since f (x, y) = (x + y)2 ~ 0 for all (x , y), t.he extrema must be minima.
=
=
5. By the chain rule a/ lox 2x exp (1 + x2  y2 ) and a f/ ay 2y exp(1 + x 2  y2). Setting these partials equal to 0, we find that (0,0) is the on ly crit.ical poi nt . To classify this critical point, recall that the exponential fu nction is monotonic. Looking at the function along the % axis, set y = t o get f (x, 0) = exp(l + x 2 ). Thus, x is obviously a m inimum. If we look at the y axis , we set x = to get f (O,y) = exp(l y2). The reader should be convinced that y = is a maximum. T herefore , we conclude that (0, 0) is a saddle point .
°
°
=°
°
9. At (0, Q1.50s~+ y2) = l. T he cosine function cannot exceed 1, so (0,0) is a maximum. At (v7r/2 , V7r/2) , cos(x 2 + y2 ) = cos(7l') =  1, so (v;J'i, v;J'i) is a m inimum since the cosine function is never less than  l. Now, consider the critical point (0, ft). We calculate cos(x 2 +y2) = 1, so this critical point is also a m inimum. Note: This exercise cannot be done by the maxmin test for functions of two variables (Theorem 6) because the second partials all vanish at the critical points and so t.he t est is inconclusive.
a
10. The partial derivatives are 0/1 ox = in y and f/ ay = 1 + x cos y. Setting t hese equal to O. we see tha t the critical points are ( 1, 2n7r) and (1 , (2 n + 1)7r), where n is an integer. Since 0 2 f I ax2 = 0, we cannot use theorem 6. Consider the case when x = 1. Our fu nction becomes f(y) = y+sin y. The graph of f( y) shows t hat when y = (2 n+ 1)7l', there is an inflection point there. Therefore , we conclude that the point (1 , (2n + 1)7l') is a saddle point since, along t he I, th D line x = 1, the critical point is neither a maximum nor a minimum. Si mil arly, if x f( y) = Y  siny has inflection points at y = 2n7l'. Therefore, the cri tical points (1,2n7r) are also saddle points.
=
49
HIGHERORDER DERIVATIVES; MAXIMA AND MINIMA
fly)
fly)
2n
11:
11:
y
y
fly) = y + siny
.try) = y  siny
14. First, we compute of/ax = (y cos xy)/(2 + sin xy) and of/ oy = (x cos xy)/(2 + sin xy). Since the denominator is between 1 and 3, we only need to solve y cos xy and x cos xy == 0. From ycosxy = 0, we get y = or xy = (2n + 1)11"/2, where n is an integer. From x cos xy = 0, we get the additional solution x = 0. To classify the extrema, we look at f(x, y) . Since 1 $ sin xy :::; 1, we have In(l) :::; f(x, y) :::; In(3) . So when xy ::: 311"/2,11"/2, ... , (4n + 1)11"/2, we have f(x, y) = In(3) and these points are local maxima. Similarly, when xy = 11" /2,311" /3,711" /2, ... , (4n + 3)11"/2, we have f(x, y) In(l) and we have local minima. Now, look at (x, y) (0, 0). If x = y, then In(2 + sin x 2 ) is a minimum when x = y = 0. On the other hand, when x y, we have In(2 + sin(x 2 )) In (2  sinx 2 ), which is a maximum when x y 0. Therefore, the point (0,0) is a saddle point .
=°
°
=
= =
=
=
=
z
x
(31t12) 112
z =In(2 + sin i
)
(91tI2) 112 x
z = In(2  sin i
)
18 . We shall find and classify the critical points in terms of k. First, set the partial derivatives equal to to find the critical points. We have
°
of = 2x+ky=O ax
and
of =2y+kx==0. oy
If k f:. ±2, the only critical ponit is (0,0). We shall return to the case when k = ±2 later . The second partial derivatives are
02f ox 2
= 2,
02f o2f oy2 = 2, and oxoy
=k .
CHAPT ER .
50
The discrimin ant is D = 4  k 2 . If k 2 < 4, i.e.,  2 < k < 2, then D > 0 and sine 0 2 f /ox 2 > 0, the second derivative test tells us t hat (0, 0) is a local mini mum. If k 2 > 4, i .E k > 2 or k <  2, t hen D < 0; therefore, (0,0) is a saddle point if k 2 > 4. For t he case k = 2, we have
The graph of f is a parabolic cylinder that opens upward and whose vertex is the line x on the xy pl an'~' Similarly, fo r the case k =  2, we ge t
=
In this case, the graph of f is a parabolic cylinder that opens upward and whose vertex is th Line x = y on the xy pl a ne. W hether k = 2 or k =  2, f attains t he minimum value of 0 a t i ;: vertex. In summary, (0, 0) is a local min imum if Ikl < 2 and (0,0) is a saddle poin if Ikl > :. Minima occur a t points where x = y if k = 2 and they occur at poin ts where x = y if k = _
19. (a) We calcula te the partials: of/ox = 6x( y  x 2 )  2x (y  3x 2 ) and o f /o y = (y  x 2 ) (y 3x 2 ) . Setting o f/oy equal to 0, we ge 2y 4x 2 = 0, which im plies y = 2x 2 • Substitutior into o f/ ox giv s  6x( 2x 2  x 2 )  2x (2x2  3x 2 ) =  6x 3 + 2x 3 = 0, which implies th at x = . and so y = O. Therefore, (0 , 0) is a crit ical point. (b) Let g(t) = (at , bt ). T hen f (g(t) ) = (bt  3a2 t 2 )(bt  a2 t 2 ) = b2 t 2  4a 2 bt 3 + 3a 4 t l Differen iation gives f ' (g(t )) = 2t b2 1 2a 2 bt 2 + 12a 4 t 3 = t (2b 2  12a 2 bt + 12a 4t2 ) which impIieo that t 0 is one of t he solutions. The second derivative is f" (t) 2b2  24a 2bt + 36a4t 2 and 2 f"(0 ) = 2b ~ 0, independent of b (or a), except possibly at b = O. For b = 0, f (g (t) = 3 a4 f~ which has a minimum at t = O. Thus, t = 0 is a rela tive minim um of f (t) along any straigh line t hrough t he origin. (c) Look at the par abola y = 2x 2. Then f (x, y) = (_ X2)(:z;2) = _ x 4 < O. T hus, along this curve, f(O ,O) is a maxi mum.
=
=
23 . Given a vol ume V, suppose t he dimensions are x x y x z , then V = xyz and t he surface area is S 2xy + 2yz + 2x z . We want to mi nimize S. Solve for z, we get z = V/ x v . T hen S = 2x y + 2V /x + 2V/ y . Taking partials, we get
=
oS ox
oS oy
2y  2V(1/x 2)
(I
2x  2V (1/ y2)
(21
Setting (1) equ al to 0, we get y = V/ x 2 . Substitu te this into (2) and set it equal to 0: 2x  2V/(V/x 2)2 = 0, which is equivalent to x  X4/ V = 0, or x 3 = V , or x = V I/3 . T h n y = V/x 2 = V/ V 2 / 3 = V I/3, and z = V / x y = V / (V I / 3 V I / 3 ) = VI/ 3. All t hree dimensions are equal; hence, the box is a cube.
=
28. We compute the parti al derivative: of/ ox = an x " I and a f /oy cn ynl. Setting these equal to 0, we see that the critical point is (0, 0) . The second partial derivatives are 0 2 f / ox 2 = n(n  1)ax n  2 , 0 2 f/o y2 n(n  1) cyn2 and 0 2 f/8 x 8 y O. We cannot apply t he second derivative test since all of the second partial derivatives vanish at th origin. When n is even. both x n and yn are positive if (x, y) :j:. (0,0) ; both x" and yn equal 0 only at th origin . We concl ude th at t he origin is a maximum if (J and c are both negative and n is even . The origin is a minimum if both a and c are positive and n is even . For all other cases, we have a saddle point at t he origin since the origin is t he only critical point .
=
=
:31. We want to find the extrem a on the disk . F irst , check to see if an xtrema occurs inside t h disk. Take the partials of f and set t hem equal to 0:
oj  = 2x + y = 0 ox
and
of ay
= 2y + x = o.
HI GHERORDER DERIVATIVES; MAXIMA AND MINIMA
=
51
=
=
Thus (x , y) (0,0 ) is &. rela tive extrema and f(O, 0) 0. On the boundary, we let x cos B, 2 Y sin B. Then f( x , y) = f ( B) 1 + sin Bcos B. Differentiation gives f' (B) C05 B 2 2 sin B. Setting f' (B) = 0, we get cos 2 B = sin B, which means that B = 11'/4, 311' / 4, 511'/ 4, or 711'/4 on t he interval ~ B ~ 211'. For B 7r/ 4, f (x, y) ! (1/ V2,1/V2) 1/ 2 + 1/ 2 + 1/ 2 3/ 2; for B 37r/ 4, ! (x,Y) = ! ( 1/ V2, 1/V2) 1/2  1/2 + 1/2 1/ 2; for B = 511"/4, !('I: , y) !( 1/ V2,  1/ V2) 1/ 2 + 1/2 + 1/ 2 = 3/ 2; and for B 71r/ 4, ! (x, y) ! (1/ V2, 1 /V2) 1/ 2  1/ 2 + 1/ 2 1/2. Therefore, the absolute maxima oc cur at (1/V2, 1/ V2) and ( 1/ V2, 1/ V2), with the maximum value of 3/ 2, and the absolute minimu m occurs at (0 , 0), with the mini mum value of 0.
=
=
°
=
=
= = =
=
=
=
= =
= = =
=
34. First, it is always wise to check for extrema inside the region . We set the part ial derivativ s qual to 0: 8!/8x = Y = and 8!/8y x 0. Thus (0, 0) is a loc I extremum , and at (0,0) , ! (x, y) = 0. However, we have not checked the points on the boundary. For he line segment y 1, we have 1 ~ x ~ 1. On this line segment !(x , y) 1 · x = x , so the minimum for this line segment occurs at ( 1, 1) and the maximum occurs at (1,1). Similarly, we can analyze the ot her three line segments. Therefore, at two of t he corners , namely (1,1) and (1 ,  1), we have ! (x , y) = 1. And at the other two corn rs, !(x, y) = 1. So (0 , 0) is not an absolute extremum ; t he maxima are at (1 , 1) and (1, 1 ) and the minima occur at (1, 1) and ( 1, 1).
°
= =
=
=
°
39. Suppose \7 2 u = and u achieves its mini mum on D\ 8D. Let un (z, y) = u(x,y)( l / n )e"' , then \7 2u n = ( l /n) eX < 0. Thus, Un is strictly superharmonic, and by Exercise 37, can have a
minimum only on 8D, ay, a.t Pn = (Xn , Yn ). We have un (x n , Yn ) = U(XY' Yn )  (l/ n) exp(xn ) ~ ,Yn) e1 /n, since all X n ~ 10n 8D. If (xo, Yo) isapointinD,then un( xO,Yo) > un (xn , Yn), or e e u(xo, Yo)   > u(x n , Yn )   . n n
Estimating the lefthand side upward (since el n > 0) , we have
U(X n
e u( xo, YO ) > u (Xn , Yn )   . n
Since D is closed and bounded and u(x, y) is continuous, there must be a point q = (xoo, Yoo) on 8D such th at the above inequality holds in an arbitrarily small neighborhood of q. Hence, u(xo , Yo) ~ u (xoo , Yoo ) and u h as a minimum on the boundary 8D. 42 . First , notice that h
=
(y/ 2) tan 0 and d A
=
(y/ 2) sec O.
Thus, we want to maximize
= xy + ( / 2)yh = xy + (y2 / 4) tan B.
We are given P = 2x + y + 2d = 2x + y + ysec B. Note that Pis constant. Solving for x, we ge x (1/2)( P  y  Y sec B) and our area function becomes A(y, B) = (P y  y2  y2 sec B) / 2 + (y2 tan B)/4. Taking the partial derivatives, we get
=
x
8Al oy = P/ 2  y  y see B + (y / 2) tan 0 and
8A/8B = (_ y2/2 ) sec 0 tan 0 + (y2 / 4) sec 2 B = (y2 / 4) sec B(  2 tan B + sec B).
=
=
Setting 8A/ 8B = 0, we get y = O ar 2tanB seeB, i.e ., sin O = 1/ 2, i.e., B 11'/6 . The solution y is impossible for this geometric problem. When 0 1r/ 6, we have sec e 2/va and tan B 1/va, so 8AI8y PI2  y  2y/ va + y/ 2va. Setting 8A/ fJ y 0, we get
=° =
=
P (
y
2
1
= 2 1+ va  2va
Therefore, the maximum area is
=
)1 2vaP va 5.
=
=
CHAPTER
52
P y  y2 _ y2 sec e 2
where
+
P..j3
y ;=
2V3 
y2 tan e 4 7r
5
and
B=
6'
3.4: CON STRAINED EXTREMA AND LAGRANGE MULTIPLIERS
G OALS 1. Be able to find the extrema if one or more constrai nts are given.
2. Be able to find minimum and maxi mum distances in geometric problems. 3. Be able to an lyze the critical points of a function with one constraint by using t he border _ Hessian . 4. For economics problems, be able to explai n the significance of A.
STUDY HIN TS
fl S means "restricted to." For example, if g(x, y) means we want to consider the fu nction 9 = 2 + y.
1. Notation. T he not ation
gl(x
= 2)
= x + y,
2. Method for fi nding constrained extrem a. Note that t here is an alternative method which
th ~
equations (3) rather than (2). First , rewrite the constraint so that the righthand side .• zero; for example, x + y 2 becomes x + y  2 O. Then consider the function h(x , >.) = f( x) + >. . (constraint). Solve 8h / 8x; = 0 and 8h/ 8A = o. In example 2, w analyze h(x,>.):::; x 2  y 2 + >.(x 2 + y2 _ 1) and in example 3, h(x, A) = x + Z + A(X2 + y 2 + z2  1).
=
=
>.; only the values of tht:" variables. Sometimes, solving for A in terms of the variables, thus eliminating >., is the righ thing to do.
3. Solving equati ons. In general , we are not interested in the value of
4. Cautions. Remember , all of your equations must be solved simultaneously. Solving one equa
tion alone does not complete the job of ,finding an ext remum. 5. Generalization. If there is more than one constraint, then "Y f (xo )
... + >'n "Ygn(XO)'
= >'1 "Y g1 (xo) + >'2 "Y g2(XO)
The righthand sides of equations (2) will have the form
t
>'i
g;i, in plact'
i=1
J
of >.. : :' and there will be extra const raint equat ions . If you prefer to use equations (3), Ie J
h(x , >.) = f (x) + I: Ajgj (X) and solve 8h/ 8xj = 0 and 8h/ 8>., = 0 simultaneously. 6. B ordet'ed Hessian. This is a Hessian with a "border ," which are the additional top row and
the additional 1 ft colu mn. T he entries from left to right or top to bottom are 0,  8g/ axl 8g/ 8X 2, ... ,  8g /8 xn , where 9 is t he const raint . Note that all entries except for the border are second partial derivatives. Xo . (a) If the k X k subm atrices of the bordered Hessian are aL negative for k ~ 3, then Xo is a local minimum. (b) If the signs alternate: positive, negative, positive, .. . , starting with the 3 x 3 matrix, then X o is a local maximum. (c) If t he pattern does not satisfy (a) or (b) , and the submatrices are not all zero, then Xo is at a saddle point .
7 . Classifying critical points
53
HIGHERORD ER DERIVATIVES; MAXIMA AND MINIMA
8. Extrema on a region. T he method of Lagrange multipliers is only good for locating extrema on a boundary. Don't forget to analyze the critical points inside the region.
9. Geometry. As in exam ple 8, section 3.3, we can analyze the square of the d istance rather th an the distance itself. 10. Economics. Isoquants are curves showing all possible combinations of capital and labor which
produce the sam output. In t hese examples, ). tells you how much more can be produced with one extra unit of capital or labor . Note that). has significance only a t the optim al point.
SOLUTIONS TO SELECTED EXERCISES 1. Use the method of Lagrange m ultipliers. We have \If(x, y, z) = (1 ,  1, 1), and the const raint is g(x ,y,z) = x 2 + y2 + z2  2, so )'\lg(x , y, z ) = ).( 2x , 2y , 2z ). Thus, Vf = >'Vg gives us 1 = )'2x , 1 = )'2y a nd 1 = )'2z. So we have x = z =  y = 1/2),. Su bstitute this into the constr aint: {1/ 2>.)2 + (1/2).)2 + (1/2).)2 = 3/ 4).2 = 2, or ). = ±(1/2) J3/2. For ). = + (1/2)J3/2, we have (x,y, z) = (J2/3,  J2/3,J2/3), and for)' = (1/2)J3/2, we have (x, y , z) = ( J2/3, .j2f3,  J2/3). These are the two extreme points and the maximum is v'6, while the minimum value is  v'6.
3. We want to find t he extrema of f(x , y) = x subj ect to x 2 + 2y2 = 3. Use the method of Lagrange multipliers. From the constraint , let g(x, y) = x 2 + 2y2  3, so V f = (1,0) and Vg = (2x ,4y). We want t o sim ultaneously solve Vf = ).Vg and the constrai nt equation:
>.2x My 3.
(1) (2) (3)
From (2), we get y = O. From (1), x = 1/ 2), . Substituting for x a nd y in (3) gives us (1/ 2).)2 = 3, so 1/2>. = ± J3; therefore, x = ± v'3. At (v'3, 0), f( x, y) = v'3 and at (v'3, 0), f (x , y) =  /3. We concl ude that the m aximum occurs at (v'3,0) and the minimum is at
( y'3, 0). 8. On S, y is restricted to be cos x , so f (x, y ) = x 2 _ y2 = x 2  cos 2 x. Applying onevariable methods , we cal culate f'( x ) = 2x + 2cos xsin:!.' . The derivative van ishes when x =  cos x sin x =  (1/ 2) sin 2x. This is a transcendental equation and can be solved by graphi sin 2x / 2 cal m ethods. The graphs of y x and y only intersect at the origin , so (0 , 0) is an extremum. Since x 2 ~ 0 and 0 ~ cos 2 X ~ 1, we conclude th at (0,0) is a minimum.
=
=
13. We want to m inimize the surf ce area of th cylinder subject to t he constraint of the volume. That is, we want to minimize S(r , h) = 2rrl' h + 2rrr2 subject to rrr 2 h = 1000 cm 3 . Use the met hod of Lagrange multipliers. From the constraint, we get g(r, h) = rrr 2 h  1000. We compute th following first partial derivatives: 8S/ 8r = 2rrh + 4rrr , 8g /8r = 2rrrh, 8S/ 8h = 2rrr ) 8g/ 8h = rrr2 . Now we want to solve the fo llowing system of equations: 2rrh
+ 4rrr 2rrr rr7· 2h
1000.
(1) (2) (3)
CHAPTE R 3
54
Factor out rrr from (2) to get 2 = Ar or A = 2/r . Substitution into (1) and factoring out 2rr gives = (2/ r )rh = 2h , or h = 2r. Substitution into (3) gives rrr 2 h = rr7,2 . 2r = 2rrr 3 = 1000, so r = 10/(2rr)1/3 and h = 2r = 20/(2rr)1/3. To check that our result sat.isfies the const.raint we calculate 2 100 20 rrr h = Tt (2rr)2/ 3 (2rr)l/3 = 1000.
h + 2r
Therefore, the desired cylinder has height 20/(2rr)1/3 cm and b ase radius 10/(2rr)1/3 cm. 18. Use the method of Lagrange multipliers. Let
= x + 2y sec B + A( XY + y2 tan B
f(x , y, A)
A).
Then of/ax = 1 + AY; of/oy = 2 sec B+ A(X+2y tan B) ; a nd OflaA = xy+ y2 tan B A. From af/ox 0, we get A = 1/y; whereas from of/oy 0, we get
=
=
A=
 2sec B x + 2y tan B
Hence , 2y = (x + 2y tan B)/(sec B) , so 2y(1  sin B) = x cos B. Thus, x = 2y(sec B  tan B). Substitute this into a fI OA = 0 to get 2y2(sec Btan B)+y2 tan B = A. T hen y2 (2 sec Btan B) A , so
=
y
2
= 2 sec BA
Acos () 2  sin B.
tan B
25 . (a) Use the method of Lagrange multipliers on t he auxiliary fu nction h(x, y, A) A(2x2 + y2 1). We compute the following partial derivative : hx
hy
= = 0 or A = 1. h).,
= x + y2 (1) (2) (3)
1  4XA == 0 2y  2YA = 0 _(2x2 + y2  1) = O.
From (2) , we get either y For the case where A = 1, we get x = 1/4 from (1) and then y = ±V778 from (3) . W hen y = 0, we get x = ±1/V2 fro m (3) . Thus , the critical point.s for the constrained function a re located at (1/4, ±V778) and (±1/V2, 0) . (b) We let the const raint be g( x , y) = 2x2 + y2  1, so h(x, y, A) = f(x , y)  Ag(X, V). Then the bordered Hessian (T heorem 10) is
og ax
og oy
og ax
02h ox 2
a2 h oyox
og ay
02h oxoy
02h oy2
0
/H/=
At (x, y, A)
 4x
2y
 4A 0
0 2  2A
= (1/4, ..ft78, I), we find /H/=
(01: , y) = (1/4 , we have
SO
0
4x  2y
0 1
Jr72
1 4 0
j7fi 0 0
= 14> 0,
V778) is a relative m axi mum point. At t he point (x, y, A) = (1/4, V778,1). I 0 1 Vf72 IHI = I 0~2
~4
~
= 14 > 0,
55
HIGHERORDER DERIVATIVES; MAXIMA A ND MINIMA
so (x, y) = (1/4, ftl8) is also a relative maximum point. When (x, y) = (1/V2, 0) , equation (1) tells us that>. so at the point (x, y, >.) (1/V2, 0, we see t hat
= Ji78,
°
IH,I:o: J8 I
 078),
=
° °
2 + V2
J8 V2
° °
= (2 + \1'2)( 8) < 0.
Thus, (1/V2,0) is a relative minimum. Similarly, when (x,y) at the point (x,y,>.) = (1/V2, 0, Ji78), we get
°
IHI = J8
° 2  °V2
J8
V2
° °
= (2 
= (1/V2,0), >. = +Jl78, so
V2)(8) < 0 .
Thus, (1/V2,0) is also a relative minimum. Evaluating the function f(x,y) at the critical points tells us that (1/4,±ftl8) are absolute maxima, (1/V2,0) is an absolute minimum, and (1/V2, 0) is only a relative minimum. 26. With a hyperbola, there is only a minimum distance from a point. With a parabola, there is also only a minimum. There can be no maximum distance from these geometric figures because both figures extend to infinity. 31. Let the price of labor be p and let the price of capital be q. We want to optimize Q given the constraint S::;;: pL + qK == B. We compute the partials of Q and S: {)Q/{)K = AaK,,l L1", {)S/ {)K q, {)Q/ {)L A(l  a)L " K", {)S/ {)L = p. Use the method of Lagrange multipliers to get the following system of equations:
=
=
AaK,,l L 1"
>.q
A(l a)K" L"
>.p
pL+ qK
B.
From (1 ), we get q = (A/>')aK"lL 1,,, and from (2), we get p Substitution into (3) gives us (A/ >.) (la)K" L 1" + (A/ >.)aK" L1" B or >. = (A/ B)K" L 1". Substitute for >. in (1) and (2): AaK,,l L 1" A(l  a)K" L"
From (4), we get K
=
= (a:
= (A/>.)(l  a)K"L  ". = B or (A/ >.)K" L 1 " =
(A/ B)K" L i "q (A/ B)K" L 1"p.
= aB/q, and from (5), we get (K, L)
(1) (2) (3)
(4) (5)
L = (1  a)B/p. Thus, the point
, (1 pa)B)
optimizes the profit .
3.5: THE IMPLICIT FUNCTION THEOREM GOALS 1. Be able to determine if an inverse function exists near a point x. 2. If an inverse exists, be able to find a derivative by implicit methods.
STUDY HINTS 1. Advanced material. The theorems presented in this section are usually proved in more advanced courses . You should be most concerned with understanding the statements of the theorems .
      



CHAPTER J
56
2. Notation. DxF is used in th is section. It is just another notation for " F with respect to x although as in Chap ter 2, DxF m ay be a m atrix if F is vectorvalued . 3. Local Theorems. T he theorems introduced in this section may not apply if th range or domain
is too large. 4. Special implicit fu nction theorem. If o f / oz =F 0, then z can be wri tten in terms ofx at a given point (xo, zo), and the derivative of z = g( x) is
_  DxF(x, z) D 9 (x ) of . a.; (x, z) Note that we can differentiate z even though we don 't have a formula for z. Don't forget the minus sign in the derivat ive. 5 . Commonly used formu la . When z is a fu nction of x and y, we get t he formul a dy dx
oz/ox  oz/oy'
It looks almost like division of fr actions, except fo r the minus sign . Again , don 't forget the minus sign . 6. General implicit function theorem. In general , z m ay be a vector. We form a matrix with the top row being the parti al of Fl wi th res pect to Z j, j 1, .. . , m (alm ost like grad ient) I, ...m . If the determ inant of thi Similarly, the 0 her rows consist of th part ials of Fk, k
= =
m x m matrix is nonzero, then z is a function of x and a derivative exists.
7. Jacobian. T his is the de erminant of the m x m m atrix descri bed in item 6 above. It is denot d
8. Inverse functio n theorem. As stated in item 6 above , if J f (xo) =F 0, then z can be written in terms of x. It may not be easy to express z in terms of x , but it is possible in principle. g. Example 3. T his is a t ypical problem . St udy it carefully. Note that when more than one function is given , you will n ed to solve a system of simultaneous equations to fin d a partial
deri vati ve.
SOLUTION S TO SELECTED EXERCISES
=
2. Let F( x, y , z) xy + z + 3xz 5  4. Since we want to know if we can solve for z as a funct ion of (x, y), we need to know t hat o f /oz does not vanish near the desired point , so o f/oz 1 + 15x z4. ear (1,0, I), Fz 16 =F 0, so F 0 is solvable for z as a function of (x, y) .
=
=
=
Therefore, oz ax
 Fx Fz
+ 3z 5 1 + 15x z 4 y
and
Oz _  Fy _ oy Fz
x 1 + 15xz 4
.
= (1 , 0), z = 1, so oz/ox =  3/ 16 and oz/oy = 1 /16. 7. Let Fl = y + x + uv and F2 = uxy + v. Then we want At (x, y)
~
/ OU = I OFI o Fdou
oFI/ ov o F /ov 2
The entries of the determinant are a FI/ou We see that at (0,0,0 , 0),
I =F 0 = v,
at (x, y , u, v) aFi/ ov
= (0, 0,0, 0).
= u, o F 2 / o u = xy and oF2 /8v = 1.
HIGHERORDER DERI VATI VES; MAX IM A AND MINIMA
so we may not b able to sol ve for tI, v in terms of x , y near (x, y , u , v ) = (0,0 , 0,0) . To check directly, the firs t equation gives us ltv =  (y + x ), so v = ( y + x)/u . Combining this with the second equation , we get ux y = (x + y)/u, or 1.1 2 = (x + y)/ xy. For (x , y) near (0,0) , either there is no solution for u small, or t here are 2 solutions for u. 10. (a) Using the definition of a(x , y)/ o(r , 0) and computing the part ial derivatives , we get
a( x, y) _l ax/or 8(r , O)  ay/or
ox/ ao oy/oO
I_I cosO  sinO
r sinS [_
rcos O
 r.
At ( 1'0 , ( 0 ) , r = 1'0 . (b) By the inverse function t heorem, we can form a smooth inverse fu nction (r( x, y) , O(x,y)) as long as r I 0. As a direct check, solve fo r r and 0 in terms of x and y: x 2 + y2 = r and 0 = arctan(y/x) . Since we have written r and () in terms of x and y, the result above is confirmed . Note t hat if x 0, then 0 7r/ 2 or 37r/ 2, depending on the sign of y. If, in addition, y 0, then r 0, a.nd B can have any value, so we cannot find an inverse, as we did above.
J
=
=
=
=
12. Let F l = xy2+XZU +yv 2 3 and F2 = u 3 yz+ 2x v u 2 V 2  2. T hen oFt/f)u = xz , oFt/ov = 2yv, oF2/ol.l = 3u 2 yz  2uv 2, oF2/ov = 2x  2u 2 v. At (x,y, z ) = (1,1,1 ) and (1.1, v) = (1, 1),
Since L\. I 0, it is possible to solve for ov/ ay, we use the chain rule:
tJ, tJ
in terms of x, y, z near the given point. To compute
and
1 3 OV oF2 = 3u201.  yz + u z + 2x  oy oy oy
AU 2 av 2 2u  v  2v  u oy oy
= O.
At (x , y, z) = (1,1, 1) and (u , v) = (1, I) , those equa tions become 2+ ou/ay + 1 +2(ov/oy) = 0 and 3(8u /oy) + 1 + 2(ov/oy)  2(01l./oy)  2(av/oy) = 0, or au/oy + 2(ov/oy)  3 and ou/oy  1, so ov/8y 1.
=
=
=
SOLUT IONS TO SELECTED R EV IEW EXERCISES FOR CHAPTER 3 2. (c) Compute the partials of l and set them equal to 0:
°
2x + 1 i = 2xy + 4 y 3 = 0.
(1)
(2)
°
From (I ), y2 =  2x (so x ~ 0). Substitute into (2) to get y(_y2 ) + 411 = y3 = or y = 0. T hus, x = also and (0,0) is the critical point. T he reader should verify that the discriminant Dis 0, so this test is inconclusive. However, l(x, y) x 2 + 2x y2 + y4.  x y2 = (x + y2 )2  x y2, so 1 (1: , y) > 0 for all x < 0, and for x positive, (z + y2 )2 < xy2 implies x 2 + y4 < _ z y2 but this is impossible since Z2 + y4. is always positive and _ x y2 is always negative. We can conclude that the minimum value of l( x, y) is 0 at (0, 0).
°
4. The Taylor expansion is
=
CH APTEP
58
where all of the partial derivatives are evaluated at (3': 0, Yo). We compute th partial derivati and evaluate at (xo , yo) = (0,0) as follow:
f (x, y) f ,, (x, y) fy(x , y) f" x(x, y) f"y (x, y) fy y(x , y)
e"Ycosx; f (O , O) = 1, ye"y cos x  eXY sin x ; fx (O,O) = 0, xe xy cos z; fy (0,0) = 0, 2 xy y e cos x  2ye"Y sin z  eXY cos z; Ix;c(O, 0) =  1, e"Y cos x + zye xy cos x  xe xy sin z; f;c y (0, 0) = 1, x 2 exy cosx; f yy (O , O) = 0.
Thus, the secondorder Taylor expansion is Z2
f (x, y) =1
2
+ x y.
7. Th critical points are tho e where the first parti al derivatives vanish . T hus, we need to so
of
y1T cos (11'z )
ax
of oy
sin (11'x) =
= 0,
o.
The equ ation si n(11'x) = 0 implies that x is an integer. Since the sine and cosine functions a never 0 at the same angle, y11' cos (11'x ) = 0 implies that y = O. Thus, the critical points a (n , 0) , where n is an integ r. Next, we calculate the second parti 1 derivatives. They are 02f
ox 2 82 f
11' cos (11'z ),
ozoy f)2 f
oy2
=
O.
=
At the cri tical points (n ,O), we have fxx (n, 0) = 0, f;cy (n, O) = ( 1)" 11' and fyy (n ,O ) Using T heorem 6 from Section 3.3 , we compute D = [I,,;c(n, O)][ly y{n , 0)]  (fxy(n , O)P o (( _1) n11')2 = _ 11'2 . Since D < 0, we conclude that the points (n, 0) are saddle points. 11. Use the method of Lagrange multipliers with g(x,y) = x 2 + y2 constraint eq uation, we get
og og
= 1.
From
Vf = >'Vg and
=
t
8x>' = 2x>., 8y>'
= 2y>. ,
l. If x and yare both nonzero, t hen the first two equations tell us that  sin (x 2  y2 ) ar sin (x 2  y2) must both equal >., so >. must be 0; therefore, x 2  y2 must be 0 also. Notice th x 2  y2 cannot be a multiple of 11' because x 2 + y2 1 means both IxI and Iyl must be less th or equal to 1. From x 2  y2 0, Z2 + y2 1, Ixl ~ 1 and Iyl ~ 1, we get four critical poin t
=
=
=
If x = 0 and y # 0, then x 2 + y2 = 1 tells us that y = ± 1 and consequently, >. = ± sin ( l
Thus, (0, ±l) are criti al points . Simila.rly, when y = 0, z = ± 1 and>' = ± sin (l ). Therefor


59
HIGH EROR DER DER IVATIV ES; MAXIMA AND MINIMA
(±1,0) are also critical points. At the points (±,ff, ± Vt), f has the value 1. At (±1,0) and (0 , ± l) , f has the value cos(I). Thus, f has a maximum value of 1 at four points and a minimum value of cost 1) at four other points. 13. By the method of Lagrange multipliers, we get he foll owing system of equations:
x
>. >.
x+y
1.
y
Substituti ng into the last quation gives us 2>' = 1, or >. = 1/2. Thus, x = 1/2 and y = 1/ 2. However, in this case , it would have been much easier to solve for x and substitute to get z = x(1  x ) = x  x 2 . By onevariable calculus methods, x = 1/2 is the critical point, and the maximu m value of z subject to x + y = 1 is 1/ 4. 16. (b) By the implicit function theorem ,
of/ 8x  of/ 8y '
dy
dx In this case , F( x, y)
= xS 
sin y + y4  4 = O. Thus,
18. (c) T he rectangul ar parallelpiped (box) is symmetri ,so if x is a coordinate of a point (x , y, z) on the corner of the box, then the di mension corresponding to that coordinate has to be 2x. W want to maximize V (x , y, z) = (2x )( 2y)(2z ) = 8xyz subject to x 2/ a2 + y2 /b 2 + z2/ e2 = 1. Use the method of Lagrange multipliers:
oV/ 8x 8V/o y 8V/ 8z 2
x /a
2
= 8yz = 8xz
2x>./ a2 2y>"/ b2 2z>. /e 2
= 8xy
+ y2/h + z2 / eZ 2
1.
Solve (1) for x: x = 4a 2 yz/>.. Substitute into (2): 8(4a Zyz / >.. )z = 2y>'/b 2, or z2 Substitute fo r x in (3): 8( 4a 2 yz / >.) y = 2z>.. / c2 , or y2 = >..2 / 16a2 e2 . Then
P lug these results into (4) to get
>.2 16a2 b2c2
i.e ., xyz
>.2 >.2 + 16a:lb2c2 + 16a 2 b2 c2 = 1.
= abe/ 3v'3. T herefore, the maximum volume is V = 8xyz
abe
8abe
= 8 . 3V3 = 3V3 '
(1)
(2) (3) (4)
= >.2/ 16a2bz .
CHAPTE R
60 22 . First check for extrem a on the interior of the circle ofradius J2. We have f( x , y)
= xyy+ x 
_
so set its partial derivatives equal to O. We have
of ax = y + 1
of = x  I , By
and

so the point (1 ,  1) is an extremum. Since (1) 2+(_1)2 = 2, this point is within the constrain To find ot her extrema (if any), we use the method of Lagrange multipliers: y
+1
= =
xI
x
2
2x"\ 2y..\ 2.
+ y2
First consider the case when ..\ = O. Equations (1) and (2) gi ve us (x, y) = (1, 1), which had considered earlier. If ..\ :j; 0, then (1) and (2) added together yields x + y = 2"\(x + y) or (2,,\  1)(x + y) T hus , either ..\ = 1/2 or x = yo For x =  y, equation (3) gives us the points (1, 1) ~_ ( 1, 1). If >. = 1/2, then equation (1) reduces to x = Y + 1. Substit ution into (3) yiel (y + 1) 2 + y2 = 2 or 2y2 + 2y  1 = 0 or y = (1 ± /3) /2 . Therefore , two other possib extrema occur at ((1 + V3)/2, (1 + V3)/2) and ((1  /3)/2 , (1  /3)/2). Evaluating at of these possible extrema, we get
=
f
f
e
+2v'3,  1; v'3 )
(

'
2'
1 /3 1 v'3) 2 ' 2
2'
1
f( l ,  1)
0;
f(I , I)
4 .
Thus, the maximum points are ((1 + v'3) /2 , (1 + V3)/2) and ((1 /3)/2, (1 /3)/2) wi maximum vlaue of 1/2. T he minimum point i (1,1) with minimum value of 4. 25. We use the implicit function theorem. We need to show that the determinant
of
of
au ov oe oc au av is not 0 near (x, y,
1.1,
v) = (2, 1 , 2, 1). T he determinant is
I
 31.12 2v 4u 121.1 3
I = I 812
2 12
I=  144 + 16:j; O. ou/ox, "
Since the determinant is not 0, 1.1 and v exist as functions of x and y. To compute implicitly differentiate the given equations with respect to x. Keep in mind that 1.1 and v functions of x and y. We get
201.1 2x  31.1 2y 
OV
ox + 2v ax aU 3 0V 4u ax + 12v ax =
0 O.
To make the calculation si m pler, we can plug in (x, y, 1.1, v)
au
av
= (2, 1, 2,1).
4  12 ax + 2 ax
0
au + 12 ov ox ax
O.
2  8
Then
~
61
HIGHERORDER DERIVATIVES; MAXIMA AND MIN IM A
Solve this simple system of two equations by your favorite method. You should get 13/32.
au/ox =
30. (a) Using the given formula, we write
s ;: f(m, b)
= =
+ (3  2m  b)2 + (3  4m  b)2 (46m + 16b) + ((m + b)2 + (2m + b)2 + (4m + b)2).
(1  1 . m  b)2
19 
The problem is to find m and b which minimize f(m, b), so we take derivatives:
of
am
_ 46+2(m+b)+4(2m+b)+8(4m+b)
=
46 + 42m + 14b,
and
of
16 + 2(m
ob

+ b) + 2(2m + b) + 2(4m + b)
16+14m+6b.
Next, we set them equal to 0:
0 O.
46+42m+14b 16+14m+6b
Solving this system of 2 equations, we get b = 1/2 and m = 13/14. The reader is encouraged to verify that this is indeed a minimum point. T herefore, the bestfitting straight line to the points (I, I), (2,3), (4,3) is Y = 13x/14 + 1/2, as shown in the graph below.
x
33. If y
= mx + b is the bestfitting straight line, we must have, in particular,
Performing this differentiation, we get
"
22)Yi  mXi  b)
.=1
= 0,
which implies that the summation has to be 0, or the positive and negative deviations cancel.
TEST FOR CHAPTER 3 1. True of false. If false, explain why.
(a) If y is a differentiable function of x, then
z is a differentiable function of x and Y, and oz/oy i= 0, dy oz/ox dx  oz/oy'
CHAPTER 3
62
= w + xyz + y2 has no critical p oints. (c) For any f(x,y,z), we have o4 f loxo y 2 oz = o4flozoyoxoy. (d) A fourt h order T aylor series for f (x ) y, z ) w) = x 3 + y 4  z2 + w (b) The function g(w, x, y, z)
as
f
W
z 2 is exactly the same
itself.
(e) Suppose D is a closed region on the xy plane. If f (x y) is cont inuou and has a minimllI!' on D, t hen f (x , y) also has a m aximum on D. 2. Use Taylor 's theorem to calculate a fi rstorder and a secondorder approxim ation for xy2z 3 a
th e point (1. 1, 2.03,0.98).
3. Consider the surface described by z crosssection in the plane y = 4.
= 5x 3 y 
4. F ind all of the critical poin ts of h(u, v, w , r) 5. Let
f
be a function of x and yand of lax
2x y2. Discuss the concavity of the surface '~
= v 3 + w 3 + u 2 + 7"2  3v  12w + 4r  8.
= 3x 2 yex . Wh ich of t h following, if any, can Ix
aflay? (a) x 3 2y+cosy
(b) 3x 2 +y8 (c) 3e Y +x3 6. F ind the minimum and m aximum values of x
+ yon
7. F ind the m in im um and m axim u m values of x 2

x
the cur ve where 2x 2 + y2
+ y2 
= l.
Y on the followi ng sq uares and thei!
interiors: (a) The square with vertices a t (0,0), (1,0), (1, 1) a nd (0,1), but not including the x or axes. (b) T he square with ver tices at (0, 0), (2,0), (2,2) and (0,2), bu t only t he border on the ~ a nd y axes is included in the region .
8. Let a particle move in a potential field in R 2 given by V( x, y) Find the critical points of Vand classify t hem.
= x 2 + 4x y + y2 
2x + 4y + E
9. Consider the following system of equ ations :
x 2u
+ yv + uv 2 u + x v + yu
o k,
where k is a constant. If .T and y may be wri tten in terms of u a nd v, com pute o x l o v at th given point and for the given k.
(a) (u, v, x, y) = (1 , 2, 0, 1); k (b) (u, v, x, y) = (2 , 1,0, 2); k
= 2. = 6.
10. Farmer Jones owns a golden goose farm and he wants to maximize p rod uc tion of gol den eggs He knows that more geese will lay m ore eggs, but t oo m any geese will inhibit egg layin g as t h geese instinctively will not overpopu late. On the other hand , too m any coyotes will prey upo most of the golden geese and too fe w will cause the geese to overpopulate . Farmer Jones h& determined that th e golden egg production is propor tional t o
E(g , c)
= 10gexp( O.lg) 
2cexp ( 0 .2c),
where 9 is the goose popul ation and c is t he coyote popu lation. (a) F ind the critical poi nts of E. (b) Classify the critical points of E.
63
4
VECTORVALUED FUNCTIONS
~.l:
ACCELERATION AND NEWTON'S SECOND LAW GOALS 1. Given a path, be able to compute the velocity vector , the acceleration vector and the speed at a given point . 2. Be able to use Newton's second law and Kepler 's law.
STU DY HINTS 1. Velocity, acceleration and speed. Reca.ll that the velocity vector's components are the first derivatives of the components of the path with respect to time. Speed is the length of the velocity vector. The second derivatives make up the acceleration. Beware that t he derivative of speed is not the accelerat ion . Note t hat speed is a scalar, while velocity and acceleration are vectors. 2. Differentiation rules. No tice how the differentiation rules for paths are comparable to the rules you learned in your onevariable calculus course. 3. Ne wton }~ second law. T his states that F should remember this.
4. R egular path. If c'{t)
1= 0, t hen
= rna, where F is force and a is the acceleration . You
c{t) is a regular path and the image curve looks smooth.
5. K epler 's law. T his law relates an orbiting body's period to its radius. Vector alculus is used to derive the equation. It is probably not necessary to remember the equation, but ask your instructor to be sure.
SOLUTIONS T O SELECTED EXERCISES 2. T ake the first derivative of each component to get the veloci ty vector; that is, c'{t) == v{t) = (t cos t + sin t ,  t sin t + cos t, via). The acceleration is composed of the second derivatives, so c"{t) = a(t ) = ( tsin t + 2cost ,  tcost  2sint,0 ). Therefore, v (O) = (O ,I,vIa), a (O) = (2 ,0,0) and the tangent line is 1(>..) = c(O) + >"v(O) = 0 + >"(0, 1, via) = >"(0 , 1, via). 5. By the sum rule, we may add vector fu nctions first and th n differentiate , or we m ay d ifferen tiate first and then add the derivatives. By adding first,
! [Cl (t) + C2(t)]
~ (e t + e t ,sint + cost , t 3 (e t
 e  t, cos t  sin t,
2t 3 )
 3t 2 ) .
On the other hand, d dt [Cl (t)
+ C2(t)]
d
dtc dt)
d
+ dt C2 (t)
(e t , cos t, 3t 2 ) + (_e t ,  sin t,  6t 2 ) (e t  e t , cost  sin t, 3t 2 ), which is what we got by adding first and then differentiating.
CHAPTER
64
10. We wan F(O) = ma(O) , where m = 1. T he accelerat ion vector is given by a(t ) ( cos t ,  4 sin 2t ), so a (O) = (1 , 0) . Thus, F( O) =  i gcm / s 2 =  O.OOli newton.
= r l/(t )
13. To show that the speed is constant, we differentiate it with respect to time and show that deriva.tive is zero. Since speed is IIvll = .;v:v, it is more convenient to work with the squ of the speed. (If th square of t he speed is constant, then the speed must be a constant, t We calculate d d dv
dt II v ll2
= dt (v· v) = 2v·
dt
= 2v . a,
but the acceleration a is perpendicular to the velocity Vi therefore v . a (d/dt)ll v (t )W = 0, so Il v (t)1I is constant.
=
O.
Th
17. Integrating each component of c'(t) = (t,e t ,t2) gives us c(t ) = (t 2/2+ A,e t + B,t 3/ 3+ where A, B and C are constants. When t = 0, we have e(O) = (A, 1 + B, C) = (0,  5, 1), so A = 0, B = 6 and C = 1. Therefore, the path is c(t ) = (t 2 / 2, et  6 t 3 / 3 + 1). 19. (b) Let X = 2x, so X 2 + y2 = 1. Recogn izing this as a circle, we let X = cost and y = sint, so x = X / 2 = (cost )/ 2. OUf
path e(t ) is described by (x,y) = ((cost )/2,sin t) . From the
original equation , we recogn ize 4x 2 + y 2 = 1 as an ellipse with
xintercepts at ±1/2 and yintercepts at ± 1.
y
1/2
x
20. By the cross product rule in the box preceding Example 1 of the text ,
d [mc (t) x v(t)] = m dt d [c(t) x v(t )] = m [~ dt dt x v (t ) + e(t) x ~ dt ] .
But dc /dt
= v(t), dv / dt = a(t) and v (t) x v (t) = 0, so the m[O + c(t ) x a(t )].
xpression reduces to
If k is a constant,
k (u x w)
= ku x w = u x kw,
so
m [c(t ) x a (t)] = c(t ) x matt ) = c(t) x F(c(t )). If F(e(t )) and c(t) are parallel, their ross product is 0, so the angul ar momentum i con This is t he case of planetary motion; since
F(c (t))
=
GmMc (t) Il c(t)113 '
we see that F(c(t )) is a multiple of c(t), so it is parallel to c(t) .
4.2: ARC LENG TH G OALS 1. Be abl to compute the arc length of a given segment of a path.
STUDY HINTS 1. No tation. Often
5
is used to denote a path in space, rather than c.
2. A rc length. T his is j ust t he length of a curv . Lengths may be added together, so we may pute t he ar I ngth of curves which are not differentiable at finjtely many points by sum:
the lengths of the pieces .
65
VECTORVALUED FUNCTIONS
3. A rc length formula . You may find it easier to remember that arc length is the integral of speed (n ot velocity !) . T his m akes sense because arc length gives the distance traveled . In any case , you should know that the for mula is
L(c)
=
I
tl
Ile'(t) 11 dt =
to
Itl
J (X'(t))2
+ (y'(t))2 + (z'(t))2 dt.
to
4. Integration tricks. (a) Due to t.he nature of the arc length formula , you m ay want to memorize the for mula
T his formula may be derived by trigonometric substitution if you do not wi h to memorize it . Ask your instructor if this and similar form ulas will be provided on an exam . (b) Look for perfect squares. Radicals can be elim inated from the integrand if a p r£ ct square occurs. 5. Positive lengths only. If you compute a negati ve or a zero arc lengt h, then you m ade a mistake. Arc length is always positive. 6. R iemann sum derivati on . If the derivation does not make sense now, you should return to this discussion after R iemann sums are explained more thoroughly in C hapter 5. Understand ing the derivation of fo rmulas gives much more insight into the t h ory. 7. Parametrization warning. If you need to parametrize a curve, be sure the curve is traversed once and only once. T he orien ' at ion is also importan , and this will b come especially signif icant in chapter 7. T he bad consequences of an incorrect parametrization are illustrated in example 1.
S OLUTIONS TO SELECTED EXERCISES 3. We compute e'(t)
= (3cos3t,  3sin 3t,3Vt), so
lie' (t) II = [(3 cos 3t) 2 + (3 sin 3t )2 + (3vt) 2P/ 2 = ';9 + 9t = 3v'l+t. T hus, the arc length is
1
1
3.Jf+t dt
o
= 3 · 2 . (1 + t )3/2 11 = 2(23/2 3
1)
= 4h 
2.
0
6. Given e(t) = (t , t sin t , t cos t) , we co mpute e' (t) = (1 , sin t +t cos t, cos tt sin t). T hen lie' (t ) II = [1 + (sin t + t cos t) 2 + (cos t  t sin t)2j1/2 ';2 + t 2. We want t.he arc length on the interval [0 , 1t'], so we need to compu te:
=
fo re J2+t2 dt. This m ay be integrated by he method of trigonometric substitut ion: One would let tan B = An alternative is to use the formula given in the integration tables (See study h int 4 above .):
V2t , aq,d t hen perform an integr ation involvingsec 3 8. This is left as an exercise.
i J2+t2 o
re
dt
=
'12 [tJi2+2 + 2ln(t + Ji2+2)] Ire0 ~ [ 1t' J 1t'2 + 2+ 2 In(1i + ~)  21nh] .
(HAPTE
66
= (2t , t 2 , logt), we compute c' (t) = (2, 2t, l/t) . Then Il c'(t)11 = (4 + 4t 2 + l/t 2)1/2 = [(4t 2 + 4t4 + 1)/t2F/2 = (2t 2 + 1)/t = 2t + l/t. Since c(l ) = (2,1,0) and c(2) = (4, 4, log2), we want the arc length on [1,2]' so we n
9. Given e(t)
ed
compute
1 2 ( 2t +
~) dt = (t
2
+ 10gt{ = 3 + log2 .
=
=
12. (a) Since T(t) . T(t) = II T (t)112 1, we can differenti ate both sides of T( t ) . T(t) l. By product rule fo r dot products (see the box preceding example 1 in section 4.1), (d/dt)( T T(t)) (d/dt )( l ), i.e., 2T (t) . T'(t) 0, which implies that T (t) . T ' (t ) = O. (b) Beginning with T(t) = c'(t)/lle'(t)ll, we differentiate with respect to t, using the qUOl rule:
=
=
'( ) = !!.. ( ~) = c"(t)ll c' (t)1I 
T t
dt
c'(t)(d/dt)lle'(t)11
lie' (t) 112
Ilc'(t) II
.
Recall that Ilc'(t )112 = c' (t) . c'(t), so d lle'(t)11
dt
= d y'c'(t) . c'(t) = 1 (e' (t ) . e'(t))1/2(2e' (t ) . e"(t)) = e'(t) . c"(t) . dt
2
Ile'(t)11
Su bsti tution yields
, T (t)
e" ( t ) ,
= lIe'(t )1I21Ie (i)11 
c' (t) . e" (t) , lI e'(t) 11 3 e (t)
=
e" (t) II e' (t) 112  (c' (t) . e" (t) )e' (t) lIe'(tlIl 3
17. (a) From the definitions of k and N in exercises 13(b) and 14,
dT II' ( ) II T' ( s) ds = T '() s = T s IIT'(s)11 = kN . T he vectors T, N and B have un it length and form a righthanded system of mutually ort onal vectors, so we have T x N
= B,
N x B
=T
and
B x T
= N.
Differentiate , using the product rule fo r cross products , to get: dN
d
ds = ds (B
dB
x T)
= ds
x T
+B
dT
x
dB
ds = ds
dT
xT 
ds
x B.
Using the fact that dB /ds = TN (Exercise 15) along with the results derived earher i exercise and then factoring out a constant from t he cross prod ucts, we get T( N x T)  k( N x B )
= T(  B ) 
kT
= kT + TB.
Finally,
d
dB ds
dT
dN
= ds (T x N) = ds x N + T x ds = kN x N + T = k N x N  kT x T + TT x B = TN = Tx T = o.
since N x N (b) Let w = WI T and W3. We have
+ W2N + W3 B.
dT /ds
= =
w
x ( kT
+ TB)
We shall use the results from part (a) to find scalars
x T
= wt{ T
x T) +w2(N x T) +w3(B x T)
0  w2B +W3aN
= kN .
67
VECTORVALUED FUNCTIONS
Here, we have used the facts that (1) B = T x Nand (2) T, Nand B form a righthanded system of mutually orthogonal vectors, and so there exists scalars a and b such that B x T = aN and N x B = bT. Thus, W2 0 and aW3 k.
=
=
Similarly,
dN/ds
= wdT x N) + w2(N x N) + w3(B x N) w1B + 0  w3bT =  k T + rB. w xN
This gives us
Wl
= r,
bW3:; k
and
a
= b.
Finally,
dB/ds
w x
B=
(T x B) + w2(N x B) + w3(B x + 0 + 0 = TaN = rN,
Wl
wl(T x B)
which gives a ::: 1, and this implies that b == 1 and W3
= k.
B)
Therefore,
w == rT+ kB.
4.3: VECTOR FIELDS
GOALS 1. Be able to sketch simple vector fields. 2. Understand the relationship between flow lines and a vector field.
STUDY HINTS 1. Vector field. This is a mapping from ]Rn to ]Rn . Note that the dimensions of the spaces are equal. Each point x in the domain is assigned a vector. To depict a vector field, We draw the assigned vector originating from the point x. 2. Scalar field. A scalar field differs from a vector field in that each point x in the domain is assigned a scalar, not a vector. An example is the annual rainfall at each point on the earth's surface. The wind velocity at any instant of time is an example of a vector field. 3. Gradient vs. vector field. All gradient fields are vector fields, but not all vector fields are gradient fields. For example, the vector field i + xj is not a gradient field because you cannot find an f such that of/ox = 1 and of/ oy = x. If a vector field Pi + Qj is a gradient field then oP/ 8y = 8Q / ox. This statement comes from the fact that {j2 f / oxoy = 0 2 f / oyox for a wellbehaved function f. 4. Flow lines. This is the path a par ticle would take if it was free to move along the vectors in the field. T hinking of the vector field as velocity, the flow lines would show displacement . A formula description of a flow line can be obtained by integrating each component of a vector field (or solving a system of differential equations). Flow lines c(t) must satisfy the equation
c'(t) = F(c(t)).
CHAPTER
68 SOLUTIO N S TO SELECTED EXE RC ISES 1. At each point (x, y) , we sketch the vector (2, 2) oriO"i
nating from (x, y). Alternatively, we may draw a small multiple of the vector field. T he orientation of the vectors must be maintained at 45 0 fro m the positive x axis , and all of t he vectors must have the same ma.g nitude.
/' /' /'
/' /' /'
/'
/'
/'
/' /'
/'
/'
/' /' /'
/'
5. At each point (x , y), we sketch the vector (2y, x ) orig
inating from (x , y). T he magnitude of the vectors in creases as (x, y) moves away from the origin .
/'
/'
/'
/'
/'
y
10 . At each point (x , y), draw a li ttle arrow in the direction (x ,  y ), then connect the little arr to ge t flow lines. Al tern atively, one can solve the system of differential equations dxldt = dYI dt y and write y in terms of x, then plot th flow lines y C I x fo r various consta C. Our compu tergenerated sket.ch is shown below.
=
=
x
13. If c(t) is a flow line of F , then c' (t) = F(c (t)) . The lefthand side is c' (t ) = (2e 2t , l i t , lil t f::. O. On the righthand side , we have F( c(t)) = (2e2t , lit,  1It 2 ) since x = e2t and z = We got the same result for bo th sides of t he equation, so c(t ) is a flow line of F .
69
VECTO RVALU ED FUNCTIO NS
18. From the ch ain rule, we have dV(c(t)) dt
= \7V(c(t)) . c'(t).
We are given that c(t) is a flow line of F = \7V, so c'(t) = F == \7V and
dV~~(t)) = c'(t)
. c'(t) =
llc/(t)112 :::; 0.
Thus, the derivative of V :::; 0, which shows that V is a decreasing function of t. 20 . First , we compute
At each point (x,y), we draw the vector \7V. Notice that far from the origin, the denom ina tors of  \7V, (x 2+y2) 2, will be much larger than t he numerators . Thus , the magnitude of  \7V is very sm all at points far from the origin . A few computations reveal that  \7V is very large in m agnitude near the origin . For example,  \7 V(0.5, 0) == (4, 4) and \7V(O. l , 0.1) = (50,50). A sketch of  \7V is shown below at the left. To sketch the equipotential surface V( x , y) = (x + y)/(x 2 + y2 ) 1, we rearrange and (x + (y Therefore, complete the squares to get (x 2  x + + (y2  Y +
i) =
t)2
=
t)2 = t. the equ ipotential surface V(x , y) = 1 is a circle of radius V"f, centered a t (~, t). This is shown
i)
below at the right .
y
,
,
2
/
/ 2
1
2
0
x
2

x
,
4.4: DIVERGENC E AN D CURL GOALS 1. Given any vector field , be able to compute its divergence.
2. Given any vector field in 1R 3 , be able to compute its curl. 3. Be able to explain the physical significance of the divergence and the curl. 4. Be able to manipulate expressions involving the cross product, the dot product and the del operator .
CHAPTER 4
70
STU DY HINTS 1. T he operation \1 . This operator , called "del," tells you to assem ble the vector of partial deri vat ives: (%x, %y, % z). 2. Divergence. Note th at the divergence is a scalar , not a vector. Know the t wo notations: \1 . F = di v F . You should know t hat the divergence is a rate of expansion or, if it is negative, compression. T he term incompressible means that div F = O.
3. Curl. Note that the curl is a vector , not a scalar. Know the two not ations: \1 x F ::: curl F. You should know t hat the curl is associated with rotations and that the term irrotational means that curl F = O. 4. Valid space f or curl. Note t hat curl is a property of ]R3. We do not attempt to take the curl of vectors in dimensions higher t han t hree. T he twodimension al vector xi + xyj is taken to mean xi + xyj + Ok if a curl is desired .
5 . Laplacian. The expression \1 2 f means \1 . (\1 f) , which is a scalar.
=
=
6. Theorem s. T he facts t hat \1 x (\1 f ) 0 and V . (\1 x F) 0 are useful ; it is nice to commit t hese to mem ory. If you need to know one of t hese facts and you forget t hem, you can always do t he comp utation . 7. Formulas in JR3. Note that t he cross product or curl occurs in some of the basic identities of vector analysis in the t able preceding exam ple 15; therefore, it is assumed that the formulas are used in ]R 3 and not in higher dimensions . 8. Basic iden tities of vector analysis. T he for mulas in the table preced ing Example 15 are useful for developing the theory of vectors; however, you should not memorize the t able. You can refer back to the table as needed . It will become obvious which formulas are most important as you refer back t o them frequently.
g. Exercise 30. These fo rmulas are referred to quite frequently in the examples. You should do this exercise even if it is not assigned .
SOLUTIONS TO SELECTED EXERCISES 2. T he divergence is \1 . V
= tx (yz ) + :y (x z ) + : )x y) = O.
7. Recall t hat flow lin es are defined by
c' (t)
= F(c(t)) .
In this case , we have c' (t) = (dx / dt,dy/ dt) = (y, O) .
ince dy / dt = 0 , we know y is some const ant, k l .
Then dx/dt = y = k l , so X = kI t + k 2 , where kl and k2 are constants. Thus, the flow lines have the form (kIt + k2 , kJ). If kl is positive, flu id flo ws from
left to right as t increases , and if kl is negative , fluid
flows from right to left, as shown in t he sketch. We com pute the divergence: \1 . F = 8(y)/8x +8(O) /8 y =
O. The sketch shows that flu id appears to be neither expandin g nor contracting , which is consistent with our calculations.
t
• •... .=r=; y
1
....
....
.... ...
11. The divergence is
\1. F
= :x (sin (xy)) 
: y (cos( x 2 y))
= ycos(xy) + x 2 sin(x 2 y).
..
x
71
VECTORVALUED FUNCTIONS 14. The curl is i
j
k
xz
0 oz XY
() a ax oy
\7x F
yz
= i (~(Xy) oy
~(xz)) oz
 j
(~(Xy)  ~(yz)) ax oz
+ k (:x (xz)  :y (Yz) )
+ (z 
(x  x)i  (y  y)j
z )k
= O.
17. The scalar curl is the coefficient of k when \7 x F is computed for a vector field F in lR 2. Here,
we have
\7 x F =
i
j
a ox
a oy
Sill X
cosx
k
a
=
oz
[:x (cos x) ; y (sinx) ] k = (sinx)k.
o
Therefore, the scalar curl is  sin x. 22. First, we compute that \7 J = (y + z)i
i
\7 x \7J
+ (x + z)j + (y + x)k.
o
o
J
k
ax y+z
oy x +z
az y +x
[ ~(y oy + x) (1  1) i  (1 
Then
o
~(x + Z)] i  [~(y + x)  ~(y + z)] j oz ox . oz + [:x (x+ z) ;y (y + z)] k l)j + (1  1) k = o.
25. We know that the curl of a gradient is the zero vector. Thus, it suffices to show that \7 x F We have
i j k 000 ax oy oz y cos x x sin y 0
\7 x F =
= (0  O)i + (0  O)j + (sin y  cos x)k
i=
O.
i= o.
28. We refer to the table of basic identities of vector analysis, which precedes Example 15. (a) From identity (5), div (F + G) = div F + di v G = 'V . F + \7 . G, which sums to zero, according to the given hypot hesis . Therefore F + G does have zero divergence. (b) From identity (8), div (F x G) = G· curl F  F·curl G, which , in general, does not equal O. As a simple example, let F = xyi  z yk and G = i. We have \7 . F = \7 . G = 0 and div (F x G) = z. 30. (a) We have \7(l/r) = \7(1/ Jx 2 + y2 + z2) . Begin by findi ng (8/ox)(1/r) or
a ( ox
Jx2
1 )
+ y2 + z2
2x xx = 2( x 2 + y2 + Z2)3/2 = (x 2 + y2 + z2)3/2 = ;:J'
By symmetry, (%y)(l/r) = _ y/ r 3 and ({)j oz)(l /r) = z/r 3 • Putting these together, V'{l/r) = (xi + yj + zk)/r3 =  r / r 3 . In general, \7 (rn )
= \7((x 2 + y2 + z2)n/2)
CHAPTER 4
72
and
(O /O X)( X2
+ y2 + Z2r/2) = (n/ 2)(x 2 + y2 + Z2)n/21)2x = nx1'n  2,
so by symmetry,
\7 (1'n) = nrn  2(xi + yj
+ zk ) =
n1' n  2r.
Finally,
= \7(log( J
\7(logr) and
a
(log( Jx 2 + y2 ox
+ z2 )) =
so by sym metry, \7(logr) = (xi + yj (b) Using the results of part (a),
\7 2 (1/1')
x2
+ y2 + z2)) ,
a
1
J x 2+y2 +z2
.  ( J x2 ox
+ y2 + z2) =
x 2' l'
+ zk)/1' 2 = r/1'2 .
\7. \7 (l / r ) = \7. (r / 1'3 )
=
= 
[: x C JX 2 +:2+ Z2 )3 ) +:y
CJX2+~2+ Z2)3)
C + ~2 +
+ :z
J x2
z2) 3) ].
The first partial derivative is
x
(x 2 + y2
+ z2)3/2 _ ~Jx 2 + y2 + z2 (x 2 + y2 + z2 )3 2 x + y2 + z2 _ 3x 2 (x2 + y2 + z2)5/2 .
:x Cx2 + y2 + z2).3/2) By symmetry,
a( oy
and
a( OZ
Then
\7.
y (x 2 + y2
(x 2 + y2
)
x 2 + y2 + z2 _ 3y2
= (x2 + y2+ z2)5 / 2
)
x 2 + y2 + z2  3z 2
= (x2 + y2 + z2)5/2 .
+ z2)3/2
z
+ z2 )3/2
.2x
(r) = 3(x 2 + y2 + z2)  3(x 2 + y2 + z2 ) = O.
1'3 (x2 + y2 + z 2)5/2
Simil arly, part (a) tells us that in the general case,
\7 2 (1'n) = \7 . \7(1'n)
= \7 . (nr n  2r) .
We compute
: x « x 2 + y2
+ z2) n/ 21
. x ) = n [( x 2 + y2
+ z 2 t/ 2  1 + 2x2 (~
_
1) (x 2 + y2 + z2) n/ 22] .
Again, by sy mmetry, \7 . (n1' n  2r)
n [3(x2 + y2 + z 2t/ 2 1 + (x 2 + y2 + z2)(n _ 2)(x 2 + y2 n1'n z(3 + n  2) = n(n + 1)1'n2.
+ z2r/ 22j
(c) The identity \7 . (r/r 3) = 0 follows immediately from part (b) since n = 1 in that case. For the general case , use part (b) again to compute \7 . (1'''r). Note that we only need to divide
73
VECTORVALUED FUNCTIONS the general result by n and ch ange n  2 to k (Why?) . T hen \7 . (rkr ) = (k + 3)rk = (n (d ) By a dired computation, we get
i
0 8x x
'\7 x r =
J
0 8y y
+ 3)rn.
k
8 az z
= 0.
Using the fact that the urI of a gradient is the zero vector, the calculation j ust completed, and t he general case in part (a), we get
\l x (rnr ) = rn ('\7 x r ) + \l(rn ) x r = 0
+ m· n  2 (r x r ) = o.
32 . (a) 1
j 8
3x y
x3 + y3
o ox2
curl F = \l x F
k(3x 2

k
oy
o oz o
= o.
3x 2 )
(b) Note that if F = \l I, then 3x 2 y = (%x) /(x, y) and x 3 each equation:
f( x, y) =
+ 11
= (%
y)/(x, y). Integrate
J
3x 2 y dx = x 3 y+g(y),
where 9 is a fu nction of y only, and
where h is a function of x only. In bo th cases, I (x , y) must be the same , so compare both sides. If we let g(y) = y4 / 4 and h(x) be an arbitrary constant , then f(x,y) = x 3 y + y4 / 4 + C satisfies \l f = F.
33. (c) Let F( x, y) = e'" cos yi  eXsin yj , where the j component is the real part of ex  iy and th j component is the imaginary part . We calculate div F = (0/ ox)( eX cos y)  (%y)( eXsin y) = eX cos y  eX cos y = 0 and i
curl F
j
o ox
o oy
k
a
8z
eX cos y e'" sin y 0 k(  ex sin y + e" sin y) = o. Thus, F is both incompressi bl and irrota tional.
SOLUTIONS TO SELECTED REVIEW E XERCISE S FOR CHAPTER 4 2. The velocity vector is
vet)
= e'(t) = 2ti + ( 2t sin(t
))j + 4t 3 k. The acceleration vector is 4t cos(t )lj + 12t k. When t = ."fi, we have v( y'ir) = 2
+ [2sin(t ) 2y'1ri + 47r."fik and a(y'ir) = 2i + 47rj + 127rk. The speed is the length of the velocity vector. When t = ."fi, the speed is [47r + 1611"3]1/ 2 = 2/11" + 4~ . The equation of the tangent lin IS given by l(t) e(y'ir) + tv (y'ir) = (r.  1,  1, r. 2 ) + t(2."fi, 0, 4r.y'ir). aCt) = e" (t ) = 2i
2
2
2
2
=
7. We use t he equation F = rna = me". Here, e" = (2,  sin t,  cos t) . At t = 0, a (O) (2,0, 1), so F(O) = rn(2, 0,  1).
_• ....a. _____ .
CHAPTER 4
74
9_ From x 2 = .; = Zll , we get y = X 2 / 3 and z = X 2 / 5 . Let x = t, so the path of the curve is c(t) = (t, t 2/ 3, t 2/ 5). Here, we have 1 ~ t ~ 4 and c' (t ) = (1 , it1/3, i t 3/S). The arc length is
t VI + i9t
1
4I1C'(t)11 dt =
11
1
2/3
+ i. t 6/ 5 dt. 25
13. We set the given equation equal to t and solve for x, y and z: x I = 2y + 1 = 3z + 2 = t. From x I = t, we get r = t + 1. Similarly, 2y + 1 = t yields y = (t  1)/2 and 3z + 2 = t yields z (t  2)/3. Therefore, we get (.7:, y, z) (t + 1, (t  1)/2, (t  2)/3) .
=
=
15. We want to show that c'(t) = F (c(t) ). Here, x = 1/(1  t) , Y = 0 and z = et /(1  t). We ~et dx / dt = 1/(1 t) 2, dy / dt = 0 and dz/ dt = (2e t  t et )/(l t)2. T hus, dx / dt = .:z:2 , dy/ dt = 0 and dz /dt = z (l + x) = let /( 1 t )][l + 1/(1 t )l = (2e t  tet) / (l  t )2.
+ /y (y2 ) + f. (z2 ) = 2x + 2y + 2z .
18. We compute div F = \l . F = fx( .:z:2) i
\l x F = I f}.:z: .:z:2
8u y2
8z z2
i
8 ay z
{}x
y
8x
f}z
8x
= f}y/ f}x + az/ 8y + 8x / 8z = O. k 8
J
a
[8  (z 2)  8 (x 2]. ) J + [{} (y2)  8(r 2] ) k = o.
1
Z ) 
21. The divergence is \l  F
=I

[aya( 2 8za( y2)] '
=
\l x F
k
888
=
curl F
j
Also,
= ({}X_ {}Z ) 8y
8z
{}z
ay
The curl is
i + ( f}y _ {}x ) j + (8Z _ 8Y ) k = ij k. {}Z ax (}x 8y
x
25. We compute
\lJ
= ( ~~ ) i + ( ~~ ) j + ( ~~ ) k = [2.:z: exp(z2 ) + y2 sin (zy2)]i + [2xy sin(xy2 )li + Ok.
Next, we compute I
\l x \lJ
=
8x 2x exp(x 2 ) + y2 sin(xy2 )
[:Y (0) 
k
J
{}
:z (2xYSin( X
8 ay 2xysin(zy2)
y2))] i +
[:z(2x
8 8z 0
exp (z2 ) + y2 sin(xy2 ))  : z (0)] j
+ [:x(2x y sin (x y2))  : y (2xex p (x 2 ) + y2Sin(Xy2 ))] k
=
(0  O)i
=
o.
T hus, for J (x, y)
+ (0 
= exp (x
2
) 
O)j + [2y sin(xy2 ) + 2x y3 cos (xy2 )  2y sin (xy2)  2xy3 cos(xy2 )] k
Cos(xy2 ), we have \l x \l J
= O.
J exists, then we know that F = \l1 = ({}f/{}x )i + (al/ ay )j + (8f/8z)k . Thue , we have {} 1/ 8x = 2xye Z , 01/ oy = eZ x 2 and 81/ oz = x2ye~ + z2. Integrating aI/ox = 2xye Z with respect to x gives us I(x, y, z ) x2ye~ +A(y, z ), where A is some function involving yand Z only. (You can check this result by computing (}f/o x. ) Similarly, integrating (}f/ ay with respect to y gives us l (x, !I , z) = z2ye~ + B (x, z ), where B is some function involving x and z only. Next, in tegrating oj/ 8z with respect to z gives us I( x , y, z ) = z2ye z + z3/ 3 + D(x , y), where D is some function involving x and y only. Comparing the three resulte for I(z,y , z ), we see that A(y, z ) = z3 / 3 + C, B(x , z ) = z3 / 3 + C and D(x , y) = C, where C is an arbitrary constant. Therefore, f (x, !I, z ) = z2ye~ + z3 / 3 + C.
28 . (b) If such
=
75
VE CTORVALUED FUNCTIO NS
32. (a) At the intersection, we know that y = 1, so t he intersecting curve has the equation x 2+( 1)2+z2 = ;3, or x 2+z 2 = 2. Eq uivalently, we have x 2/2+z 2 /2 = 1, or (x/V2)2+(z/V2)2 = 1. Since we know that cos 2 t+sin 2 t = 1, we get x/V2 = cost and z/V2 = sin t, or x = V2cost and z = V2 sin t. To get one revolution of the intersecting curve, we let t vary from 0 to 2r.. Thus, a parametrization of the in t ersecting curve is x = V2 cost, Y 1, z V2 sint,
=
o < t < 2r..
=
(b) From Sect ion 2.4, recall that an equation for a tangent line is l(t ) = c(to) + (t  to)c'(to). Here, c(to) = (1, 1,1), where c(t) is the param etriza tion given in part (a). We want (V2 cos to, 1, V2 sin to) = (1, 1, 1), so t = r./ 4. Differentiating each component of the parametrizat ion, we get cl(t) = (V2 sin t,O,V2 cost) . T hus, the tangent line is
l(t)
= (1,1,1) +
(c) In par t (b), we computed c'(t), so Illc'(t)11
f 2rr
io
(t  r./4)(  1, 0,1).
= V2. Thus,
1
the arc length is
2rr
=
Ilc'(t)11 dt
0
h dt
= 2hr..
As expected, this is the circumference of a circle of radius V2. 36. (a) T he direction of the rotation vector w is the same as the axis of rotation, in this case, the positive z axis, or k direction. vVe're given the magnitude as 4, so w = 4k. ,( b) When r = !5v'2 ~ i j), the velocity is
v
=w
x r
=
0 5V2
j 0 5V2
k 4 0
= 20hi
20hj .
(c) The vector r is the vector from the axis to the point , so r is the vector from (0,0,5) to (0, 5V3, 5), or r = .5V3j. For the point (0, .5V3, .5), we get J
v
=w x r =
0
o
0 .5V3
k 4 0
= 20V3i.
T EST FOR CHAPTER 4 1. True or false. If false, explain why.
(a) The curl of any vector field in
]R3
is another vector field .
(b) One revolution of the path c(t) = (cos2t,sin2t) has arc lengt h (c) If band c are vectorvalued func t ions of t, then d(b x c)/dt
f;rr Ilc'(t)11 dt.
= db/dt x c +
dc/ dt x b.
(d) Two particles which have t he same acceleration must travel at the same speed. (e) Suppose G(x, y) is the velocity field of a gas. If G(x, y) = xi, then the gas is expanding and if G(x, y) = yi, then the gas is not expanding . 2. Let c(t) = (2t, t  3, t 2 + 1) be a possible flow line for a velocity vector field . W hich of the following, if any, could be such a velocity vector field?
(a) F(x, y, z)
= (2, x 
2y + 7, x).
(b) F(x,y,z)=(yiz ly+5,1, 2y 6).
3. Suppose the velocity vector fields represent the flow of a fl uid. At what poi nts in does the flui d lack rotation?
]R 3,
if any,
'


76
CHAPTER 4
(a) F( x, y, z)
= 4x yi + 2x 2j + k.
(b) F(x , y, z ) = xyi
+ zj + xk.
4. A particle travels counterclockwise along the ellipse 9x 2
+ 4y2 = 36.
(a) Express the distance traveled going from (2,0) to (0, 3) as an integral of the form J + b cos 2 t dt for constants a and b.
va
(b) Suppose t he part icle had flown off on a tangent line at (l ,3 V3/2) . Where will the particle hi t the xaxis? 5. For a given function , g(t), let a particle's posi tion be described by 2g(t)i
+ [g(t)j2j 
k.
(a) What force is acting on the particle if its mass is 3 units? (b) Wh at conditions must be imposed on g(t) for the force to be O? 6. Let F(x , y, z) = xi + yj gas expanding?
+ (z2 + y) k
be the velocity vector field for a gas. Where , in ~3, is the
7. In R 4 , a path is described by y = 2x, z = 4x 2  6 and w as an integral in y for 0 ~ x ~ 2. Do not evaluate.
=e
Z
•
Find the arc length of the path
8. Which of the following, if any, is the curl of some vector field G such that F
= curl G?
(a) F(x, y, z ) = 2i + 3j  2k.
(b) F (x, y,z ) =xi + 2j  zk.
(c) F (x , y,z )= yi + yj + x 2 k. 9. (a) Let v(t) = (sin 2 t)i respect t o x?
+ (t 4  t 3 )j + 3k.
What is t he deri vative of v( x 3
(b) Let c(t) be a path in ]R 3 and let p(t) be a scalar function. acceleration of p(t )c(t) .

x2
+ 3x 
4) with
Find a formula for the
10. At the annual camel races, camel A 's position is given by (cos 5t, 2 sin 5t) and camel B's position is given by (cos 4t , 2 sin 4t). (a) Show that the camels are running on the same path . (b) Suppose camel A 's mass is 500 . What force is being generated by camel A at t = 7r/5? (c) Camel B saw an oasis at t = 7r/ 5. It ran off on a tangent line. What is the equation of the tangent line? (d) Suppose camel B had continued to run on the track. How far behind (in distance) was camel B when camel A had com pleted one lap? Leave your answer in the form of an integral.
77
5
DOUBLE AND TRIPLE INTEGRALS
5.1: INTRODUCTION GOALS
•
1. Be able to calculate double integrals over rectangles.
2. Be able to use Cavalieri' principle.
STUDY HINTS 1. Some notation. (a) The cartesian product of two intervals in lR 2 is a rectangle. If a ~ x ~ b
and c ~ y ~ d, then it is denoted [a, b] x [c, d]. (b) Sometimes dx dy or dy dx is abbreviated dA, the "differential of area."
2. Review. Before continuing, you should re iew integration techniques for one variable. It is essential that you remember how to integrate by parts and by subst itution .
lID
3. Geometric interpretation. If f(x, y) ~ 0, then the double int gral f( x, y) dA is the volume under the surface defined by the graph of z f (x, y). Recall that with one variable, f(x) dx is the area under the curve y = f(x).
=
I
4. Computing a double integral. Just as with partial d rivatives , all but one variable is held
constant in each step. We integra te from the inside and work our way to the outside. For example,
i
bi df(x, y)dy dx = Ja,b [F(x ,d)F(x,c)]dx,
a
c
where Fis an antiderivative of fwhen x is held constant. Now , we com pute the second xintegral using onevariable methods. 5. Cavalieri's principle. This principle is used in most onevariable calculus courses to derive volume formulas. These fo rmul as are often referred to by names: disk or slice method.
2:7:=0 f(Ci)( Xi +!  Xi ), ci can be chosen anywhere in [Xi , xi+d . In a Riemann sum, we take the limi t as n + 00, so in most cases, ! (c;) is almost independent of Ci because X i +! and Xi eventually get close together.
6. Riemann sums. Be aware that in the sum m ation
SOLUTIO NS TO SELECTED EXERCISES 1. (b) First hold x constant and integr ate with respect to y, then integrate with respect to x:
t (ycos x +2)dy dx ior/ Jo 2
r/ (12' io 2
cos
)
x+ 2
dx
= ( 2'1 sin x + 2x ) 17f/2 0 = '12 + 'fr.
78
CHAPTER 5
2. (b) We can change t he order of integration . Thus , we hold y constant and integrate with respect to x firs t:
r/ (y cosx + 2) dx dy Jor Jo i
2
Jot
1"'/2 ]
[(y sin x + 2x ) x=O dy
1\y + dy= (y; + 7r I: = ~ + Y)
7r)
7r.
As expected, we get the sa me answer regardless of the order of integration.
3. By Cavalieri 's principle , the volume of a solid is b
V
=
1 a
A(x ) dx, •
where A (x ) is the crosssectional area cut out by a plane. Note that at the same height, a crosssection of the lefthand side is a circle of radius 7' and a crosssection of the righthand side is also a circle of radius r . Since A (x ) is equal in both cases, the volumes must be equal.
5. With the setup in figure 5. 1.12, we slice W vertically by planes to produce triangles Rx of area A(x ) in the figure. T he base b of the triangle is b = .Jr2  x 2 and its height is given by h = b tan (j = .Jr2  x 2 tan (j, Thus A(x ) = ~bh = ~ (r2  x 2) t an (j . Hence, the volume is
j
r _rA(x) dx
=
jT 21 (r
2
21 t an (j
( 2r 3
_r
32r

8. Note that when y is in the interval [1 ,0], then
Jl (Iyl
cos
~x )
r
3
2 1 ( r 2 x '3 x ) x )tan(jdx= 2(tan(j) 3 )
= 32r
I
r
3
tan (j.
Iyl = y,
and so
2
dydx
= Jor jO1 (YCOS 7r4X) dy dx O 2 r [(_y2 7rX)I Jo 2 cos 4" y=l
1 2
o
1dx (1 7rx) dx = 2. 7rX  cos 2 4
sm 
I
7r
12
2
0
7r
4
10. Since f( x, y) ~ 0 for all points in R, ffR f( x, y) dy dx is the desired volume. It is
1 2
1 \1+ 2x + 3Y)dY dx =
12
[(Y+ 2X Y +
12 (2X+~)
3f )I~=J dx
dx=
5. 2: THE D OUBLE INTEG RAL OVER A R ECTA N GLE G OALS
1. Be able to com pute a double integral over a rectangular region.
2. Understand Fubini's theorem .
(x2+ 52X)
I:
11 2
DOUBLE AND TRIPLE INTEG RALS
79
STU DY HINTS 1. Definition. The definition of the integral is more importa nt fo r theoretical rather than com putational work . It is defined to be a limit of Riemann sums:
J1
! (x, y) dA
= nl~n;,
R
n 1
L
f ( Cjk )
~x ~y.
j ,k=O
Although it is not required, it is usually convenient to use a regular partition, i.e., a partit ion with ual spacings. 2. Properties. Many of the properties that hold for single integrals also hold for dou ble integrals. Some of these include the integrabili ty of any continuous or even piecewise conti nuous fu nction .
3. Warning. As in onevariable integration, [fJRfdA] [fJR g dA]
=1= JJR!gdA, in general. In other words, the product of two integrals does not usually equal the integral of the product.
4. New terminology. T h nam s of the proper ties are known as lineari ty, homogeneity, mono
tonici ty and additivity. You should know the statements even if you don 't know the names.
5. Fubini's theorem . Th is tells you tha.t, for most reasonable functions, you can int grate one variable at a t ime and the or der of integration does not matter. 6. Double integrals and volumes. Recall that if !(x , y) 2: 0, t he double integral is simply the volume of the region between the graph of !(x, y) and the xy plane. If f (x, y) < 0, then we subtra t the correspond ing volume between the graph of !(x, y) and the xy plane.
SOLUTIONS TO SELECTED EXERCISES 1. (b) We compute t he do uble integral as an iterated in tegral:
Note that we chose to integrate in x first be ause that integral is simple (integrat ing with respect to y first would requ ire integration by parts). 2. (b) T his is the integrated integral
fa 1[ (a~2 + (by + C)x) [=al dy fa (~ + by + c) dy = [b~2 + (~ + c) yJI~ 1
a
b
"2 +"2 + c. 5. We will use t h fact t hat J cf(x ) d:c = c J f( x ) d:t for any cons tant c. If we integrate in x first, then g(y) is held constant in the fir t st p, and so
J
k[J(IX )g(y)]dx dy
=
ldib ld lb
f (x)g(y) dx dy
[g( y)
b = ld [ i
1
! (x ) dx dy ,
1
f(x )g(y) d:c dy
CHAPT ER 5
80 where
J: I( z ) dz is a constant in
y. Factoring out the constant integral gives us
[i d
[ib I(X)dXl
n, so t he double integral represents
7. The fu nction x 2 + y is positive over
111\x2 +y) dydx
g{Y)dyl .
11[
=
(
x3
3
+ 3X ) 11 2
J
J J J
=~ + ~ =~. 3
0
2
6
x 2  Z2
=
tan 2 () 2
x' sec 411u (xsec 0dO)
y
,>
2
x3(1  tan B) dB X4 sec 2 B
=
J
2
2 cos 0  sin (J dB z
cos 2B dO = sin 2B + C. x 2x
From the triangle, we have sin e = y/ ';'2 z::+y2 ~ and CO! 0 = z / sin 2B 2x
dx
x sec 2 OdO and
11. Begin by substituting y danO, so dy J x 2 + y2 = X sec e. We get
y . .. d (x~" + y 2)"
dx
y; )
1 16 .
Thus, the volu me is
x2  y2
[=J = 11(x2 + ~)
+
2
(YX
the desired volu me.
+C =
sin e cos e x
+C =
~ . __ y x Vx2+ y2
x
J z2 + y2. Then
x J x2 + y2
+C=
Y +C X2 + y2 '
Evaluating, we get
[I x 2 Then
111x2 
1 o
_ y2
y2
(2
2)2
x +y
0
y
(x 2 + y2 )2 dy =
10
dydx
=
Xl
11
+ yl
y2 (x2 + y2 )2 z2 
J
1
0
SO
dx
2
J J11 (1 y2
1
y=0
2~dz x +1
For the second integral, we substitute z = ytan , This gi ves :;:~:;:: dx
11
= x2 + l '
l
11"
= ysee 2 ¢d &nd J
tan  y2 ( sec 2 #) y4 sec4 ¢ Y tan y4 sec 2 ¢
1
= tan  1 xlo= 4 '
J
xl
= 11 (tan 2 ¢ 
+ y 2 = y see . 1) d
y4 sec 2
d .
Using the method above, t his becomes  x / {x 2 + y2 ) + C. Evaluating, we get X2  y2 ~ dx 
1
1
o (xl
Finally,
111 1
+ y2) 2
x2  y2
o
( 2
0
x +y
2) 2
dx dy =

11 0
 x 11  1
~ x 2 + y2 .: =0  1 + y2 . 1  2 dy 1+ y

=  tan 1 y I01 =
Fubini's theorem does not apply in this case because (x 2  y2 )/(Z2 the entire domain , namely, at (0 , 0) .
11"
 4 .
+ y2 )2
is not bounded on
81
DOUBLE AND TRIPL E INTEGR ALS
5.3: TH E DOUBLE INTEGRAL OVER MORE GENERA L R EGI ONS GO A LS l. Be able to compute a double integral over an elementary region on the xy plane.
STUDY HINTS
=
=
l. Region types. A ysimple region is bounded by the lines x a and x b and two curves which are functions of x, for a ~ x ~ b. Such a region is called ysimple because the curves that form the boundary can be described by yas a "simple' function of x. Similarly, an xsim ple
region is bounded by the lines y = c and y := d and two curves which are functions of y, for c ~ y ~ d. Such a region is called xsim ple because the curves that form the boundary can be described by x as a "simple" function of y. A sim ple region may be classified as both ysimple and xsimple regions. See figure 5.3.3 of the text . 2. Classifying regions. For purposes of integrating, it is more important to be able to recognize a region type rather than being able to name it. Your primary concern should be learning how to perform double integration.
3. Simplifying complicated regions. Most plane regions may be broken up into regions each of which is ysimple or xsimple. For example, the region at the left is divided into six peices, each of which is either ysi m ple or xsimple. In fact, many of the subregions are both ysimple and xsimple.
4. How to integrate. It is best to use the iterated integral
J: J:12(~/ f(x, y) dy dx
for ysim ple
regions. For xsimple regions, it is best to use the iterated integral JcdJ:12(~/ f(x, y) dxdy.
5. Choosing integration limits. When you perform multiple integration, be sure that the limits of integration do not include any previously integrated vari ables. In particular, no variable should appear in the limits of the ou termost integral. For example, the integral of (x + y)2 over the region D defined by 9 y ~ x 2 , 0 ~ x ~ 1, should be written as
:s
not
i
x'
1
10 (x + y)2 dx dy.
Sketching the region often heJps in choosing your limits. Also, it helps to read the limits of integration in this example as follows : "while y ranges from 0 to x 2 ; x ranges between 0 and I" or "for each x between 0 and 1, y ranges between 0 and x 2 ." Draw a picture and think about it. 6. Definition of the integral. As in the last section , this is important prim arily for theoretical
purposes. Up to this section, we only know how to integrate over rectangles, so we cover a region D with a large rectangle and let f(x, y) = 0 outside of D. 7. Double integrals and area. If f( x, y) = 1 on a region D , then
JIDf(x, y) dA is the area of D.
·
~.
CHAPTER 5
82 SOL U TIONS TO SELECTED EXERCISES l. (b) The iterated integral is
1 [1 + 3x
1
1
=
dy ] dx
2
2x
12 ( 13x+1) 12 + y
y=2x
1
(3x
1\
x
y = 3x + 1 dx
1  2x ) dx
+ 1) dx =
( ~2 + x ) I: = ~.
At each Xo in [1,2], the region extends from 2xo t o 3x o + 1,
so we get the region shown in the sketch . This is a ysi mple
region because it can be described by
1~ x
~
2
and
2x
~
y
~
x
3x + 1.
2. (b) First, we will sketch the region. At each Xo in [1 , 1], the region extends from 2'lxol to Ixol, so we get the region in the sketch . The integration is most easily done by di\riding the region into two parts, so
l1
1X '
1
1
eX +Y dy dx =
 1  2/x l
1x
10i X 2x
 1
Note that when x is in [1 , 0]' the sum is
1 0  1
11 1
=
eX +Y dy dx .
2x
0
Ix I =:
1 0
X
( ex+yl ) dx y=2x
x
eX +Y dy dx+
1
x . The first in tegral of
(1 e3x ) dx
(x  e~X) [1
2 1 "3  3e3 '
The second integral of the sum is
11( o
) dx eX +Y IX y=2x
=
11
(e 2x  e X ) dx ::::
0
Therefore, the entire integral is e2 / 2 + l/e  1/3e 3 (e) The iterated integral is
11 [l>x + n
ym)
dX] dy
t Jo
t Jo

(e,2 + e2X
11 =  +   . 2
0
e 2
1 e
3 2
5/6.
Y
[ ( _x n+1 n +1
x)
+ xym) I
1dy
X= y2
(_ y'n_+_1 + ym+1 _ _y 2_n +_2 _ ym+2) dy n+1 n+1 yn+2 ym+2 y2n+3
ym+3 ) ( (n + l)(n+ 2) + m+ 2  (n+ 1)(2n+ 3)  m+3 1 1 1 1
(n + 1)::(n+:::2) + m+2  (n + 1)(2n + 3)  m+3' To sketch the region, note that for every Yo in [0 , IJ , x extends from x = y~ to x
= Yo.
11
10
83
DOUBLE AND TR IPLE IN T EGRALS
y
x
4. The equation of an ellipse with serniaxes a and b is x 2 / a 2 + y2/b 2 = 1 or x 2 /b 2 + y2 / a 2 = l. In either case, the areas are equal. We will use the first equat ion , find t he area of the region in the ·first quadrant , and t hen multiply by 4. The region is descri bed as
The area of a region D can be computed by A = ffD dx dy.
In this case, we have
11
1 A=
4
a
a
2 o b ..j1 x /a dydx =
a
1" ( l y
a
b..j1x o/a»
1 g2 a
dx =b
y=O
1 2 dx. a
a
1
Let u = x/a to get b f01 a..jl  u 2du = ab f0 .Jl=U2 duo The integral f01 ~ du is the area of a quarter of a circle of radius I, which is If/4. So A/4 ablf/4. T herefore , the area of the ellipse is A = ablf .
=
7. The region D, shown in the sketch , can be described as
 V3/2 ~ 3
x
y~
V3/2
and
0 ~ x ~ _4y2
+ 3,
so
J.JDr 3
3 j ,;3/2 r + x 3y dx dy ,;3/2 JO ,;3/2 (x4 1 4y 2+3 )' dy = j ,;3/2 y 4y O
x y dx dy =
=
Let u is 0.
j ,;3/2
4
x =O

(_ 4y 2+ 3)4y dy . ,;3/2 4
= _4y2 + 3 and we get an integral whose limits of integration are °and 0, so the integral
12. From section 2.1, we know that an equ ation of the form z2 = a (x 2 + y2) is a cone. To get t he t ip at (0,0, h) as shown in the figure, the equat ion becomes 2 2 z2 = h a( x +y2) . W hen z = 0, we get h = a( x + y2 ).
2 If a = h/r2, we get r 2 = x + y2 , a circle of radius
r. Thus, the equation of the cone is z = f(x, y) =
Jh  (h/r 2)( x 2 + y2). T he region of integration is
the cone's base, which is described as
z
       ~.~. y
x
__a_
_
~_~
_
CHAPTE R 5
84 Therefore, the volume of the cone is
J rJ~ V/ :2 (X2 + y2) h
 ..!r'x'
r
dy dx .
Fortunately, we are only asked to set up the integral. We can "simplify" the integral by using t rigonomet ric substitut ion or by finding the following equation fro m an integral table:
 x2 + j Ja2x 2 dx= ~Ja2 2
2 0
2
sin 1
(~) a
for a>O.
By using the equ ation above, the volum e of the cone becomes
J:r [(rJh  V; x 2) sin  (~r21_ x2) 1dx. 1
Changing the volum e integral to polar coordinates, a technique introduced in chapter 6, makes t he calculation easy. 14. Let D be a ysimple region ; t hen D is described by a ~ x ~b
and the integral
IID f( x )g(y) dA
¢ 1(X) ~ yS4>2(X)
and
becom es
l
b
a
l¢2(X)
f(x)g(y) dy dx .
¢ , (x)
Now, we integrate in y. Let G be an antiderivative of g and note that f( x ) is constant in y. We get
l
a
b[
f (x)
i ¢2(X) g(y) dy1dx = l bf(X)[G(4>2(X)) ¢,(x)
G(4)l(X))] dx .
a
G depends on x, so we can't consider it to be const ant and we cannot factor it outside of the integral sign . Thus , in general, f( x )g(y) dy dx is not t he p roduct of two integrals.
IID
5.4: CHANGIN G T HE ORDER OF IN TEGRATIO N
GO ALS 1. Be able to evaluate a double integral by changing the order of integration. 2. Be able to st ate a nd understand t he mean value t heorem for dou ble integrals.
STU DY HINT S 1. Rati onale for changing order. Recall t hat Fu bin i's t heorem allows you t o change the order of in tegration . Somet im es , changing the order of integrat ion simplifies a problem. Try doing ex am ple 1 without changing t he order of integration and compare the efficiency of both m ethods. (You will p robably need to use trigonom etric substitution.) At other times, a double integral can only be com puted if the order is changed . 2. B egin ning the change of mYier. It is useful to sketch t he region of integration from the given lim its before choosing new limit s. 3. Mean value theorem for int egrals. If two conditions hold: (i) elementary region , then the conclusion is
JIn for som e point (xo, Yo) in D.
f( x , y) dA
f
is continuous amd (ii) D is an
= f(x o, yo) . area(D)
DOUBLE AND TRIPLE INTEGRALS
85
4. Mean value inequality. If m is the minimum value of ! (x , y) on D and M is the maximum, then
m· area(D)
Jl
~
~M
!(x, y) dA
. area(D).
This allows you to estimate the value of a dou Ie integral .
SOLU TIONS TO SELECTED EXERCISES 1. (b) Fir t, we recall that cos 2 B = (1
r/ ros 2
+ cos 2())/2
and compu te the integral as written:
9
io io
cos() drdB
= iT
4 From the graph, we see that if we choose an region can also be described as O ~ r ~ l
then B extends from 0 to cos 1(ro) . Thus, the
TO,
and
O ~ B ~ cosl(r).
r
r
a
Therefore, changing the order of integration gives us
11COSl(r) cos BdOdr = 11( sin() ICOs (r)) dr = 11si n(cos 1(r)) dr .
1 a 1
0
0
9=0
0
From the triangle, we see that sin( cos 1(7))) =~. We recognize the integral s he area of a quarter of a circle of radius 1, so 1
1
iT
~ dr = . o 4
2. (c) The region of integration is sketch here. T here is no obvious function whose derivative is exp (x 2 ) , so we try changing the order of in tegration. The region can be expressed as a ysimple region:
o~ x
~
0~ y
2 and
Y
4
~ 2x.
Thus , the integral becomes
rx ioio (2
Now let u
exp (x 2 ) dy dx =
r [ yexp (x )1y=o io 2
2X
]
dx .
=
eti
2 x
= x 2 and we get
12
2x exp ( x
2 )
dx
=
14
eti du
I: =
e4

1.
86
CHAPTE R 5
3. This formula is an application of the mean value inequa.lity. T he region D is [11",11"] X [11" ,11"], eo A (D ) is t he area of D, which is 411"2. The funct ion I(x, y) = esin (x+ y ) is largest when sin(x + y) is largest, i.e., when x + y rr / 2 ± 2n1l" , where n is an integer. Similarly, I (x , y) is smallest when x + y  11" / 2 ± 2n1l" . Over the region D, we have 1 ~ sin (x + y) ~ 1, so m li e and M = e 1 . Therefore, substituting everything into t he mean val ue inequality gives us
=
=
=
~ ~ 4!2 J L
l( x , y) dA
~ e.
t·
6. The area of the triangle is The function I(x, y) = 11(y  x + 3) is smallest when y  x + 3 is largest on D. Notice that y ~ x in D, so y  x + 3 is largest when y = x , and so m = 1/ 3. Similarly, we see that M occurs at (1,0) where 1(1, 0) = Now, the mean val ue inequality gives us
t.
~~2JlI(x,Y) dA~~
~ ~JL/(x,Y) dA ~ ~ .
or
10. We need to divid e the triangle into two parts: one part for x in
[0,1] and the other part for x in [1,2]. When x is in [0 , 1], we have x ~ Y ~ 3x, so the integral over this region is
[ 1 [3'"
J J., o
eX Y dydx
11( I::J ee  y
11 2
11
e ", ) (  2+ x
2 [4  "
1 1
J.,
eX  Ydy dx
~ y ~
(2.2) dx
+ l )dx
(_ e 2e
For the second region, we have x
x 1
0
1
= 2e 2 + 2·
4  x, so the integral over t hat region is
1 e 2 (
X

_
4 YI 
1
2X
(
Y
_e
2

4
+X
X )
dz =
11 = " )
12 1
1\e
2X

4
+ l )dx
1
=
1
1
2e 2
+ 2·
Adding th e two integrals together, we get 1 + 1/ e2 . 13 . First, we sketch the region. For a.ll Yo in [0,1], x goes from Yo to /2  y~ . To describe the region as an xsimple region , we have to subdivide the region at x = 1. When xis in [0 ,1]' we have 0 ~ y ~ x . W hen x is in [1,)2], we have 0 ~ y ~ )2  x 2 . Thus, we have
y
2 112
Jo[1
[1~ I( x , y) dx1dy = Jot [Jro I (x, y) dy] dx + Y
1v'2 [[~
In many cases, the lefthand side will be easier t o work with .
1
Jo
1
I( x , y) dy dx .
x
87
DOUBLE AND T RIPLE INTEGRALS 15. T his proof requires using the chain rule. Let
=
G( x , u)
l jd x
f(u, y, z) dz dy .
We want to find dG/dx (the "total derivative ," if you will ) when u
dG _ aGI dx  ax u= x
~~
= x.
By the chain rule ,
+ aG I au
u=x'
:x J: [t
is simple; it is equal to f( u, y, z) dZ] dy. From the fundamental theorem of (one variable) calculus, we simply take the "integrand" (the dz integral here) and replace y by x (because we are integrating with respect to y) . At u = x,
aG ~ uX
I
jdf( x , x, z) dz.
=
u=x
c
~~ i s a little trickier. First,
l
a a Jc{ d f(u,y,z)d z dy a
aG au
au
l
X
X
j
d
auf(u,y, z )dzdy,
a c
assu ming that f is a nice function such that the order of integration and differentiation can be interchanged. Now we want to evaluate everything at u = x, so we simply replace u by x, and get
aGI ~ uU u =x
=
lj X
a
c
d
() ;;f(x,y,z) dz dy .
uX
Add the two results and you've got it .
• 5: THE T RIP LE INTEG RA L GOALS l. Be able to compute a triple integral over geJleral regions in space.
2. Be able to change the order of in tegration for computing triple integrals .
STUDY HINTS l. Notation . We use dV for the differential dx dy dz since it represents a volume.
2. P r·operties. Many of the properties of the triple integral are the same as those of the double integral. T he triple integral may be considered as a t hreefold iterated integral and we integrate from inside to outside. Fubini's theorem still holds and we can still integrate all piecewise continuous functions.
3. Balls. Example 3 demonstrates one way to describe the unit ball. Another way to describe it IS
Yet another description is
The solu t ion of example 3 uses a ysim ple description of D. Using an xsimple description would generate three more descriptions of the unit ball.
CHAPTER 5
88
4. How to integrate. As with double integrals , knowing how to set up the limits is important.
Drawing a picture of the region is helpful for finding the limits of integration, although in many instances it is easier said than done. 5. Triple integrals and volumes. If
I(x, y, z) = 1,
then fffw
I dV
is the volume of W.
6. Integration trick. A timesaving device is shown in example 4. If we realize that an integral is
the area of all or part of a circle, then we don 't have to go through the process of fi...nding an antiderivative. 7. Factoring out integrals. Some special t riple integrals can be easier to compute if you use the
following fact: If the limits of integration of the innermost integral do not involve a certain variable and the limits of integration for that particular variable are constant, then that variable may be integrated separately and multiplied by the remaining integral. For example,
11 1 1
2
(1 3
3
:&
I( x)g(y)h(z ) dzdydx =
(1 1'2XI( x)g(y) dy dX) 1
h(Z)dZ)
because the variable z does not appear in any of the limits of integration and its limits of integration are the constants 2 and 3. We cannot integrate X' separately because it appears in the limits of integration for y.
SOLUTIONS TO SELECTED EXERCISES 2. It is easier to integrate in x before integrating in y:
JJl exYydV
111111
=
1111 o 0
eX'Yy dx dydz
=
1111[
(e=:Y)
11[
(le Y)dydz=
(y+ey)1
0
1] y=o
yl:=J dydz dz=1
11
e a
1 dz;;:::. e
ote that you would have had to use in tegration by parts if you had integrated in y before integrating in x. 6. It is useful to make a sketch of the region. T he elliptic
cylinder runs parallel to the y axis and its intercepts
on the coordinate axes occur at x = ± 1/ .j2 and z =
±l. The ball is a sphere of radius 2 centered at the
origin. To describe t he region inside t he cylinder and
the ball as an elementary region, let x vary between
1 / J2 jnd 1/J2. Next, for any given xo, we let z go
from  1  2x5 to J I  2X6. Finally, for any given
(x o, zo), we let y vary between two surfaces of the ball ,
namely, y =  J4  x5 Z6 and y = J4x5z6.
Therefore, t he desired region is descri bed by
1/V2 ~ x
~
1/ V2, 
Jl 2x2 ~ Z ~ J l 2X2 ,

\1"4 
X2  z2 ~ Y ~ J4 _X2  Z2,
Another possible solution is 1 ~ Z ~ 1,
 J(l z2 )/2 ~ z ~ J(1 z2)/2,
 J4  x 2 
z2
~ Y ~ J4  x 2  z2.
is a pa.raboloid opening upward. The surface Z = 10  x 2  2y2 is a paraboloid opening downward. When we equate the two surfaces, we see that they intersect where x 2 + y2 = 10  x 2  2y2 or 2X2 + 3y2 = 10, which is an ellipse. The ellipse can be described by
9, The surface z
= x 2 + y2
 V5 ~
x ~
V5
and
 J (10  2x 2 )/ 3 ~ y ~ J(10  2;2)/3.
89
DOUBLE AND TRIPLE IN T EGRAL S
As a triple integral , the desired volume is
Vsjv'(lO2XJ) /3 1 10X2 y2 j V5
v'(102X ~) / 3
dz dy dx
=
l
Vs jv'(1o2X~)/3
l
Vs
 Vs  y "(lO2x::')/:3 :
x'+y'
[ (lOy  2x 2 y _
Vs
y3)1v'(lO2X~)/3 l
dx
y(lO 2x 2 )/3
_4_ j Vs
3V3
(10  2x2  3y2) dy dx
Vs
[10v'2~ _ 2v'2x 2 /5 _ X2] dx.
=
From the integral table, we have J ~ d:c (x/2)J5  x 2 + ~ sinl (x/V5) 1 5 J x 2 J5=X2' dx = (x/8)(2 :c 2  5)J5  x 2 + 28 sin (x/V5) + C. Therefore, we get
+C
and
[10..;2 ( ~/5 _ x 2 + ~ sin l (~)) v = _4 3V3 2 2 V5 £  2v2
V5 (x_( 2x2  5)Vr;;.; sm ( x) )] 5  x + 25. 1
2
8

v5
8
1
V5
251l'V2 501l'. = =
V3
y6
12. We will co mpute the volume of the region in the first 0 tant where x, y and z are positive. Due to t he symmetry of the region , this result multiplied by 8 gives the correct answer. In the first octant, we let x go fro m 0 to a. Since the region lies within the cylinder x 2 + y2 ::; a 2 , one boundary is y = v' a2  x 2 • Since we are only looking at the first octant, we have 0 ::; y ::; v'a 2  x 2 . Finally, the z values of the region go from the xy plane to the cylinder x 2 + z2 = a. 2. Thus, 0 ::; z ::; v'a2  x 2 . The volume of the region in the first octant is the triple integral
l al~l~ a
dz dydx
=
a
0
Therefore, the volume of tb entire region is 8(2a3 /3) 17. The plane x
+ y = 1 can be written
as x
=1
y. T herefore, W can be described by
0 ::; y::; 1, O::; :c::; ly Therefore, the desired integral is
r r1 l1Y x 10 10 0
= 16a 3 / 3.
and
0 ::; z ::; 1l'.
1 7rl 1 (cos z ) ( x331 1
Y
2
cosz dxdydz
o
0
)
dydz
x=o
Alternatively, one can integrate in z first:
11 l 1
1 Y 
1f
x 2 cos ;;dz dxdy =
l
1 Y
1
1

( x sin z [=J 2
d:cdy = 111 1Y Od:cdY=0.
CH APTER 5
90 21. Evaluate as an iterated integral:
112X1X+Y
1o
0
x'+y 2
dz dydx
t J Jo
=
{ 2X
X
[ I +Y
o
z z=x 2 +y'
11 [((x  x
2
4x ( 3
1 { 2X
1J ~ ) [:al
J dy dx =
)y + y; 
0
o
(x+yx2_y2)dydx
r dx = Jo
1 (
4x
2
+3) 14
dx
3_ 7X4) 11 = ~ 6
6·
o
22. (a) Starting with the interval 0 ::; x ::; 1, sketch the area D
which extends from y = 0 to Y = Xo for each x o in [0, 1]. Then over each point (xo, Yo) in D, extend t he region W from z = 0
to z Yo.
z
&
=
y
(1,1,0)
x
(b) We will now fi nd an equivalent integral of the form
f f fw! (x, y, z) dx dy dz . First, note that z goes from 0 to 1. Now , for each Zo, Y goes from Zo to 1. This gives us a triangular area in the yz plane. Finally, for each (Yo , zo) in the triangular area, x goes from Yo to 1. Thus , we describe Was follows:
o::; z ::; 1 Thus,
and
z::; y ::;
111 1
X
1 and
111 1
Y
!(x, y,z )dz dy dx =
y::; x ::; 1.
1
1
! (X,y , Z)dxd Y dz.
As a quick check, we can see t hat both regions have the same volume by letting !(x , y , z) In this case, the volume is
i.
26. We note that x 2 + y2 + z2 ::; 1 describes the unit ball
centered at the origin. If z ~ 0, then the region W is
the upper hemisphere as depicted in the sketch . We are given t hat Ixl ::; 1 and Iyl ::; 1, but this does not
provide any extra information. To find t he limits of integration, note that x varies between  1 and 1, so 1 ::; x ::; 1. For each x, t he variable y extends fro m Vf=X2 to +Vf=X2. Finally, for each (x, y) in the unit disk , the variable z goes from 0 to + x 2  y2. Thus, W can be described by
VI 
1 ::; x ::; 1,
and
z
,, \
"
)..     ~~
y
 ~ ::; y ::; ~ x
o :S z ::; VI  x 2 
y2.
T herefore ,
jii
W
! dV
1
=
1 1~
 1 ~
lVIX,y2 0
= 1.
!(x , y, z) dz dy dx.
91
DOUBLE AND TRIPLE INTEGRALS 27. The volume, as a triple integral, is
11 [I D
f( X' Y )
dz
]
dx dy
0
and integrating with respect to z yields
JL
!(x , y) dx dy,
which is the double integral of f over D.
SOLUTIONS TO SELECTED REVIEW EXERCISES FOR CHAPTER 5 2. Evaluate this as an iterated integral:
6. The region is sketched here. As an x'simple region, it can be described as 0 ::; y ::; 1 and 0 ::; x ::; y'l , 0 we get
[( X + y)3IY2 ] dy
1 1
Y
a
3
(1,1)
x=o
x
9. Evaluate as an iterated integral:
l 1 [(yz + x;2 ) [=Jdydx 1
X
111 (y2 x~2) 11[(1+ ~) y; [=J 11 ( ~3 ~4 ) = (~; X
+
=
dydx
dx
+
dx
+
;~) I: = 6~ '
CH APTER 5
92 13. From onevariable calculus, we expect t.o get
l
y
b (f( X)  g(x)] dz
if 1 ~ 9 for z E [a, b]. The region D can be described as a ~ z ~ band g(x ) ~ y ~ I (x ) because it is a ysimple region. Therefore, the area of Dis a
ll b
a
f
. (X) dy dx =
g(", )
lb [yl
f(X)]
a
dx =
!I =g( 3; )
b
x
Ib[/(x) g( x) ]dz . a
15. (b) As a ysimple region, the unit circle can be described by
 l~x ~ l and
 ~~ y~ ~.
Thus, we have
11v'1=X'2 x 11 ~
2 y2
dy dx =
111[~ 3 1 ~ 1dx == 211x (1 3 y=_~ 3 1
2
2
Use the trigonometric substi tution : ~
J
x 2 (1  x 2 ) 3 / 2 dx =
X 2 )3 / 2
dx.
= cose , x = sin e and dz = cos edo, 80
J
sin 2 Bcos 3 B(cos e dO) =
J
(sin2 e  2sin 4 B+ sin 6 e) de,
=
which was obtained by using the identity sin 2 B + cos 2 B 1 and expanding. Now, use the identities: sin 2 e = (1 cos2e)/2 and cos 2 e = (1 + cos 2e)/2. We get
J[
1  cos 2fJ _ 1  2 cos 2e _ cos 2 2B 2 2 2
J( ~ ~ 16
~x ~1  x2
+
1  3 cos 2B + 3 cos 2 2fJ  cos3 2e] dB 8 2
cos4B + cos 20 sin 20) dO= .i. 16 8 16
_ sin 4B _ 64
3
sin 2e + C. 48
Now , we use t he double angle identities and the following which we get from the triangle
e == sin 
1 :1: ,
sin e == z
and
cos e = ~.
Substitut ing and evaluating at ± 1, we get
I
sin  z _ [ 16
4x ~(164
Z2
x2)
_
8z 3 (1 Z2) ~lll 48
Therefore, the original integral becomes (2/ 3)(11"/16) == 7r/ 24.
I
_~ . 16
93
DOUBLE AND TRIPLE INTEGRALS 20. As the integral is written, the region is described as 0 ~ x ~ 2 and
 3v'4  x 2 / 2 ~ y ~ 3v'4  :z:2/ 2, which is a ysimple region . When we
rearrange y 3v'4  x 2 /2, we get y2 / 9+ x 2/ 4 1, which we recognize
as an ellipse, so our sketch is as shown . We evaluate as follows:
=
r j3"f4X 2
=
2
/2
5
(
3~/ 2 v'2+X
io
+ y3)
dy dx
l' [(vA.
~') C:~:~<J d.
+
12 15~ dx =
x)3/21: = 20V2.
 10(2 
24. As written, the region de cribed by 0 ~ y ~ 1 and 0 ~ :z: ~
3y is an xsimple region . The region is sketched as shown.
We evaluate t he double integral as follows:
t [3 io io
y
e +Ydx dy = X
11 ( X 1::0) Y
e +
dy
4l1
t (e 4Y  eY)dy = ( e _e ll Jo 4
)1
1
= e4 _e+ ~. 4
0
4
27. The region is sketched h reo Note that is consists of the following two regions:
o ~ :z: ~ 1 1 ~ :z: ~ 2
0 ~ y ~ _ x2 + X
and
 x2 + X ~ Y ~
and
o.
We evaluate as a sum of integrals:
tlX2+X
y
io x
[2j O !(x , y) dy dx + il _X 2+x I(x , y) dy d:z: .
0
The first integral is:
11 + + 1
X2 X + (:z:2
2xy2
2) dydx
l' [(x'Y+ 2xt
+ 2Y)
C+<] dx
T he second integral is :
[ 2j O (x 2 + 2xy2 i1  x'+x
+ 2) dy dx = [2 [ ( x 2y + 2x
(1)
12 (~:r/ _x
8
il
\
3
+ 2x6  2:z:5 
2x7
(1) ( 12 + 7
Add the two integrals together to get
x
6
3 
Ii .
y3
+ 2V)
0 ] dx
1
y=x'+:r
~4 + X3 
5 X4 x 15 + 4
3 2x
2X 2 + 2X ) dx
 3 +x
2)110 =( 1) (  584 105 
27 ) 70 .
.....
94
CHAPTER 5
31. T he function f (x, y) = x 2 + y2+ 1 has the value r2 + 1 where r is the distance from the origin. Thus, on D, 1 ::; f( x , y) ::; 5. The area of Dis 411", so by the mean value inequality, we have the desiIed result.
35. Dividing thIough by x2, the integrand becomes (l/x)/[1 + (z / x )2]. Let u
= z/ x, so 3; du = dz
and the integral becomes
[1 [Y t / ../3 (
10 Jo 10
r ( tan
t
1 ) 1 + u2 du dx dy
Jo Jo 11"
11 (
6
xI
Y
1
)
11/../3)
u u=o
= 161"
dy
x=O
0
dx dy
11
[11 = "611" J[1o 10 dx dy
y dy
0
= 161" ( y22110 ) = 1211" .
TEST FOR CHAPTER 5 1. True or false. IT false, explain why.
(a) The area between the parabola y = 4  :r 2 and the x axis can be expressed as the double integral
[ j Jo 4 _ X2
2
dx dy.
2
(b) If g(x , y, z) is integrable over a region Q in 1R3, then
JJk
g(x,y,z )dx dydz
=
JJ
kg (x,y,Z ) dy dZdX.
(c)
tj4 jS_ y3 (z + Jo y2
y)3/2 dz dy dx
 2
15
= (j4  2
y3
(z
y2
+ y)3 /2 dz dY)
.4.
lID
(d ) For any integrable function f (x, V), the double integral f (x, y) dx dy is the volume of the region bounded by t he sUIfaces z = f( x, y) and z = 0 over D.
(e)
1
2jl (X2 + 4x o  1
+ .JX)(y3 _ y2 + 4y) dydx
[1 1 2
1
1 ( 3;2
[1 1 2
+ 4x + .JX) dy dX ]
1
1 (y3
2. Change the order of integration and rewrite the triple integral Compute its value .
 y2 + 4y) dy dX] .
101:2Io ydz dy dx in five ways. 1
1

= x 2 + 1, z = 0, y = 0 and y = x 3 for 1 ::; x ::; 2. 1 4. Evaluate the double integral I/  COS(I ) Ic1os ( I _ y 2) y dxdy. 3. Find the volume of the region bounded by z
I
5. An am ateUI magician's hat is described by z::; 9  x 2

y2
and z ~ O.
(a) Use Cavalieri's principle to calculate how much milk he can pour into his hat ifhe wants to bake a magic cake. (b) Express the volume of the hat as a double integral in t he form (c) Use parts (a) and (b) to quickly calculate I~3(9  t 2)3/2 dt.
II f(x, y) dx dy.
95
DOUBLE AND T RIPLE INTEGRALS
6. Evaluate the double integral JJT (0,0) and (0, 2) .
Ix  yl dy dx,
where T is the triangle wit h vertices at (2 ,0),
7. (a) Suppose f : W t ~ is continuous and W is an elementary region in IR . Le t V( W) be t he volume of W, m the minimum value of fon Wand M the maximu m value of f on W. Make a statement relating m, M, V and the triple integral of f over W. (b) Show that
v'22 :S 16 e
l 1l"/4 1
0
Y
a
0
cos(xy) 7r +z dx dz dy :S 8 ' eY
8. Eval uate J04JolJ3z exp (x3 ) dx dydz. 9. Eval uate JJD x dx dy where D is the region shown.
T he bOWldary on the left is y = 4  x 2 . All of the
other boundaries are straight lines.
x
(4. 3)
10. A cakeeating monster saw a pyramidshaped cake which he easily devoured. T h cake had a square crosssection and at a height h, the side of the square had length 8  h. T he cake monster's son could only eat the top ~ of a. similarly shaped cake. (a) Use Cavalieri's principle to calculate the cake monster's minimum stomach capacity. (b) How t all is t he rem ainder of the cake that the son started eating?
.
.
97
6
THE CHANGE OF VARIABLES FORMULA AND APPLICATIONS OF INTEGRATION
6.1: THE G EOMETRY OF MAPS FROM]R2 TO JR2 GOALS 1. Be a ble to determine whether a m ap is onetoone. 2. Be able to graph a region D which is the image of a region D* under a mapping .
STUDY HINTS 1. No tation. The equality D = T (D* ) expresses that fact that T t akes a point in D* and maps it to a point in D, and t hat all such points of D are obtained in this way. T is called a mapping or a transformation. T hinking of T as a fu nction, this is similar to y = f(x) for f : JR t JR.
2. Onetoone. If two distinct points always get mapped to distinct points, i.e., distinct points never get m apped to a single point, then the function is said to be onetoone . A domain may need to be chosen carefully for this property to hold. (See example 3.) For integration, this is a nice property ; otherwise, the double integral over D may not equal the double integral over DO.
3. Onto. If every point of a range D can be obtained fro m some point in the domain D" , then the m apping is onto. Onto does not imply onetoone because t wo points of D* may get mapped to one point of D. By t he sa me token, onetoone does not imply onto.
4. Finding D f rom D* . In many cases, the range D can be determ ined by simply mapping the boundary of D* . Then decide whet her the map takes D* to points inside or outside the bound ary of D. For exam ple , let D be the unit disk. If
then T m aps the uni circle to the unit circle, and T maps a point such as (~, ~) to (2,2), a point outside t he circle. So T maps t he inside of the circle to the outside [and is not defined at (O, O)J .
5. Important exercise. The result of exercise 8 is used in the examples of section 6.2. It states that if T (x) = Ax, then T is onetoone if and only if det A =f:. O. The result is true for all n x n determinants. 6. L inear transformations. If T( x ) = Ax + b and A is a 2 x 2 m at rix, t hen T takes parallelograms to parallelograms. If D is a parallelogram, one can often obtain it from the unit square D* with such a T.

CHAPTER 6
98 SOLUTIONS T O SELECTED EXERCISES 2. The transformation is given by the following equation :
T( x·, y* ) = Ax =
[1//2 1//2] [ X: 1//2 1/V2 y ].
We compute det A = 1 i 0, so by theorem 1, the linear m apping T maps vertices to vertices. The origin of D* gets mapped to T(O, 0) = (0,0) in D. Similarly, the points (1,0), (1, 1) and (0,1) of D· get mapped to (1/ V2, 1/V2), (0, /2) and (1/V2, 1/)2) , respectively, in D. It is easy to show that the length of each side of the transfor mation is 1 and the dot product can be used to show that the angles are 7r /2. Thus , T rotates the unit square by 45 degrees. y
••
y!
T
1
x
x*
4. The four vertices (0,0), (1,0), (1,1) and (0,1) of D should be mapped from the four vertices (0, 0) , (~, ~) , (152 , \6) and (t, 152 ) of D·. We want to find a matrix A so that
T( x*, yO)
= Ax = [~ ~] [ :: ] .
=
=
When C~, 1) gets mapped to (1,0) , we get 8a/5 + 4b/5 1 and 8e/5 + 4d/ 5 0, so d = 2e. When Y52 ) gets mapped to (0,1) , we get 4a/5+ 12b/5 = and 4c/5+ 12d/5 = 4c = 1, or a = 3b and c = d = t. Substituting a =  3b back into 8a/5+4b/5 = 1 gives us  4b = 1 or b and so a ~. T herefore,
(i,
i,
= t,
°
=
1. , 4'x 1 *+ '12 y*) . = (34'x * 4'y
T (x • , ~I *)
7. Since sines and cosines appear, we will try to use the identity sin 2 t + cos 2 t = 1 to eliminate the parameters and obtain a recognizable fo rm: x 2 + y 2 = p 2 sin 2 ¢ (cos 2 B+ sin 2 B) = p2 sin:! ¢ and x 2 + y2 + z2 = p2(si n 2 ¢ + cos 2 ¢) = p2 . We recogn ize that x 2 + y2 + z 2 = p2 is a sphere of radius p; since :s p :s 1, D is the unit ball. The map T is not onetoone. For example, (0, 7r, 7r/2) and (0 , 7r /6, 7r / 5) both m ap to the origin. Also, (I,O,7r/2) and (1, 27r , 7r/2) both map to (1, 0, 0). Since p always gets mapped to the origin , we want to elimin a.te that point . Also, si nce B = and () = 27r give the same mapping, we want to eli minate either one of those points. Hence , T can be made onetoone by using the following intervals: ¢ E [0 , 7rJ, B E (O ,27r] and
°
°
=°
pE(O ,l].
10. Using the hint and applying T to the parallelogram described by q T( q)
= p + .xv + /,W,
we have
= Aq = Ap + >.Av + /lAw .
Since A is a 2 x 2 matrix and p, v, w are all vectors in JR2, then Ap, Av and Aw are all vectors in JR 2. Since A is nonsingular (det A i 0), Av cannot be a scalar multiple of Aw if v is not a scalar multipleofw. T his is because if Av aAw , a scalar, then Av aAw = A(v  aw ) 0 , which implies t hat either det A = of v = aw . Thus, the image of T is a parallelogram in JR 2.
°
=
=
TH E CHANGE OF VARIABLE S FORMULA AND APPLICATIONS OF INTEGRATION
99
6.2: THE CHANGE OF VARIABLES THE OREM GOALS 1. Given a transformation T, be able to compute its Jacobian.
2. Be able to use the J acobian to change variables in do uble and triple integrals. 3. Be able to state and use the change of varia bles formula for polar, cyl indrical and spherical coordinates .
STUDY HINTS 1. Review. You should review the definitions of polar, cylindrical and spherical coordinates from section 1.4. 2. J acobian. The Jacobian is a determinant, not a matrix. In this section, we introduce the
notation
a(x , y) a(u, v)
and
o(x,y,z )
o(u, v, w) ,
which are the determinants
l
ax/au ax/av ay/au oy/ov
and
ax/a u ox/ov ax/ow ay/au oy/av By/ow az/au 8z/Bv Bz/8w
respect ively. 3 . Changing variables . As stated , the change of variables theorem requires th at T be onetoone.
However, the theorem still holds if T is not onetoone on the bou nd ary of D· , a sit uation th at occurs in a number of examples . A similar theorem holds for triple integrals.
+ y 2 occurs in the integrand, try using polar or cylindrical coordinates , which have the Jacobian r. If x 2 + y2 + z2 occurs in the integrand, try using spherical coordinates, which has t he Jacobian p2 sin 1jJ . These two Jacobians should be memo rized .
4. Useful change of vari ables. If x 2
5. Integrating over ellipsoids and ellipses. If you are integrating over an ellipsoid with equation
(x /a) 2 + (y/b)2 + (z/e) 2 = 1, the subsitutions u = x/ a, v = y/ b and w = z/e may be made to transform the ellipsoid into a sphere. Spherical coordinates may now be used to integrate over u 2 + v 2 + w 2 = 1. Similar methods can transform an ellipse into a circle a nd then polar coordinates may be used. 6. Limits of integration. W hen you set up the limits of integration, remember that ¢; is mea sured from the "north pole," not fro m the "equator ." Also, recall that () is measured in a
coun erclockwise direction.
SOLUTIONS TO SELECTED EXERCISES 2. We know that x = u + v and y = u  v. Adding the two equations gi ves us (x + y)/2 = u and sub traction gives (x  y)/2 v . By mapping the boundaries of the triangle in the x1Tpl ane , we ge t another triangle in the uvplane as shown .
=
y
v
(~.~)
u
CHAPTER 6
100 The Jacobian is the absolute value of
a(X,y )
la(u, v)
I_I ax /a u  ay/au
ax /av ay/av
so the Jacobian is 2. In this case, we integrate x
J o
+ y = 2'11
t/2 [I  V
[
JD (x +y)dxd y =
1_11  1
i"
Jo 4
1  1
as follows:
{ 1/2
(2u)(2)dudv= 4
1  2,
Jo
{ IV
J"
u du dv
(u11V) dv =2 11/2 (1  2v) dv= 2
1/2
1
2
o
2(vv 2)1
0
ti = v
1/2
1
= 2'
0
Don 't forget t hat the Jacobian is the absolute value of the determinant. A direct calculation gives us
1111 o
(x+y) dx dy
y
=
11  + 0
[( x2 2
3. (b) We compute th at x  y
xy) 11
0
x=y
= 2'11 
1dy =11(12 + y  32y2 ) dy = ( 2y + y22  y3) 11 = 1 2 2 0
0
3v and the Jacobian is
I
a(x, y) =
Ia(u, v)
14 a I= 12. 2 3
Therefore , the integral becomes
12111\2U  3v) dvdu
=
1211 [ ( 2UV 12
3~2 )
[=J
du = 1211 ( 2U
~)
du
(u 9;)1: = 12 (1 ~) =  42 . 2
_
Notice that we did not even need to know what D looks like to perform the integration. By theorem 1 of section 6.1 , this linear transfor m ation maps the rectangle D· to a parallelogram D. We map the vertices and then connect them to get D. v
y
(4,8)
(4,5)
2
u
x
7. We compute t hat x 2+ y2 = (u 2 _v 2 )2+4u 2 v 2 = u4 + 2u 2 v2 + v4 = (u 2+ v2)2, and so ';x2 + y2 = u 2+ v2. In this case, the Jacobian
IS
2u  2v 2u
1= 1 a(u, v) 2v la(X'Y)
I= 4u
2
v
+4v2 .
T herefore,
J1
D.
4(u2 2 u
+ v22 ) dudv = 4 +v
Jj
D.
dudv .
This is just 4 times the area of D* , which is a quarter of a unit circle, so the answer is 7r .
1 u
TH E CHANGE OF VARIABLES FORM ULA AND APPLICATIONS OF INTEGRATION
101
11. The transformation takes (x,y) to (rcosO,rsin O) , so
the Jacobian is
y
I I c~s f)O
8 (x,y) = 8(u, v} 1
Sill
Recall that t he area is t he double integral In polar coordinates, this becomes
J10.
2
rsinO 1= r . rcos f)
fID dx dy.
2lrl1+sinB rdrdO
1
r drdO =
o
0
f21r
(r2 11+Sin B)
dO
Jo 2 r = O ~ 1 271: (1 + sin 0) 2 dO.
x
Use the half angle formula to get
11271:
20
( 1 + 2· sm
0+ I COS O) dO 1 (30  2 cos 2
22
21r
II rJ 
sin20 )1 40
_


311" 2
.
14. When D = [0, 1] X [0 , 211"] gets mapped. to a unit circle, we know we are using polar coordinates. Substitute u = 1 + r2 to get
17. Here, we want to take the uni t square and find a transformation which maps it 0 the giv n region R, which is a parallelogram. Note that we can map (0 , 0) 0 (1,0) , (1,0) to (~ , ~), (0 ,1) to (0 ,1) and (1,1) to (~, ~ ) . The transforma ion should be linear, so we should get (x, y) = (au + bv + c, du + ev + f) . When (u, v ) = (0 ,0) gets mapped to (x, y) = (1 , 0) , we see that c = 1. T he mapping of (1, 0) to ( ~,~ ) gives us a = ~,and he mapping of (0 , 1) to (0 , 1) gives us b = 1. Similarly, we can determine d, e and l The transformation is (3u/2  v + 1,3u/2 + v), so the Jacobian is 8 (x , Y) 1 8(u,v)
I= II 2
11
I
= 3.
y
v
(YzX)
1
Also, we compute x
~,~
u
+ y = 3u + 1 and
1
x  y
=1
x
2v, so the change of variables gives us
CHAPTER 6
102
Since the integrand can be factored into separate factors which contain only one variable and the limits of integration are all constants, we get
(1 (3u + 1)2 dU) (1 e 1
3
1
1
1)31:) ( e~;V I:) 3(64; 1) (e1_~ e = 221 (e _~) .
dV)
2v

3 ( (3U;
1
)
20. The Gaussian integral tells us that
1:
exp(x
We substit ute u
1:
2
)
dx
=,fo.
= 2x and du = 2dx to get
exp( _ 4x
2
)
lim
dx
J
a+oo
aexp(4x
2
)
dx
== lim
a
21 a+oo lim
a+oo
Jaexp( _u _
a
2
)
du
J2a exp(u 2a
2
)(du/2)
= ,fo 2 .
Since the function exp( 4x 2 ) is symmetric about t be y axis, we deduce that
1
00
exp (4x2) dx
=~
1:
exp(4x
2
)
dx
= V;.
23. We will use spherical coordinates. The region S can be described by a
~
p ~ b, 0
~
B ~ 211"
and
0
~
¢
~
11".
The Jacobian for spherical coordinates is p2 sin ¢, so the integral becomes
l 11 b
lr
a
0
2lr
0
p2 sin...:.... ¢ !......3
dB d¢ dp.
P
Since all of the limits of integration are constant, we can "factor" the integral as follows:
(l
b
~) (1'" sin ¢ d¢)
(1
2 ". dB) = log
(~) . 2 . 211" = 411" log ( ~) .
26. T he Jacobian for cylindrical coordinates is r. (a) After sketching B, we see that B lies over the unit disk (see figure at left, below) and z extends from 0 to y'x 2 + y2 = r. T herefore ,
1"'11 12"'1 2
f f l zdV
=
1
r
rz dz drdB=
1
o
0
r3 d7' dB 2
21 1"'1 (r;2[=J
drdB
2 12".(r411) dB = 11 ". dB = 11" . 8 8 4
=.
0
r=O
0
103
T HE CHAN GE OF VA RIABL ES FORMUL A AN D APPLIC AT IONS OF INTEGRATION
z
y
y
x
x
(b) The bound ary x 2 + y2 + Z2 ::; 1 is the same as r2 + z2 ::; 1 (see the sketch of W above). Solving for 7', we get r ::; ~. Also, the integrand is l/vx2 + y2 + z2 = 1/v'r 2 + z 2. T herefore, the chan ge of variables gives us
r t 2
7<
10
Let u
= 1'2 + z2
r~ I' drdzd(). 11/210 vr2 + z2
in the inner r integral to get
2"1111dudzd() 1 o 1/2 z ~ 2VU
1
=
27<11 ( v'ul 1 )
1 o
u= z~
1/2
{ 27r [ (
10
z 
z2 ) ll
2
z =I/2
dzd()
= 1 27< a
fl (1 z) dzd() J 1/2
1d() = "81 10{27< d()
1T
='
4'
27. We want to fi nd a mapping of the unit square B* to the rectangle B. We want to map (0,0) to (1 , 0), (1, 0) to (4, 3), (1, 1) to (3, 4) and (0 , 1) to (0,1). We want a linear transformation which maps a parallelogram to a parallelogram . O UI desired m apping is (x,y) = (au + bv + c,du + ev + f) . Mapping (u,v) = (0,0 ) to (x,y) = (1,0) gives us c = 1. Mapping (1,0) to (4, 3) gives us a = 3. And mapping (0,1) to (0,1) gives us b = 1. Similarly, we get d, e and f We get (x, y) = (3u  v + 1, 3u + v). T he J acobian is
IO(X'Y)I=1 3 8(u ,v)
3
 1 1
1=6. Y
(3,4)
v
(4,3) 1 1
u
1
oX
Therefore, the integral gets changed into
..
104
CHAPTER 6
30. In the first octant of JR3, the spherical coordinate B goes from 0 to rr/2. T he desired region of integration is sketched below. Recall that the J acobian for spherical coordinates is p2 sin ¢, so the desired integral is 7f/ 21tan1 (2)
1o
1.../6 (1)  /sin¢dpd¢dB
a
P
7f / 2 jtan I (2)
7f/4
1a
1.../6 p sin ¢ dp d¢ dB a
7f/4
7f/21tanI(2) ( 2
~ sin ¢
1a 3
7f/4
p=O
7f/21tanI(2)
1 a
7f/4
From the triangle below showing ¢ 3 (
1
1.. /6 ) d¢ dB
sin ¢ d¢ dB
=31
7f/2 (
a
 cos ¢
ItanI(2))
dB.
1>=rr/4
= tan 1 (2), we get the follo wing:
1 )
r
vis + Vi.1 o
/2
dO
3rr ( 1
=2
1 )
Vi  v'5 .
z
2
y
1
6.3: APPLICATIONS GOALS 1. Be able to use double and triple integrals to compute averages, centers of mass . moments of inerti a and gravitation al potentials.
STUDY HINTS 1. Averages. In general, an average is defined to be the integral of fd ivided by the integral of 1. Hence, the two average formulas given in this section are
IffII' fdV fffll'dV
and
ffD f dA ffD dA '
2. Cente1' of mass. This is a weighted average. In general, the center of mass of a region in JR n is (Xl,X2,X3, ... ,X n ), where Xj is fR x j c5 dV
fR c5 dV and dV = dXl dX2 ... dX n and R is the region. Since c5 is the density, fR c5 dV is the mass. If the region is in JR2 use double integrals, and if the region is in IRs use t riple integrals.
THE CHANGE OF VARIABLES FORMULA AND APPLICATIONS OF INTEGRATION
105
3 . Moment of intertta. A good way to remember Ix = IIIw (y2 + z2) 8 dV is to notice that the integrand lacks an x term. Similarly, Iy and l z lack a y and a z term , respectively.
4. Gmvitational potential. Its formula is V =  OM m/ R. Although you do not ne d to under stand the physic 1 t heory behind the discussion the mathematics shou ld be clear to you . You probably will not need to reproduce the discussion for an exam . 5. Geometry. Recall that if the integrand is 1, then the triple integral IIIw dV gives the volume of W. Sim il arly, the double integral lID dA gives the area of D.
SOLUTIONS TO SELECTED EXERCISES 3. The fo rmulas used to compute t he center of mass (x, y)
for a region in the plane are
_ IIDxtS(;z;,y)dxdy = lIDtS(x, y) dx dy
x
_ lID ycS(x, y) dx dy y  lID 8(x , y) dx dy .
and
In this case, we sketch the region D and see that it can
be described by
o ~ x ~ 1
x 2 ~ y ~ x.
and
x
T he numerator of x is
[11" x(x +y) dydx
Jo
rl
T he deno minator is
11[(Xy + y; )[=".] 11(3;2 _ ~4 )
( l "(X + y) dydx
Jo
r'
x3 _
Therefore,
dx
dx =
X3 _ (
X4 _
2
X
5
)1
10
4
1
0
18 120
65 x = i! . We leave it to you to calculate y = 126'
6. Th mass of the plate is lID 8 dA, which is
[21<
r
2
Jo Jo (y2 sin 4x + 2) dydx
)1" {27< ( 3 ) = Jo[27< (y3 '3 sin 2 4x + 2y y =O = Jo ~ sin 2 4x + 21r
dx.
Recalling t hat in 2 t = (1  cos 2t )/2, we have
1r3
{ 2"
'3 Jo
1  cos 4x 2 dx
{21r
+ 21r Jo
The area of the plate is 21r x 2r.
= 4,,2
(x sin 4X ) dx = '3 '2   8  a 1r3
2
1 "
+ 41r
2
1r4
'3 + 41r
2
.
Thus the average density is mass/ area, or r. 2 /12 + 1.
7. (a) Let 8 be t he density of the box, then the mass i· IIfw If dV , which is
t /2 Jot Jo[2 dz dy dx = If (1 ) '2 (1)(2) = 8.
tS Jo
=
~.
 CHAPTER 6
106
(b) Again, the mass is [1/ 2
t
[2
10 1010
JJJw6 dV , which is
2
(x + 3y2 + Z +1 )dzdydx
1 10 [ (( x 2 + 3y2 + l)z + z:) [=J dydx t l2[1 t/2 [ 1 ] 10 10 2(x 2 + 3y2 + 2) dydx = 2 10 ( (x 2 + 2)y + y3 ) Iy=o dx 1
1/2
2
1
1/2
o
(X3 + 3x ) 11/2 ~ 2 ( 1 + 3)
(x 2 + 3) dx = 2 
3
24
0
JJJwf(x, y, z ) dV / JJJw dV.
11. The average value of f over W is
111 2
4
2
37 =  . 12
The numerator is
6
2
2
sin 7rZ cos 7rX dz dydx .
Since f(x , y, z ) can be factored so that each factor contains only one variable and the limits of integrat ion are constants , we get
(1
(1 dy) (1
2
4
cos
2 7rX
dX)
6 2
sin 7rZ dZ ) .
Using the halfangle fo rmulas, we get
[1
2
C c~s +
[1 d [1 4
27rX) dX]
y]
6
C
c;s
2rrz) dZ] 6
=
21 [( x
27rX) + sin ~
102] [ Yo14] 21 [( z  ~ sin 27rZ) 10 ]
1 = 21 (2)(4)2(6) = 12.
The denominator is just the vol ume of W, which is 2(4)(6) = 48 , so the average is !~ =
i.
14. T he moment of inertia of Waround the y axis is given by Iy =
f f fw 6( x2 + Z2) dx dydz.
Since W is the ball of radius R, we will use spherical coordina.tes. Since x 2 + y2 + z2 = can rewrite the integrand as 6(x 2 + z2 ) = 6(p2  y2). T hus , the moment of inertia is
6
[21f
r
p 2,
we
[R
2 (p2  p2 sin ¢ sin 2 B)p 2 sin¢ dp d¢ dB 10 10 10 [ 21f r [R 3 2 6 10 10 10 (p4 sin ¢  p4 sin ¢ sin B) dp d¢ dB
(j 1
21f
1" [(P55 1~=J
3
2
(sin ¢  sin ¢ sin B) ] d¢ dB
5
To finish the in tegral , use the identities si n 2 t (1  cos 2t) /2 . The above integral becomes
2
2
+ cos 2 t = 1 and the halfangle formula sin 2 t
r [sin¢  sin¢(1  cos ¢)sin
510 10
3
sin ¢ sin B) d¢ dB.
2
B] d¢ dB
=
T HE CH AN GE OF VAR IABL ES FORMULA AND APPLICATIONS OF INTEGRATION 5
r
(
COS
3
107
I" 1
6R Jo21r [(  cos if> + sin 2 B cos if>   3 fjJ) ) ¢= o dO 56R51 21r (
5
6R5 5
0
4.
6R51 21f
2 )
2  3'SlD B d8 =5
[2B_ ~ e + sin 2B ] 121r 3
3
0
[2  3'2(1  cos2B) ] dB
86R5 rr 15
0
19. From t he discussion in the text, we know tha t the gravitational potential acti ng on a mass m lying ou tside ofC .M.W . is  GmM/ R , where G = 6.67 X 10 11 N·m 2 / kg 2 , m is t he mass of the smaller object in kg , M is t he mass of the planet C.M.W. and R is t he distance between the object and the planet's center. When t he object lies outside of a spherical planet , the gravitational potential acts as if the mass M is concentrated a t the planet's enter. Given the planet's density as J, we use spherical coordinates to calculate M
=
'rJw J(x,y,z)dxdydz = Jr21r Jor Jr
JJ
o
Since 6 is independent of ¢ and
X 108
o
6(p ,fjJ, e)/s in <jJ dpd¢ de.
e, we have
Note that 10 12 and 10 8 are insignificant when added to a number with the magnitude of 10 17 . Therefore , the potential is
v=
(6.67 x 10 11 )(4.71 x 1019) m/ R = (3.04 x 109 )m/ R (meter/ second)2
.4: IMP R O P ER INTEGRALS
G OALS 1. Be able to defi ne and compute an im proper in tegral.
STU DY H INTS l. CompaTis on. Onevariable integrals are called improper if the fu n tion becom s infinite at
so me point or if the limits of integr ation are in fini e. The difference here is that a function m ay become infinite on a curve or surface . 2. Computation. To find an improper integral, compute a limiting value rather than substit uting the limits directly. Fubini's theorem is valid for improper integrals of positive functions. 3. f becomes infinite. We remind you t hat f m ay become infinite betwe n t he li mits of integration. In this case, we need to divide the region at the points where fis infinite. For exampl ,
1o1 ]1Yx dy dx =11 ~Y dy dx + 1111 ~Y dy dx . 1
 1
0
0
 1
0
0
• .. .&
(HAPTER 6
108
See what happens if you evaluate the antiderivative at y = ±1 and do not consider y = 0; you would get the wrong answer.
SOLUTIONS TO SELECTED EX ERCIS ES
= O. We sketch the region and see that it is a ysimple region described by
3. Th is integral is improper at x
y
(1,1)
°
~ x ~ 1
x/2 ~ y ~ x.
and
(1,1/2)
Thus, the integral is
[11X~dydx
Jo
11 (~: [=x/ J
x/2 X
_113x _ 3x211  dx 
1(x2  X2/4) 
1o
2x
1 x
dx dx 
2x
a 8
16
3 16
0
6. (a) Assuming D is a ysimple region, it would be reasonable to think of b =
Then we can define
00
as an "endpoint."
f fD fdA by °0 1 ¢>2(X)
l
f(x, y) dy dx.
¢>,(X)
a
(b) The region D is like the one described in part (a) . Since the limits of integration are con stant, we can integrate each variable separately and then multiply the results. The integrand is factora ble as fo llows:
xy ex p(x 2
y2)
_
= [xexp(x2)][yexp(y2)],
so
JI f
[1
1
(x,Y)dA
[/00 xex p (x 2 )dX]
yex p (y2)d Y]
= (. ~)
b
11)
2 p lim (ex (x ) I ) bt oo 2 1
exp(y2) ( 2 a
4
(! _~) . e2
e
7. Fubini's theorem tells us t hat multiple integration can be performed in any order as long as
the integrand is positive. In this case,
e xy
> 0, so we have
00
/2 1
eXY
dx dy
=
100 /2 e
xy
dydx .
For the lefthand side, we compute
rJo[00
exy
dx dy
J1
1
2 (
1
2 ( / ,1
xy Ib ) dy lim ~ btoo y x=O
1) dy= 12 dy
by
lim btoo
 = logy 12
_e +
y
y
1
Y
~log2.
1
On the righthand side, we compute
1 1 00
a
2
1
= 100 _e
XY
e XY dy dx
o
:Z:
12 y=1
dx
100
=
e
X
e
2x
dx .
0
Since Fubini's theorem tells us t he two integrals are equal, we have the desired result.
THE CHANGE OF VARI ABLES FORMU LA AND APP LI CATION S OF INTEGRAT IOI\l
109
9. The integrand sim plifies to 1/(x+y). T he integral is improper whenever y == x, wh'ich occurs in our region at t he point (0,0). We compu te
1111+
_1_ dy dx
o
0
x
y
=
11[
(In Ix +
yl)1
0
which we interpret as
[In(1 + x) In(x)] dx,
dx
0
yO
11
lim
~ +o+
1_] =11
[In (1
~
+ x) In(x)] dx .
Integrating by parts, we get
E~+ { [((x + 1) In(x + 1) 
x)  (x Inx  x)]I: } ·
Using I'Hopital 's r ule, we know hat
· I1m
E+O+
In f = I'1m In/f = 0 ,
f
[ +0+
1
f
and so the integral is 2 1n 2. 11. Since x 2 + y2 + z2 appears in the integrand, try using spherical coordinates. In the first octant, we have o ~ () ~ 7r/2 and 0 ~ cp ~ 7r/2. Since x 2 + y2 + z 2 ~ a 2 , we also have 0 ~ p ~ a. Recall that the Jacobian for spherical coordinates is p2 sin e/; . Subs itute p2 = x 2 + y2 + z2 and p cos cp = z to get
l
1f/2 1 ff /2 1 Q
o
0
Substitute u
pl/2. r====p2 sm cp dp de/; d() p cos cp + p4
J
0
=l
1f/2 1 1f/2 1 Q 0
0
0
p2 sin cp
J cos cp + p3
dp dcp d().
= cos cp + p3:
3
1f /2 1 1f /2 j COS ¢+a d u SI·n '+' '+' dcp d() o 0 cos¢
l
3y'U
r/1 2
1f
2
/
Jo 0 1f/2 o Jo
~ sin e/; ( y'Ul cos ¢+a 3 2
3 )
dcpd()
u= c os¢
r/2"3 sin CPJcos cp + a3 dcp d()
1r/ r/ 2
 Jo
r/ Jo
2
Jo
2
3" sin CP Jcos cp dcp d() 2
2
[_i(coscp+a 3 )3/zl1f/ + i(cosCP)3/2'11f/2] d() 9 ¢=o 9 ¢=o
27r [(1 + a 3 )3/2 9 14. W hen 0
~
g(:I:, y)
~
_
a9/ 2
_
1]
.
f(x, y), integration over D should give us
.lin
0 dA
~
.IL
g(x , y) dA
~
.IL
f(x, y) dA.
However , since the integral of 9 is sm aller than the integral of f and exist , we m ust conclude that fiDf (x , y) dA also does not exist.
lID g(x, y) dA does not
CHAPTER 6
110
16. Since x 2 + y2 + z2 appears in the integral , we will try to use spherical coordinates. The region D can be described by
I:=;p
and
0 :=; O:=;27r
In spherical coordinates, the integrand is fore, we get
27f 1 7f
1
1/ p4
and
0 :=;¢:=; 7r.
and recall that the J acobian is p2 sin r/!. There
Joo p2 sin ¢ dp d¢ dO. ';4'
o 0 1 P Since all of the limits of integration are constant , we may integrate each variable separately and multiply the results:
{27f
f"Joo ~ sinriJ dp d
Jo Jo
1
( { 27f
Jo
=
dO
(f" . ) ( roo Jo sm ¢ d¢ Jl p2
)
dP)
27r (  cos ¢
I:) b~~ ~ I: ) = 27r(2)(I) = 47r. ( 
SOLUTIONS TO SE LECTE D REV IEW EXERCISES F O R CHAPTER 6 2. (a) This is a linear transformation of a paralellogram, so by t heorem 1 of section 6.1 , all we need to do is to map the vertices . We compute T(O, 0) = (0,0) , T ( 1,0) = (2, 1), T( 1, 1) = (2 , 4) and T(O, 1) = (0,3) . The transformation is depicted below:
y,
v,
(1,1)
(2.4)
(0.3) T •
(2,1)
r
~
u
x
(b) Using the notation shown in t he sketch in part (a), our formula is
J10 In this case, u(x , y)
f(u, v) du dv
= 2x
=
and v(x, y)
a(u , v) y)
1a(x,
1
J10.
f (u(x, V) , v(x, V) ) 1
= x + 3y,
= 1 au/ox av/ax
~~~: ~~
1
dx dy .
so
au/ay av/ay
1
= 1 21
°= 3
1
6 .
Thus, our formula is
J10
f(u, v) du dv
=6
J10.
f(2 x, x + 3y) dx dy.
4. (d) Recall that t he Jacobian for spherical coordinates
is p2 sin qS and sin 2qS = 2 sin qS cos qS. We factor out
the Jacobian from the integrand and we are left with
2p cos qS, which converts in rect angul ar coordinates to
2z. From our sketch of the region , we see that the
region of integra tion is over the disk of radius ~, cen
tered at the origin, so we have
I/V2 < x < I/V2 
and
J~£ ~ x2 < Y < J~2

x2. y
x
111
T HE CHANGE OF VARIABL ES FORMU LA AND APPLICAT IONS OF INTEGRATION
Now z extends from the cone to the sphere of radius 1. The equation of the cone is z and the upper hemisphere has the equat ion z = x 2  y2 so
/ 1
= x 2 + y'l
T herefore, the integral is converted to 1//2 ] v'1/ 2  E~ ]
l v'IX
2
Y2
2z dz dydx.
1//2 Jl/2 x' E' +y'
5. The desired region is shown here. The surfaces inter sect where z x 2 + y2 2  z2 or z2 + z  2 0 or z 1. We disregard z  2 as a sol ution since we want z ~ O. Looking at our drawing, we can describe it in cylindrical coordinates as follows . Notice t hat the entire region lies over the unit disk. Also, the region lies between z = x 2 + y2 = r2 and z x 2  y2 J2=T2, so the r gion can be described by
=
=
= /2 
=
=
=
z
=
o~ r
~ 1,
0~
y
e ~ 27l'
and
Since the Jacobian for cylindrical coordinates is r, we get
8. For convenience, we will rotate C 2 so that its axis lies on the z axis. T he volume remains unchanged. Using the res ul t of exercsie 12 in section 5.5, we substitute a = 1 since the diameter is 2 and the radius is 1. Thus, the answer is
\6 .
11. When the cu t is made by the plane x + y + z volume of the (shaded) solid below t hat plane is
r
fa a x f O x y
io io io
=a 
the
fa fOx
dzdydx =
r [( (a = io Substitute u
= a,
io io
x)y 
(a x y)dy dx
_Xl ~2 )\ Illy=o
dx
r (a~ x r dx . = io
1
y
x to get
x
i.
T hus, the volume for the entire solid, when a = 1 is If the solid is to be cut into n equal volum es, then t he volume under ;I: + y + z a should be k / 6n , where k is an integer such that 1 ~ k ~ n  1. Therefore, a3 /6 ::::: k/6n implies that cuts should be made in the planes x + y + z = (k/ n )1/3.
=
CHAPTER 6
112
15. Let u = x + y, v = y  x or x = (u  v)/ 2, y = (u + v)/2. The reader should veri fy that the J acobian 18(u, v)/8(x, y)\ is ~ . From the fig ures, 0 ~ u :S 1 and u ~ v ~ u. v
y
(1,1)
u >
(1,1)
~
x
T hen
Jl
exp
[~ ~ : J dx dy
tjU 10  u
exp(v/u) dv du 2
=~ t 2
10
u
u ( ev/u l
) du
v =u
~ (e  ~) 11 du = ~ (e ~) . U
16. T he tot al mass is the integral of the density. Since we are integrating over a sphere , use spherical coordinates. The sphere of radius R is described by
o ~ P :S R,
0 ~ r/J
~
1r
and
0 ~ 0 ~ 21r.
T he dist ance d from the origin is the same as p and the Jacobian is p2 sin ¢. T herefore , the total mass is 2" 1"lR p2 sin r/J '=3 dp dr/J dO . 1 o 0 0 l+ p Since all of the limits of integration are constants, we can integr ate each variable separately and multiply the results. We get
2
(1"dO) (1"
sin r/J dr/J)
(1 R1:2p3 dP)
= 21r [  cos¢ I:J
[ ~ 1n(I+ p3)Cl = ~1r ln(1+R2).
19. (a) Let T( x, y, z ) be the temperature at (x , y, z ). T hen the average temperature is fffe TdV/ fffe dV . By definition, d = J x 2 + y2 + z 2, so d2 = x 2 + y2 + z 2 and T(x, y, z ) = 32(x 2 + y2 + z2). T he numerator is
32 [ 1 [ 1 [ 11(x 2 + y2 + z2) dx dydz
32[:[ 1 [(~3 + + z2 )x ) [=J dydz 64[11[11(~+ y2 +Z2 ) dy dz (y2
64[ 1{ [(~ + z2 ) y+ Y:J[=J dz 1
128
3 + z 2) dz = 128 (2 "3Z+ 3Z3 ) 11 j 1(2 1
1
= 256.
T he denominator is just the volume of the cube C, which is 8, so the average temperature on C is 2~6 = (b) T he average temperature is attained wherever 32d 2 i.e., d l. Thus the average temperature is att ained at all points of the cube on a sphere of r adius 1, centered at the origin .
32.
= 32,
=
113
THE CH AN GE OF VARIABL ES FO RMU LA AND APPLICATION S OF INTEGRATION
t
t
20. The inequality y2 + z2 ~ describes the inside of a circular cylinder of radius centered on the x axis . T he inequali ty (x  1) 2 + y2 + z2 ~ 1 describes a ball ofradi us 1 centered at (1, 0, 0). We also want x ~ 1, so we get he region shown. By symmetry, we easily see that fi = z = O. For conven i nee , we will shift the region in question so that its axis of symmetry is the z' axis and we will place the ball's center at the origin . For the center of mass of the new region D+ , it is the integral of z· over D" divided by he volume of D . T h region D* is described in cylindrical coordinates by
o < 7' < !2'
0
<_ B _<
211"
0 < z·
and
< V! ~.  r~
z
.
z
y
112
1/2
The Jacobian is r, so the integral of z · over D· is
211"1 1/21 Ylrl r z" dz· drde = r 1l" [ 1/2 (r(z.) 2 1 ~) 1o 0 0 Jo Jo 2 z. =O
27r1 1/2(r 2 r 3) drde
1 o
11
2
0
dr
dB
0
2 ".
[(r2  r4 ) 11/2] dB _  7 1211" de _ 711". 2
4
128
r=O
0
64
The olu me f DO is
1
27r1 l/2 1~
o
0
r dz· drdB
0
211"1 1/2 ( rz· I~) dr dB z · =O 1o 0 211"11/2 rvh  r2. drdB = 2"
1 o
0
( !3 _ D* was shifted down
t
1 [(1r2 )3/2Il /2] (! _.;3) 3
0
v'83) Jo[2'"
dB
uni t, so for the original region,
= 211"
3
8
de
r=O
.
x is
24. (a) Since we are integrating a spherical region , we will use spherical coordin tes and the Jacobian is p2 sin ¢. We get
ffl
(:1: 2 + y2 + z2 )xyz dV
=
1"1"'l 2
R
p2(p cos {Jsin 4J )(p sin esin ¢ )(pcos ¢ )p2 sin ¢ dpd¢ de
CHAPTER 6
114
r [R
( r.
10 10 10 lcosOsinOcos¢Jsin3¢Jdpd¢JdO. Since all of the limits ofintegration are constants, we can integrate each variable separately and multiply the results. Thus, the integral is
(l
R
(1
(1 cos ¢J sin ¢J d¢J )
27r
p7 dP)
7r
COS
3
0 sin 0 dO)
C)( 0C) (
~ O.
2
( p;
sin: ¢J [)
sin 2
28. Since x 2 +y2 +z2 appears in the integrand , we will use spherical coordinates, and the Jacobian
is p2 sin ¢J. To integrate over all of ~a , we will integrate over a sphere of infinite radius. Thus, we describe ~3 by o ~ p < 00, 0 ~ ¢J ~ 1T, 0 ~ 0 ~ 21T . Therefore,
r
(Xi
JJ
[21f
p2 sin ¢J
10 10 10 {,.
!a /(x,y,z)dXdYdZ
1f
~3)3/2 dP) (1 sin ¢J d¢J) (1 (1 :2:~3/2) (  cos ¢JI:) ( oe)
00
(1 (1
(1
o
dO)
p';idp
00
1 +
Substitute u
21f
(1 +
00
41T
~\~/" dO difJ dp
p3)3/2
.
= 1 + p3 to get 41T
100 1
du 3u /
32
= 41T. hm 3 b+oo
1b
u
3/" ~
du
1
. = 41T3 b+oo hm
(Ju)I: =
81T/3.
31. (a) This improper integral is evaluated as follows:
1001 o
100 (2
x2 Y
11
x exp ( _y3) dx dy
0
11
2
)
1
exp( _y3) dy
x=O
0
00
y2 exp( _y3) dy
0
= lim
b+ oo
b
(exp ( _y3) 6
I 0
)
1
 6'
33. Notice that x 2 + y2 occurs in the integrand , so we will use polar coordinates, which we recall has Jacobian r. Consider ~ 2 to be a disk with infinite radius, so ~2 is described by
o ~ rand
0
~
0 ~ 21T.
Thus,
J~ ~
[21f
! (x, y) dA
[00
10 10
r
~. ' ~ dr dO
bl~n~(1+r2)1/21~=J dO = 10r 2
121f [
r.
dO
= 21T.
THE CHAN GE OF VARIABLES FORMULA AND AP PLI CAT IONS OF INTEGRATION
115
TEST .F OR C HAPTER 6
L True or false. If false, explain why.
(a) In polar coordinates, the volume bounded by the paraboloid z and the plane z = is given by t he integr al
°
1""l 2
Fa r(a  r 2) dr dB
00
2Jooo
(b) T he xistence ofthe improper integral J1
co:x
(x 2 + y2) with a >
=a 
°
rra2. =8
dx dy is guaranteed by Fubini 's theorem.
(c) When changing fro m Cartesian coordinates to pol ar coordinates, we have
1l1lorv'4y2
= J ,,"/6 [ 2 r2 cos B dr dB. ,,"/6lo
x dx dy
(d) The volume of the unit sphere centered at the origin with the cone z :=:; IS
[ 211" [ 1
r
lo lo l1f/4
J x 2 + y2 deleted
p2 sin if; dif; dp dB.
(e) The parallelogram with vertices at ~2, 2), (4,3), (0,5) and (2,6) can be transformed to the square [0,1] x [0 , 1] by T (x, y) + ( x+2l 10, x+;y2) and its corresponding Jacobian is 1
s· 2. Compute the integral of 1/(1 + x 2 + y2 + x 2 y2) over all of JR 2 • (Hint : Factor the denominator.)
°
3. A region in space is described by :=:; x :=:; 3,  2 :=:; y :=:; 2 and coordinates to om pute the average of z over the region .
°:=:; z :=:; ~. Use cylindrical
4. F ind the volu me of the region which lies between the surfaces z z = _( x 2 + y 2 ) 2 and which lies within the cylinder x 2 + y2 = 4.
=
10  x 2

y2 and
5. Let S be the transform ation defined by S : (x , y , z) + (u 2 , uw, V + w). Suppose this transfor mation takes D into the region [0,1] x [0 , 2] x [1,1]. Compute JJJD(y + z) dx dy dz.
6. As wit h parallelogra ms, a linear ran form a tion will map vertices of a parallelepiped to vertices of another parallelepiped. Let T be the transfo rmation T : (x , y, z) + (3u + 2v + w, 2u w , v + 2w) and let P be the parallelepiped with vertices at (0,0,0)' (3,2,0), (2,0, 1), (1, 1,2) , (5 , 2, 1), (3,  1,3), (4,1,2) and (6 , 1,3). (a) Convert JJJp (x  y  z) dx dy dz to tJvw coordinates. [Hint : Map the points (x, y, z) (0,0,0) , (3, 2, 0), (2,0,1) and (1,1,2).] (b) Evaluate the triple integral. 7. Integrate x + 2y over t he interior of the ellipse (x  2)2/4 appropriate substitu tion and then using polar coordina tes.
+ (y 
==
1)2/9 = 1 by making an
°
8. The shap of a swimming pool can be described by :=:; x :=:; 2, ~ 3Jl  x 2 /4 :=:; y :=:; 2 2 3J 1 x /4 and  J 1 x /4  y2/9 :=:; z :=:; O. Compute the volume by first substituting u = x/ 2 and v = y/ 3 and then changing to spherical coordinates. 9. Explain what is wrong with this calculat ion:
[lJY"!. dx dy = lot lo y x
In Ixll
Y _ X_Y
dy
=
[ 1(In lyl lnIY!) dy =
lo
to dy = 0.
lo
10. A nearsighted mosquito saw a t a ttoo on a man's arm which appeared to have the combined form of a rect angle, [ 2, 2] x [0,1], and a semiellipse described by x 2 / 4 + y2/9 :=:; 1, y:=:; 0. Where should the mosquito aim its probiscus to consume blood out of the tattoo's center of mass? You may assume t hat the mosquito sees a planar surface and t he tattoo has a constant density.
117
7
INTEGRALS OVER PATHS AND SURFACES
1.1: THE P ATH INTEGRAL G OALS 1. Be able to compute a path integral.
STU DY H I N TS 1. Not ation. T he path integral is denoted by
Ief ds, where ds = Ile'(t)lldt and e(t ) is a path .
2. Definition. The path integral is the integral of realvalued functions over a curve. The integrand
f only needs to be defined on c and f must be piecewise continuous on e. In addition, e needs to be piecewise continuously differentiable. T he integration of vectorvalued functions will be discussed in section 7.2. ~
3. Computation. If e(t) is defined for a
t
lb
~
band
f( c(t ))
f is defined on e, then the path integral is
lie' (t)11 dt.
4. Relation to arc length. If f = 1, then the path integral is the arc length formula of section 4.2. SOLUTIONS TO SELECTED EXER CISES 2. (b) The path in egral is defined by
lb
f(e(t))
In th is case, f( e(t)) = cos t , e' (t) T herefore, the path integral is
1
= (cos t, 
2..
1 =J f ds
3
sin t, 1) and
cos t . v'2 dt
3. (b) Let x = t, Y = 3t and z = 2t. Here, f V14. Then the path integral is
Ile'(t)11 1dt.
= v'2 sin t
lie' (t)11 = [cos 2 t + sin 2 t + 1271' 0
== O.
j
3 3 2
t dt ::::; 2v14t 3
1
1
= 52Ji4.
6. (a) Recall that in polar coordinates,
= r(O) cos 0
= r(O) sin O.
and
y
and
dy
Heie we treat r as a function of 0, and so
dx
= (r'(B) cos B
== V2.
= yz = (3t)(2t) and ds = Ile'(t)11 = [12+3 2 +22 P / Z =
(3t )(2t)V14 dt "'" 6v14
x
1P/2
r(B) sin B) dB
= (r'(B) sin 0 + r(B) cos 0) dB.
CHAPTER 7
118
Also, we know that
ds
= =
Hence,
V dx 2 + dy2 dO ([7·/(0) cos 0  r (O) sin (W + [1'1(0) sin 0 + 1'(0) cos 0]2}1 /2 dO [( 1" (0) )2(cos 2 0 sin 2 0) (i (0))2(cos2 0 sin 2 0)j1 /2 dO
+
+
+
= = Vi2+ (dr j dO) 2dO . t he path integral of f (x , y) = f (1' cos 0, rsin 0) is
1
1(x , y) ds =
1~2 f(r cos e,
l'
sin
O)V i
2
+
(~~
r
de .
(1,1,1)
°
7. (a) From t =  1 to t = 0, the path is a straight line going from to (0,0,0); from t = to t = 1, the pat h is a straight line going from (0 , 0, 0) to (1,1, 1), so act ually the same path was covered twice. Hence, the path integral is
lb
f (x (t ), y(t ))y'( x'(t ))2
211 = 211
+ (y' (t) )2 dt =
(2t4  t 4)V {4t 3)2 + (4t 3)2 dt 4t 4Y'2t6 dt
=
8h11
t 7 dt
= h.
The answer could be thought of as the area that needs to be covered if one wants to paint the two sides of a fence erected along the path x t4 , Y t4, of height f (x (t), y( t)) . (b) The arc length function is the pat h integral when f = 1, so it is
=
=
lb
f( c(r))llc'(r)ll dr ,
where a is the st arting point. To evaluate s(t ), we need to break up the interval into [1,0] and [0 , 1] . For  1 ~ r ~ 0, Ilc'(r)1I = [( 41"3 )2 + (4 r 3) 2j1 /2 = 4V2I1"13 = 4y1r 3. Then
s(t) = 
[tl4V2r 3 d1" =  V2(t 4 1) .
T hus , t 4 =  s j y1 + 1 and 4t 3 dt =  ds/ y1. When t =  1, s = 0; and when t = 0, s = y1. For r ~ 1, Ilc'(r)11 [(41"3)2 + (41'3) 2]1 / 2 4y1 r 3. We also need to add the length from  1 to 0, so we get, for t :::; 1,
°:: ;
=
=
°:: ;
s(t)
= V2 +
1t4V21"z
So t 4 = s/V2  1 and 4t 3 dt = ds/y1. When t all this together, the path integral becomes
d1'
= /2(t 4 + 1).
= 0, s = V2; and
when t
= 1, s = 0.
Putting
1V2 ( ~ + 1) ds+J~ (~ l) ds =( 2~ +s)l: + (2~  s) I~~h. 10. (a) W hen density is a constant k, we have Mass The length of the wire is
=k x
length of wire.
lbIlc'(O)11
dO ,
INTEGRALS OVER PATHS AN D SURFACES
119
so the mas is
(b) Recall th at a coordinate of the center of mass, Xj, is
Here, we have a thin wire in the yz plane. By symmetry, x so the center of mass for y, with k = 2, is f ky ds fo" a sin (J • 2a dO f= k ds 2a7r
_ a
 71'
= z = O. In
1" . 0
this case, y
= a sin 0,
0 _ 2a smOd   .
71'
Thus, the center of mass is (0, 2a/ 7r, 0) . Note that the center of mass is not on t he wire. 13. The t rick of this xercise is to use the correct coordinate system. T h intersection of the sphere and th plane is a unit circle, but in an unfamiliar orientation. T he unit normal vector of the plane x + y + z = 0 is (i + j + k) / J3, so we need to find two orthogonal vectors in the plane x + y + z = O. For example, (1,0,  1) and (1,  2, 1) are a pair. Normalization of h e two vectors yields u = (1,0,  1)/0 and v = (1,  2, 1) / V6. T he circle is now param etrized as c(B) (x, y z) cos Bu + sin Ov, 0 ~ 0 < 271'. Take the first component of c to get x = (cos B) / /2 + (sin B)/ )6. Then the mass is
=
=
1x
2
1 (~coso + ~sinor Ilc' (B)l ldB 1 (~ + ~ 1 (12 + 2 v'i2B 0+ 6 B) 2
ds
=
"
2
"
2 ..
o
You should verify that
Ilc'(B)11 = 1.
cos B
cos
20
sin B) 2 . 1 dB
sin cos
1.
ill
2
dB .
Now use the halfangle formulas to get
7.2: LINE IN TEG RALS GOALS 1. Be able to compute a line integral.
2. Be able to explain the difference between a line integral and a path integral.
STUDY HINTS l. Definition . The line int gral deals with vector fields whereas the path integral dealt with
scalar fu nctions. It is required that c be Cl and F be continuous on c (or at least piecewise continuous for both c and F) . 2. Physical interpretation. The line integral is most commonly interprete as the work performed by F along a path.
CHAPTER 7
120 ~
3. Computation. If F is defined on c and c is defined for a
t
~
b, then the line integral is
lb
F(c (t )) . c' (t ) dt.
Note that the integrand is a dot product, so the integral is a scalar . T he formula involving
T (t ), the unit tangent, may be more difficult to use since you need to compute T (t) as well as
Ilc'(t)ll · 4. Sign interpretation. Posi tive work means that t he force fi eld did a net amount of work on t he object; such is the case when the motion of the object is in the direction of the force. If work
is negative, then this amount of work is done by the object on the force field.
5. Reparametrization and orientation. If c is reparametrized and the orientation is preserved, the value of a line integral does not change. If t he orientation is reversed, only the sign of the value of the line integral changes. You can substitute for t he end points to be sure the direction is correct. Orientation is not important with the path integral of the previous section. Reparametrization allows us to break up a curve and integrate over each segment separately and with convenient parametrizations (see example 11 ). Be sure the path is traversed the correct number of times. 6. Gradients. The line integral of a gradient field depends only on the endpoints . We will see
how to make use of this fact in section 8.3 . Given this fact, we can see tha t t he line integral of a gradient along a closed path is O.
SOLUTIO N S TO SELECTED EXERCISES 1. (a) T he line integral of F along cis
11
F (c(t)) . c'(t ) dt .
We have c(t} integral is
= (t, t, t), so c'(t) = (I , I, 1) and
11 + + (ti o
tj
tk ) . (i + j
+ k) dt =
F( c(t ))
= (t, t, t ).
11+ + (t
t
t) dt
0
Therefore, the desired line
3 2 = _t
2
1
0
= 3 . 2
2. (c) We need to break up the integral into two parts. For the segment j oining (1, 0, 0) to (0, 1,0), th e easiest way to parametrize the line is to find c(t) = u(t) + tv (t ) such t hat c(O) = (1, 0,0) and c (l) = (0,1, 0) . Letting u(t ) = (1, 0,0) and v (t ) = (0 ,1, 0)  (1,0,0) = ( 1, 1, 0) gives us C1 (t ) = (1, 0, 0) + t ( 1, 1, 0) or (x, y, z ) = (1  t, t, 0) for 0 ~ t ~ 1, so dx = dt, dy = dt and dz = O. Substitute these values to get
1
(yz dx + xz dy + xydz ) =
11
[t · 0· ( dt ) + (1  t) · 0 · dt + (1  t) . t · 0]
= O.
0
Cl
For the line segment joining (0, 1, 0) to (0 , 0,1 ), use the same method . We find C2 (t) (0 ,1, 0) + t(O ,  1,1) or (x, y, z) = (0 ,1  t, t ) for 0 ~ t ~ 1, so dx = 0, dy =  dt and dz = dt. Substitute these values to get
1 ~
(yz dx + xz dy + x y dz ) = { I [(1  t) . t ·0+ 0 . t . ( dt ) + 0 . (t) . dt] = O.
h
Adding the two line integrals together, we get 0 + 0 = O. 4. (a) By definition, we have
1
F· ds =
1
F( c(t) ) . c'(t) dt .
121
INTEGRALS OVER PATHS AND SURFACES
We are given that F and e' are perpendicular, so F· e' (t) = O. T herefore, the line integral is O. (b) Again, we use the definition of the line integral. In addition , we use the definition of the dot product to get
1 1 F · ds =
F(e(t )) . c'(t ) dt
= l IIF (e(t ))lllle'(t)II' cos(O) dt =
l IIF " ds.
This result because the cosine between parallel ve tors is 0 and by definition ds = Ilc'(t )11 dt.
7. We are given that c(t) = (x,y,z ) = (t,tn, O), 0 $ t $ 1, where n = 1,2, 3, .. .. Differentiation of each component gives us c' (t) = (1, nt n  1 , 0) , so dx = dt , dy = nt n 1dt and dz = O. Therefore,
1[ydx+(3 y3  X)d y + z dz1
=
fa \ tn dt +(3t3n _t) ntn 1 dt+0) fa 1(t n + 3nt 4n  1

ntn ) dt
( 1l + nn tn+1+ ~t4n)ll 4
3 n l 4'I+n '
0
11. Given that e(t) is a path and T is a unit tangent , we have
T=~ Ile' (t) 11
by definition . Since v · v
r
= Il v 11 2, we get the line integral
r e'(t)
Jc T · ds = Jc Il e' (t )11
, . e (t) dt =
rlIIle'(t)112 e' (t)11
Jc
dt =
l ' c
li e (t )/I dt .
The last integral is simply the arc length of c(t). 14. If a vector field is a gradient, then the integral only depends on the endpoints. For a closed curve, the starting point is the same as t h ending point, so theorem 3 tells us that t he integral is O.
= Vf , where
17. From the fact that V(I/7') =  r/ r3 , we see that F theorem 3, the line integral is
f(c (b))  f (e(a))
= R1
:!
f = l/ r. According to
1  R . 1
7.3: PARAMETRIZED SURFACES
GOALS 1. Be able to parametrize a given surface.
2. Be able to determine if a surface is regular and /or differentiable. 3. Be able to compute a tangent plane for parametrized s urfaces.
STU DY HINTS 1. Parametrized surfaces. Recall that curves in the plane can be parametri zed by wo functions of one variable, namely, x f (t ) and y g(t). A curve in space is given by t hree functions of on variable. Now, we extend this idea to surfaces. We do his by letting J,', y and z be fune ions of two variables, i.e., now we hav t hree fu nctions of two variables, x = f (u , v), y = g(u, v) and z = h(u, v).
=
=
CHAPTER 7
122
2. Differentiable surfa ces. Such a surface is parametrized by (f( u, v) , g(u, v), h (u, v)) and j, g and h are all differentiable functions.
3. Review. Recall t hat a vector n which is normal to a plane gives the coefficients of the equation of that plane. See Section 1.3. 4. Tangent vectors. Like a partial derivative, we can hold u constan t to get T I) Likewise, holding v constant gives us Tu (fu, 9u, hu ).
=
(fl) , g", hI) ).
5. Tangen t plane. By holding either parameter, u or v, constant, we get a curve on the surface and if the cur ve is "nice," it will have a t angent line. Sim il arly, we can get another t angent line by holding the other parameter constant. Those two tangent lines determine the tangent plane, and t he normal vector to the plane is Tux T". 6. Regular. A surface is said to be regular if Tux T" f:. O. When the cross product is not 0, a tangent plane exists and consequently, the surface has no pointed regions (li ke a cone). Differenti ability does not imply regular (see exercise 1i ), but regularness requires differentia bility, for otherwise T u and T" would not be defined.
7. Importan t para m etrizations. Know the following parametrizations: (a) Circle of radius r: x = r cos O, y= rsine, 0 ~ (b) The hyperbola x 2  y2
f) ~
21T.
= 1: x = cosh t, y = sinh t.
(c) Sphere of radius p: x = p cos Osin ifJ , y = p sin BsinifJ, z = pcosifJ, 0 ~ 0 This is the same as the definition of spherical coordinates.
< 21T, 0 ~
ifJ
< 1T.
(d ) A surface z = g(x, y ): x = u , y = v, z = g(u , v ). 8. Normal to a graph. If z = ! (x, y), the normal to the graph is ( ~, 
%&' 1). This
is useful to know for the coming sections. Note that if we let g = z  !(x, y), then the (unnormalized ) normal to the surface is n = 'il g.
SOLUTIONS TO SELECTED EXERCISES 3. The normal to the desired tangent plane is Tux T I) ' We compu te T u = (/Jx / /Ju , /Jy/ /Ju , /Jz/ /Ju) = (2u , sin e" , ~ cos e" ). Simil arly, T" = (0, e"u cos e V ,  ~evusi n e" ). We compute i
Tu x Tv
2u
o
j sin e" ue" cos e"
k 13 cos e"
(u / 3)e" sin e"
2
((  u/ 3)e" sin e"  (u/ 3)e V cos 2 eV)i + ((2u 2 / 3)e" sin eV )j + (2u 2 e" cos e" )k (u/ 3)e" ( i + 2usine"j + 6ucose"k). To find the tangent plane, we only need to evaluate  i + 2u sin e" j + 6u cos e"k at t he given point (uo,vo) . However, we are only given (ZO, yo, zo ), so we must be clever . We know th a usine" =  2 and (u/ 3) cose U = 1 at the given point , so 2u si n e" =  4 and 6ucose" = 18. Hence, the tangent plane is (x  13)  4(y
+ 2) + 18 (z 
4. (1) A surface is regular ifTu x T" so Tu
X
T"
f:. O.
=I 2 o
1)
= 0,
or 18z  4y  x
= 13.
We calculate Tu = 2i + 2uj +Ok and T" = Oi + j+2vk,
j
k
2u 0 1 2v
I = 4uvi 
4vj + 2k,
123
INTEGRALS OVER PATHS AND SUR FAC ES which is never O. Therefore, the surface is regular. (2) Again , we calculate T ... 2ui + j + 2uk and Tv
=
T ... x Tv =
If u
= v = 2, then T ...
i 2u  2v x Tv
k 2u 4
J
1 1
= (4  2u)i  (8u
= O. Therefore,
j 01 cos v 0
+ 4uv)j + (2u + 2v)k.
the surface is not regular at (0,0, 4).
7. From t he partials of x, y and z with repsect to (cos v) i  (sin v)k. T hen
T ux T v =
=  2vi + j + 4k, so
tI ,
we compute T ... = j, and similarly, Tv
k 0 sm v
=(sinv)i(cos v)k.
By our luck, the magnitude of this vector is 1 already, so a unit normal is n = (sin v)i (cos v )k. [Note that another unit normal is (sin v)i+(cos v)k, depending on which cross product (T ... x T v or 1\ x T... you take.] To identify the surface, note that x 2 + z2 = sin 2 v + cos 2 V = 1. As v varies between 0 and 21l" , this becom s a circle of radius 1 centered at the origin on the xz plane. Since y = u, we have 1 ~ Y ~ 3. Therefore, the surface is a cylinder of radius 1 centered around the y axis and it extends between y = 1 and y = 3. 9. (a) Parametrically, the surface is described by x Ty = (hy, 1, 0) and Tz = (hz, 0,1), so 1
Ty x Tz =
hy hz
j k 1 0 0 1
=
h(y , z), y
=
y, z
=
z . In this case,
= (1, hy, h z ) .
The vector T y x T z is normal to the tangent plane, and so the tangent plane is
(x  xo)  hy(Y  yo)  hz{z  zo)
= 0,
where hy and hz are evaluated at (xo, Yo, zo).
11 . The surface is parametrized by (x, y, 3x 2 + 8xy), so Tx = (1,0 , 6x + 8y) and Ty = (0,1 , 8x ). At t he point (1 , 0, 3), T x = (1,0,6) and Ty = (0, 1,8). T he normal vector is given by n = Tx x Ty =
j k 1 0 6 018
= 6i  8j
=
+ k.
=
Thus, the desired equation is (x  Xo, Y  Yo , z  =0) . n 0; that is, 0 (x  1, y  0, z  3) . (6, 8, 1) = 6x  8y + z + 3, i.e., z = 6x + 8y  3. T hinking of the surface as the graph of t he equa tion f( x, y) = 3x 2 + 8xy, we calculate Oflox = 6x + 8y , ofloy = 8x , which become 6 and 8, respectively, at the point (x , y) = (1,0). Then the tangent line is z  3 = 6(x  1) + 8(y  0); that is, z = 6x + 8y  3. 13. (a) W hen 0 = 0, 4>(1',0) is the line segment from (0 , 0,0) to (1 , 0, 0) . As 0 increases, the line segment rotates around the z axis and the segment also moves to the height z = O. Thus , we get the sketch of a helicoid . The sketch in figure 7.4 .2 shows a helicoid for 0 ranging from 0 to 21l". T he helicoid for 0 ~ 0 ~ 41l" is similar except the graph extends to z = 41l" and makes an extra revolution around the z axis. (b) Differentiating the components of is 1
Tr x To=
cos 0  rsinO
j sin 0 r cosO
k
o 1
= (sinO,cosO, r).
124
CHAPTER 7
Bence, t he unit normal vector is (T r x T9) /II T r x T 911 = (sin B,  cosB , 7')/~. (c) Multiply the vector Tr x T o calculated in part (b) by r to get another normal vector:
(rsinB,  rcosB,r 2) = (y,x, x 2 +y2). Then the equation of the tangent plane at (xo , Yo, zo) is
Yo(x  xo)  xo (y  Yo) + (x~
+ Y6)(z  zo) = 0.
17. (a) It is obvious that the imageof~ l is the xyplane (or the uvplane, if you prefer) . Since u 3 and v3 can have any real value, and the third coordinate is 0, the image of ~2 also is the xy plane. (b) A surface is not regular is liT" x T vll = 0. For ~1' we have T" = i and Tv = j, so li T " x Tvll = Ilkll = 1. Therefore, ~ 1 descri bes a regular surface . For ~ 2' T" = 3u 2 j and Tv = 3v 2 j. If u = or v = 0, then T" x Tv = o. Bence, ~ 2 is not regular. (T his problem illustrates t hat "regularness" depends on parametrization, not necessarily the image.) (d) The answer is no. A "regular" parametrization cannot "round out" the "corners" of a graph. Indeed, by (c), a regular parametrization gives a smooth surface.
°
7.4: AREA OF A SURFACE GOALS 1. Be able to find the surface area of a given surface.
STUDY HINTS 1. Riemann sum argument. If you understand the Riemann sum argument , you should be able
to deri ve the surface area formula. 2. Form ulas. You should know at least one surface area for mula. One is
J10
A (S ) =
li T" x Tv ll du dv.
Equation (3) follows immediately from this formula, and (4) is a special case of (3). If you choose to remember equation (3), note that the integrand has the Jacobian matrix of each possible pair chosen from (x, y, z), i.e., (x, y), (y, z ) and (x, z ). 3. Oneloone. J ust as with curves, the chosen parametrization must be onetoone; otherwise, the surface may be covered more than once. See exercise 2. 4. Su r/ace area 0/ a graph. If z = !(x, y), use the parametrization x = u, y = v and z = !(tt, v). Then the surface area formula becomes A (S)
J10 /11 + /~ +
=
1 dA.
T his is important to know or you should be able to derive it from
ffD li T" x T vll dudv .
SOLUTIONS TO SELECTED EXERCISES 2. Using formula (3), we compute the following determinants :
8( x, y) 8(B, ¢) 8(y, z)
8(B, ¢)
I_I 
8x / 8B Bx/ B¢ sin Bsin ¢ cos Bcos cos 8 sin¢ sinBcos ¢ / By/8B By/ B cos 8 sin ¢ sin Bcos ¢ / _ . 2 A, (j sin ¢   sm If' cos , /
°
_
. ¢;  sm cos,
/ _
125
INTEGRALS OVER PATHS AND SURFACES
and
8(x,z) _ 1  sinBsin¢J cosOcos¢J 1 _ . 2 ", . l1
8( e, ¢J) 0  sin ¢J  sm 'I' sm
(l.
Squaring the above determinants, adding them and taking the square root gives us 4
2
liTe x T 1>11 = j sin 2 I,b cos 2 ¢J + si0 tP(cos 2 B + sin B) = j Sin 2 ¢ (cos 2 tP + sin 2 I,b ) = 1sin tP l· If we allow ¢J to vary from  7r/ 2 to 7r/2, then we get
2'1fl1f/2
1
=
Isin 4> 1d¢ dO
 1 " /2
o
[1
27r
0
r/ SintPdl,b] io 2
 'If/2
( sin¢J ) d¢J +
27r [ cOS¢J IO
 'If/ 2
+ (. cos 4» 1'If/ 2] = 47r. 0
T he answer is 47r because when 4> varies from 7["/ 2 to 7r/ 2, t he parametrization covers the upper hemisphere twice, so the surface area is the same as one compl te sphere . If we vary I/; from 0 to 27r we should get 87r , since the sphere is parametrized twice:
1 1 211" 2
1[
Isin ¢ I d¢J dB
=
27r
[111" sin tP d¢ + 12'1f( sin ¢J) d¢]
27r[ ( cos¢J)I : + costP C ] = 27r(2 +2) = 87r. If we had fo rgo ten the absolute values and simply integrated sin 1/;, we would have computed a surface area of 0 in both cases, but tills is not possib le. Surface areas have to be positi ve .
lID II Ttl
5. The ar a of of/ (D) is
8(x, y) = 8(u, v)
I
1 1
1 1= 2; 1
x Tv II du dv. Given of/(u, v ) = (u  v, u + v, u'v) , we get
8(y, z) = 1 1 1 1=  u _ v; 8(x, z) = 11 1 1= u  v. 8(u, v) v u 8(u, v) v u
Squaring t he de terminants, adding them and then taking the squar root gives us the integrand II Ttl x Tv ll = V2 2 + ( u  V)2 + (u  v)2 = V4 + 21.12 + 2v 2 = V2Vu 2 + v 2 + 2. Since we are integrating over the unit disk, use polar coordina tes. Let 1.1 r sin () , 0 ~ r ~ I , 0 ~ B ~ 27r. Then
JLh Vu
2
+
v2 +
=
2dA
111211" h
Vr2 + 2rdBdr = 2V27r 2
2V27r' i (r + 2)3/ 2
8. We choose the parametrization x = ~, y = t,
Z
1: =
11v,z
2V;7r (33 / 2 
= z, 0 :S Z
= r cos B and v =
2 3/2 )
+ 2r dl' =
i (6V6 
8).
~ 1,  1 ~ t :S 1. To compute
the sur face area integrand, we first compute
1
8(x,y~ 1= 0,
8 (y,Z) I l 18(x,Z )I 8(t, z)  , 8(t, z) 1
8(t, z }
then, II T t x Tz II =
2
J
t l dt dz 02 + 12 + Zt +
Thus , the surface area is
2+ 1111 [ffit a
 2
1
t

1 dt dz _
+1
=
t
Jt2+T'
[ffi t2 + t +
1 dt dz. 1
 2 
11 [ffi t2+  2
1
t

1 dt.
+1
We do not a ttem pt to simplify further. An alternative parametrization is to use the hyperbolic functions sinh t and cosh t; the integral only gets nastier.
(H APTER 7
126 10. Let x
= ucos v, Y = f(u ), z = usin v, a :::; u:::; b, 0:::; v:::; 2r. . The reader should verify that 8 (X,y) 1=  U!, (U) sin v, 18(u, v)
8(y, z) 1= u!,(u ) cos v, and I8(u,
v)
8 (X,Z) 8(u, v)
1= u.
1
Thus, the surface area is
=
A(S)
JIn
.ju2 + u 2(f'(u)) 2 du dv.
Since the integrand does not depend on v, the v integral can be performed, and we get the desired formula:
A(S) = 211"
lb lul.jl +
(f'(u))2 duo
We are rotating a. curve about the y axis, so consider the distance from the yaxis to the curve as t he "height," which is IxI. Thus, a crosssectional circumference of the surface at a fixed Yo is 211" 1xl· Next, describe the cur ve y = f (x ), a < x < b as a path c (t) = (t, f (t )) . T hen an infin itesimal arc length can be expressed as J1 + (f'(t ))2 dt or simply ds . T he surface area is obtained by integrating the crosssectional circumferences along the path c and the above formula reduce to A (S ) = 2r.lxl ds.
Ie
13. We are interested in the area of the surface z (x, y) = f(x, y) = 1  x  y, inside x 2 + 2y2 :::; l. First, compute
VI + r; + fJ
dx dy
= v'3 dx dy .
To compute the surface area, we need to parametrize the disc z = 0, x 2 + 2y2 :::; 1 using polar coordinates: x = rcosB , y = (r / V2) sinB, 0 :::; r:::; 1, 0 :::; e :::; h, and the Jacobian is r/V2. Our integral then becomes
2" fl
1 o
Jo
~r. v'3 dr dB =
r.V6 2 .
17. Completing squares, the equation x 2 + y2 = x becomes
(x 2  x+ t )+ y2 = i.e. , (x  t )2+ y2 = ( ~ t Th is
equation represents a cylinder whose base circle is centered
at (~, 0) with radius ~, as shown . To find the surface area
of 5 ] , we need to consider where the cylinder "sticks out"
of the sphere. Consider the positive octant. The surface
area is + r; + f3dx dy , where D is half of the base
circle (shaded), and z = f (x, y) = J1  x 2  y2 is the sphere. Since we will be integrating over a circular region, we
t,
".. ,
~,
~
x
IID ) 1
can use polar coordinates: x 2 + y2 = x is the same as r2 = r cos B or r = cos B. From the r :::; cos Band 0 :::; 8 :::; r. /2. Also , we com figure , one can see that D is described by 2 pute f x =  x /J 1  x  y2 and by symmetry, fy = y/Jl x 2  y2. So + + f~ =
°:: ;
)1 n
1/ J1  x 2  y2 , which becomes 1 /~ in polar coordi nates. Remembering that the J aco bian is r and that SJ consists of four equal surfaces, we get A(S t}
4
1f/21cos e 1o f7r/2
4Jo
0
r
v'1=T2 drdB 1  r2
= 4 1 0 7r/ 2 (
7r/2 (lsinB)dB = 4 (8+ cosB)lo
By high school geometry, we know that A (S:d (r. + 2) /(r.  2).
= 411" 
 ~
= 21T' 
(211"  4)
Icos e)
dB
r=O
4.
= 211" + 4, so A(52 )/ A(SI) =
127
INTEGRALS OVER PATHS AND SUR FACES
20. First, we are going to figure the volume of the material removed to make the hole in the sphere. Use cylindrical coordinates to describe the hole: 0::; r ::; 1, 0 ::; B ::; 21!'  ~ ::;
z ::;
~.
The limits for z were fou nd from the equ ations of the upper and lower hemispheres: z = J4  x 2  y2 and z = x 2  y2 and substituting 1'2 = x 2 + y2. Remembering that the Jacobian is r, we get
 /4 
Vhole
= ('" tl~ l ' dz drdB = t ~
Jo Jo
1
~ 1'2)3/2
2
"
(
2{4
[=J
"t
Jo Jo
dB =
2r/4  r 2 dr dB
~ (8 
1 2
3\1'3)
"
dB = 4; (8  3\1'3).
Since the volume of the sphere is 3211" / 3, th volume of the coupler is 3211" /3  (3211"/3 4rrvS) = 4rrV3. For the surface area, it suffices to calcula e the surface area of one "cap" of the hole. In rect angular coordinates, the surface area of I (x, y ) = J 4  x 2 y2 over D, the circle of radi us 1, is ffD + Ii + I~ dx dy. We calc ulate Ix =  x/ J4  x 2  y2 and Iy = y/ / 4  x 2  y2,
VI so V I+ Ii + n
4/J4  x 2
:=

y2 ,
" [

Changing to polar coordinates, the surface rea of one
cap is
12"11 h o
4 r
0
1
2
drdB = w
0
4~llr=O ]
dB
= (8 
1 2
4V3)
".
dB:= rr(16  8V3).
0
Si.nce the surface area of the sph re is 411"1'2, or 16rr , and the surface area of the two caps is 2rr(16  8V3), the outer surfa.ce area of t he coupler is 161!'(V3 1). 22 . (b) We computel:r: = y + l/( y + 1) and Iy
VI + Ii + fJ
=
dA
=x 
x/(y + 1)2. Thus
1 + (y2 + y + 1)2 + (x(y + 1)2  x )2 dA (y + l )2 (y+l)4 1
( )2 J( y + 1)4 + (0 + 2y2 y +l
+ 2y + 1)2 +
(x (y + 1)2 
xF dA ,
and the surface area is
1412 1
1
(
1
\2
Y + I,
/ (y + 1)4 + (0 + 2y2 + 2y + 1)2 + (x (y + 1)2  x)2 dx dy.
7.5: INTEGRALS OF SCALAR FUNCTIONS OVER SURFACES GOALS 1. Be able to compute the integral of a given scalar function over a given surface. 2. Understand why the integral is defined the way it is and to interpret it physically.
STUDY HINTS In t his section, we introduce dS, which stands for li T" x T v II du dv , which was discussed in section 7.4.
l. Notation.
2. Importance. This section and the next will be used extensively in chapter 8. In this section , we integrate scalar function as we did in section 7.1, and in the next section, we will integrate vectorvalued functions .
128
CHAPTER 7
3. Computation. The form ula for the scalar surface integral of a scalar function
f
is
fisfdS = flf liTu XTtl ll dudV. Here, fis usually given as a func tion of (z, y, z ) and we rewrite it in terms of t he parameters tI and v by substituting x, y, z as functions of tI, v. 4. Physical interpretation .
(a ) If f
= 1, then
we get the surface area. T his may help you to remember t he formula.
(b) If fis the mass density per unit area at each point of t he surface, we get the m ass of the surface.
5. Scalar integral over a graph. If z
fl
= g(x, y), then the formula becomes
f (z , y,g (x'Y ))V 1 +
(:!r
You should remember this or be able to derive it from
+ (~:r dx dy.
lIDf li T.. x Tvll du dv.
O. Integra ting over a plane. If S is a plane, we can simplify the integrati on formula in equ ation (5) of the text :
fis f dS fLc~ =
=
() dx dy ,
where cos () D . k and D is the unit vector normal to the plane. (Review the geometry of the dot product, section 1.2.) The region D is the projection of S onto the xy plane .
SOLUTIONS TO SELECTED EXERCISES 3 . Since we're integating over a hemisphere, it is wise to use spherical coordinates. For th hemispherical surface, we have p = a, so x = a cos () sin , y = a sin () sin 4> and z = a cos fo r o ~ () ~ 271" and 0 ~ ~ 71"/2 . Thus
II ( a sin () sin i + a cos () sin j 
liTex T¢> Il
Ok)
x(a cos ()cos 4>i + asin () cos j  asio k)11 a 2 sin . Then
f1
'{J J
zdS
{2'1f
D
a cos II T e x T 4> 11 d4> d8 =
3
r /2 sin 4> cos ¢ d¢ = 21ra
3
271"a Jo
r a ¢)1"/2 = 7I"a
Jo Jo
l2
2 a cos . sin ¢ d4> dB
( . ry sm;
5. (a) The equation of the sphere is x 2 + y 2 + z2 = 2Rz, R > O.
Upon com pleting the squares, we see t hat it is equivalent to the equati on z 2 + y2 + (z  R)2 = R2 , wbich is a spbere of
radius R centered at (0, 0, R ). The tip of the cone z2 = x 2 + y2
intersects (by design) the sphere at the origin. To fin d the other
intersections, note that x 2 + y2 + (z  R )2 R2 and x 2+ y2 z2
implies R2  (z  R)2 = z2 or z = R. Parametrize t he cone as
p cos B, y = p sin 8, z p with 0 ~ p ~ Rand
follows: x o ~ B ~ 271" . You should verify that IITp x Tell V2p. Thus ,
the area of the portion of the cone that is inside the sphere is
=
=
=
12"lR o
V2p dp dB
0
=
0
z
=
= 271"V2 p2 1R = 7I" V2R2. 2
3
0
x
y
129
INTEGRALS OVER PATHS AND SURFACES
(b) By "area of the portion of the sphere inside the cone," the authors presumably mean the area of the piece of the sphere that is the "ice cream" part of this configuration. T his is si mply the area of the hemisphere of radius R, or 21l'R2 . 7. Parametrize S usin polar coordinates: x = r cosB , y = 1'sinB, 0 ~ r ~ 1, 0 ~ B ~ 21l'. Since the su rface is described by z = x 2 + y2 , we substitute in x and y, and get z = 1'2. Then, T r x T 8 = (cos B,sin B, 21') x (1' sin B,rcosB,O) = ( 2r 2 cosB, 2r 2 sinB, 1'), and so
dS = IIT r x T elld1'dB= V41' 4 +r 2 d1'dB=1'V4r 2 +1d1'dB .
Calculat,jng
IIs ;; dS:
This integral can be done using integration by parts (or the tables) : let 1
u = r2, dv =rV4r 2 +1dr; du=2 1' d1', v= 12 (41'2+ 1)3/2 . Then t he integral becomes
8. First, integrate over the portion of the cube in the plane z = 1, which we will call 51 . We have 1 ~ x ~ 1 and 1 ~ y ~ 1, so T x x T y = i x j = k and II T x x T y II = l. Then
J1 11Jl
z211Tx x Tyll dx d y
I
1·1· d x dy
= 4.
y
1
The integral over t he portion of the cube in the plane z = 1, which we wIll call 52 , is done in the same way, so z2 d52 = 4.
J.{ is,
ow, for S3, which is in the plane x = l. Let D be the square 1 and no e that Ty x Tz = j x k = j and IITy x Tzil = l. Then
Similarly,
J.{J~
z2 dS4 =
J'J~{
z2 d55 =
J'hsf
z2 dS6 =
~ y ~
±,
1 and 1 ~ z ~ 1,
3 where S4 is the par t of the cube that is in the plane x = 1, S5 is in the plane y = 1 and 56 is in the plane y = 1. Therefore, z2 d5 is the sum of the integrals over the six surfaces, which is 4 + 4 + ~ + ~ + ~ + ~ = ~o.
IIs
CHAPTER 7
130
x 2 + y2 + z2 = R2. One can substit ute x for
y, y for z, and z fo r x, or any other perm utation , and still get the same equation. T hus, t he three integrals ought to be equal. (This is what "by symmetry" usuaUy means.) Geometrically, a sphere "looks the same" no matter how you look at it. (b) Using part (a) , we have
11. (a) The equation of the spbere is
Jis
(x
2
+ y2 + z 2) dS =
Substitute x 2 + y2
11 s
+ z2 = R2
x 2 dS =1
3
Jis
2 x dS
+
Jis
y2 dS +
Jis
z2 dS
=3
Jis
2
x dS.
and recall that ffs dS is the surface area of 5 to get
fl
(x 2 + y2 + z2) dS = 1 s 3
fl
s
411" 4. R 2 dS =R2  ·411"R 2 = R 3 3
(c) Due to symmetry of the sphere, if we integrate x 2 + y2 over the entire sphere, we should get twice the mass desired in exercise 10. So the desired mass is
~
Jis
(x
+ y2) dS = ~2
2
Jis
2
x dS
= ~11" R4.
14. By exercise 12, the average z coordinate is
Jis
A tS)
z dS.
In this case, A (S), the surface area of a hemisphere ofradius r, is 2 11"1' 2. By exercise 3, ffs z dS is 11"1'3 , so Z = 1I" r 3/ 211"1'2 = r /2 . Since there is as much of the sphere on one side of the z axis as there is on the other (by symmetry ), x = y = O. 20. If z
= g( x,y), then we can use t he fo rmula stated in equation (4), namely,
Jis
!(x, y, z) dS =
JIn
f( x, y, g(x, y) )J1 + (8g/ ox )2
Since 5 is defined implicitly by F (x , y, z)
8z 8x Given that ! (x, y, z)
8g ox
+ (og/ oy)2 dx dy.
= 0, we can use implicit differentiation to get
(oF/o x ) of/ oz
and
oz oy
og oy
(oF/oy) of/oz .
= loF/ 8z l, we get
10F IdS f·r Js oZ
=
r::
JrJD 10F8z Iy
1 i
"f
f JD JIn
[ (OF/8 x)]2 8F/ 8z
+ [(8 F/8 y )]2 dxdy 8F/8 z
18F I J (8F/8z )2 + (8F/ oX)2 + (8F/8yF dx dy OZ J(8 F/ 8zP
J(8F/8z)2
+ (oF/8 x )2 + (8F/8y)2 dx dy.
7.6: SURFACE INTEGRALS OF VECTOR FIELDS GOALS 1. Be able to compute a surface integral of a vector fu nction.
2. Understand its derivation and physical in terpretation.
131
INTEGRALS OVE R PATHS AND SURFACES STUDY HINTS
1. Notation. The symbol n( ~ (u o, vo)) is used to describe the un it vector which is normal to (a param trized surface) ~ at (uo, va ); note that ~(uo, va ) is the base poin t of n. 2. Orientation. As with line integrals, the orientation is important. T he sign of the integral changes with the opposite orientation. 3. ParametrizatlOn. As with line integrals , we can reparametrize a surface. orientation is retained, the value of the integral is unchanged .
As long as the
4. Definition . The surface integral of a vec or field F over a parametrized surface D is
~
ff~F . dS=
with domain
flF.(Tu X TtJ ) du dv .
The integral of a vector field is a scalar.
5. Generalizations. _'otice that scalar integrals (sections 7.1 and 7.5) do not depend on orienta tion. However the sign will change in the integration of a vector fie ld (sections 7.2 and 7.6) if the orientation has been reversed . Finally, note that scalar integrals involve the length of a vector and the integral of vector fields involve the dot product. 6. Reduction to calor integrals. If we know he unit vector to t he surface ~ , t hen the surface integral reduces to n) dS. Letting f = F . n, we get the scalar integral of section 7. 5. If he orientation switches, n changes sign and so does this f.
IIs(F .
7. Surface integral ouer a graph. If z = g(x, y) and F = (FI ' F2, F3 ), then a normal vector ( ag/ox ,  og/oy,l) and the surface integral becomes
fin [Fl ( ~!) +
F2 ( 
1S
~:) + F3 ] dx dy.
8. Physical interpretation. If F represents the velocity field of a flu id , the surface integral of F
over
~
gives the rate of flow of the fluid across the surface
~.
T his is known as fl ux .
9. Un it normal to unit sphere. It is useful to know that for the un1t sphere, n = r = (x , y, z).
10. Good example. Example 6 shows three ways of computi ng the same integral.
11. Summary. The formulas following example 6 give a nice summary of sections 7.5 and 7. 6. An understanding of how to use these formulas is important. If any of t he formulas do not make sense to you . it is time to review .
SOLUT I O NS TO SELECTED EXERCISES
IIs
1. As in example 4, the heat flux across the surface S is  'i1T · dS . Parametrize the surface 2 x + z2 = 2. Let x = V2 cos 8, z = v'2 sin 8 with 0 ~ 0 ~ 211". The surface can be parametri zed by S (8,y) = (V2cosO,y,V2sin8) , 0 ~ 8 ~ 211" and 0 ~ y ~ 2. Then the outward pointing normal to S is To x Ty = V2cosBi + v'2sinBk. Given T (x,y,z ) = 3x 2 + 3z 2 , we compute  'ilT =  6(.1:,0, z) =  6( v'2cos 8,0, v'2sin B). Thus
f
l  'ilT . dS
=
JL
 'ilT· (T o x Ty ) dydB
6 2""12 [ J2 cos 8) 2+ ( v'2 sin B) 2] dy dB
1 0
0
(
1212""12 dydO = 4811". If the Ty x To cross product (hence the opposite orientation ) is used, the answer would be + 4811" .
(HAPTER 7
132
6. First, we compute k
j %y x 2 + y  4 3xy
%x
\7xF =
=  2zj + (3y  l) k .
0/8z 2xz + z2
Use spherical coordinates to parametrize S: x = 4 cos Bsin
J1 J
J1
(0 ,  2z , 3y  1) . (T e x T¢ ) de d
(\7 x F ) . dS
 16 1 [(0,  8 cos
ior ior/
2
2
1'(
 16
2
(4sinBsin
r [(4 1 )1 1'(/2 ] dB 3sin3
1'(
1 6
<1>=0
21r
2
0
8. (a) The wall lies under the circle z = 4R 2, x 2 + (y  R) 2 = R2 and above the mountain x 2 + y2 + z = 4R2 . From the top view, we may parametrize the circle by 0~ 8~27r.
x = Rcos e, y R = Rsin8, Then the mountain becomes
z
=
4R2  (x 2 + y2)
4R2  [2R2
= 4R2 
[( Rcos 0)2 + (R + Rsin 0) 2] 2R z sin O.
+ 2R2 sin BJ = 2R2 
To find the surface area of the "cylindrical" wall of the restaurant, we parametrize the wall by
x = RcosO, y = R + Rsin8 , z = z 2R2  2R2 sin 8 ~ z ~ 4R2,
o ~ B ~ 27r, and the surface area becomes
271'14Rl
1 o
2"14Rl 2Rl  2R'l s in e
sin
e
liTex Tz il = R.
The reader should verify that
1o
li Te x T211dzd8.
2R~2Rl
Rdz dB
Thus, the integral becomes
= 12" R (4R2 
2R2 + 2R2 sin 8) dO
= 47rR3 .
0
(b) Parametrize the restaurant interior by x = r cos 0,
where 0 ~
1"
~
R, 0 ~ 8 ~ 27r, 4R2  (21"2 8(x, y, z) 8(1", 8,z )
y
=
+ 1" sin B!
+ 21"2 sin B) ~
os B 1 + sin B
o
l'
z = z,
z ~ 4R2 . The Jacobian is
 rsin B 0 r cos 0 o o 1
=r + rsin 8.
133
INTEG RALS OVER PATHS AND SURFACES
Thus, the volume of the restaurant is f2" fR
Jo Jo
1
2
4R
4R2 (2r2 +2r~ sin 0)
(1' +
l'
sin B) dz d1' dB
2
1
"
1 R (1' + l' sin (;1)( 4R2  4R2 + 2r2 + 2r2 sin B) d" d(;l
1 Jo
211" fR
1
a
(2r 3 + 4r2 sin 8 + 27,3 in 2 8) d1' d(;l
211" (R4
2
2) 2R4 sin 8
+ R4 sin (;I +
d8.
Since f02" sinBdB = 0 and fo sin 2 8d8 = rr , the volu me is rr~ . Thus, as long as R > 0, the volume rrR 4 is always greater than rrR4 j 2. V . dS, where V = k(6x, 2y  2R, 32z ). The roof of the restaurant (c) The heat flux is can be parametrize by x = l' cos f) J y = " sin (;I + R, Z = 4R2 for 0 ~ l' ~ Rand 0 ~ (;I ~ 2rr . We have T r = (co 8)i + (sin 8)j and T o = ( 1' sin (;I) i + (r cos 8)j , so T r x To = 1'k. T herefore, he heat flux through the roof is 27f
JIs
k
J
is (6:z:, 2y  2R, 32z )· (0 , 0, 1' ) drdB
"l
2 k 1
a
R
a
128R 2r drdB
= k1
= k
 128rr R 4k.
211"lR 32zr d1'd8
Z f
Jo
1I"
(64R2r 21R ) dB
= k(64R4 )(2rr )
r=O
The side which makes contact with the mountain can be parametrized by x = l' cos B, y = rsinB + R, Z = 4R2  (r2 + 2RrsinB + R2) for 0 ~ r ~ R and 0 ~ 8 ~ 2rr. We have Tr = {cosB)i+ (sin 8)j + ( 21'  2Rsin B)k and T o = (1'sin (;I) i + (reos 8)j + ( 2Rr cos 8)k, so T r x T 8 = (21'2 co O)i + (2Rr + 21'2si n 8)j + rk. T herefore, the heat flux through the mo untain side '  k
J
is (6x, 2y  2R, 32z ) . (2r2 cosB , 2Rr + 2r2 sin 8, r ) d1' dB {R1 27T  k Jo a (12 r 3 cos 2 8 + 4Rr zsin e + 41'3 sin 2 e + 128R2 "

32,,3
 64R 2 7'Sin e 32R2 1' ) dB d1'
k 1R(12rr,,3 + 4rr1'3 + 256rr R 2r  64rrr 3
1 R
k
( 48rrr3 + 192rrR 2r ) dr

64rr R 2 ,, ) dr
=  841rR4k .
Finally, th curved gJass wall can be parametrized as in part (a) . We have T o x T z (Rcos B)i + (Rsin B)j. The heat flux through the wall is  k
J
is (6x,2 Y  2R,32z )· (RcosB ,Rsin8,0) dzd8 (6 RZ cos 2 8 + 2R2 sin 2 8) dz d8  k 1Z1I'14R2 o 2R2  2R' sin8 k 1211' (2R2 + 4R Zcos 2 B)(2R2 + 2R z sin e) d8 211"
(4R4 + 8R4 eos 2 B + 4R4 sin 8 + 8R 4 cos 2 Bsin B) d8 1  k(8rrR4 + 8rrR 4) = 16rr R 4k. k
0
CH APTER 7
134 Adding these results, we find that the total flux is 2287TR4 k.
=
=
11. The surface S is the unit sphere, so S can be parametrized by x cos 0 sin ¢, y sin 0 sin ¢ and z cos ¢ for 0 ~ 0 ~ 27T and 0 ~ ¢ ~ 11". Differentiate each component with respect to 0 to get T o = ( sinOsin¢,cosBsin¢, O) . Similarly, T~ = (cosO cos¢, sin Ocos ¢,  sin ¢) . The normal vector for S is T o x T ¢ = (sin 2 ¢ cos 0, sin 2 ¢ sin 0, sin ¢ cos ¢). Fact or out sin ¢ to get (sin ¢)(x , y , z) = n = (sin ¢)r . Suppose F = (Fr , Fe, F¢), then F . n = F . (sin ¢r) = Fr sin ¢. Here, we make use of t he fact that (er , ell, e¢ ) form an orthonormal basis (see section 1.4). Hence
=
J1
F . dS
= 1 21r L'" Fr sin ¢ d¢ dB.
The corresponding formula for realvalued functions is
Jl!dS = 1
21rl" !sin ¢d¢ dB .
14. A parametrization of the surface ~ ( u , v) = (u, v, 0) , so liT" x T vll = Ii i x j ll = Il k ll = 1. The scalar integral is
J1
! (x , y , z ) dS
=
JIv
! (tI , v, 0) li T " x T vll dudv.
By substituting (x, y) for (u , v), we get ffD! (x , y, 0) dx dy. For a vector field F (NOTE: the subscripts here do not denote partials!),
Jis
F . dS
=
JIv
F . kdz dy
=
JIv
= (F""
Fy, Fz )
F. dx dy.
That is, only the z component of F enters into the formula. 16. (a) Parametrize the cone as fo llows: x = reos e, y = rsin O, z = (x 2 + y2)1/2 = r, 0 ~ r ~ I , o ~ B ~ 211'. T hen T o x Tr = ( rsin O, r cos B, O) x (cos O, sinB,l ) = (rcos O, r sinB,r) . The flux is the surface integral of F over S, which is
21r
r Jro (0, 0,  1)· (r cosO , rsinB,r ) drdB = Jor Jot Jo 21r
1
rdr dO=71'.
For the hot shots, note that the same amount of rain falling through the cone must also go through the disk x 2 + y2 ~ 1, z = O. The problem then could be done in your head . (b) Using the same parametrization as in part (a), the total flux through the cone is
21r
r Jor Jo
1
(Vi Vi)  2 ,0'  2
· (rcosB,rsinO ,  r ) dr dB
r r .J2 ( r Jo Jo 2 2tr
=
1
Vi Jo{21r ~ (1 2
2
cos B + r) dr dB
_ cos O) dO
= ~ V2.
7.7: A PPLIC ATIONS TO DIFFERENTIAL G EOMETRY, P H YSICS AND FO RM S OF LIFE GOALS 1. Be able to compute the Gauss curvature for a given surface . 2. Be able to compute the mean curvature for a given surface. 3. Be able to explain when the GaussBonnet theorem applies for calculating the total curvature of a surface.
INTEG RALS OVER PAT HS AND SURFAC ES
135
STUDY HINTS 1. Old calc ulations . In the process of computing the Gauss and mean curvatures, you already learned how to do the calculations for many of the ingredients earlier in the course. In fact , you should consider the calculations a review. 2. New funct ions. If your instructor expects you to memorize how to com pute E, F , G, i, m, n, it m ay be helpfu l to know that E, F and G involve manip ulations of the first partials of C). Also, note that N is dotted with the second partials of C) to compute i, m and n. 3 . Calculating Gauss cur vature. Note that E , F and G are not needed to compute the G auss IIC)u X If1 v112. curvature since W
=
4. Curvature of special surfa ces. You should know the res ults of examples 1 and 2. Example 1 shows that a plane has no Gauss or m ean curvature. Example 2 shows t hat the curvature is the same at any point on a sphere. 5. GaussBonnet theo r m. For a "spherelike" surface which is closed and bounded, we have (1/271') K dA 2. Furthermore, a surface of genus g results in he integral (1/271') f{ dA == 2  2g .
=
lIs
Ifs
SO LU TIONS T O SELECTED EXERCIS ES 1. We begin by computing all fi rst and second partial derivatives of C) (u, v) that is,
= (u cos v, u sin v, bv),
= Tu = (cosv,sinv,O); C) U = Tu = (usinv,ucosv,b);
= (0,0 , 0) ; lf1 "v = (sinv,cosv,O); and C) uu = (ucosv , usinv,O).
C)u
C)uu
Using these results, we can compute T u x Tv = (bsinv,bcosv,u); E = IIC)"W = cos 2 v +sin 2 v + F = C)U . C)v = u sin v cos v + u cos v sin v + = 0; G = II C)v 112 = u 2 sin 2 v + u 2 cos 2 V + b2 = u 2 + b2; and W = BG  F = (l)(u 2 + b2 )  02 = u 2 + b2 •
°
T hen, we calculate N = (T " x Tv) /v'W i, m and n:
f
= (b si n v, b cos v, '1.I.)/Ju 2 + b2 .
°=
1;
Now, we calculate
N . C)"U = N ·0 = 0;
N ' C)uv =(bsin2vbcos2v)/)u2+b2=b/)u'J+b2; and
m
N . C)v v
n
= (ubsin v cos v + ub sin v cos v)/Vw = O.
Finally, t he Gauss curvature is K
= in 
m
2
(0)(0)  m 2
W H
since i
= n = F = 0.
= u and y = v,
5. Let x paraboloid is lf1( u, v) derivatives of 1f1: C)u
lf1u"
so z
u2
W
and the mean curvatu re is
=:
[b 2/(U 2 + b2)]
+ b2
Cf + En  2Fm 2W
=
°
= u 2 /a 2  1I2/b 2 . T hus , a parametrization of the hyperbolic
= (u,v,u 2 /a 2 
v 2 /b 2 ). We now compute the first and second partial
= T " = (1,0,2 2u/a 2 ); = (0 , 0, 2/a
);
C)V
= Tv = (0, 1, 2v/b 2 );
C)"V = (0,0,0); and c)"v
= (0,0, 2/b 2 ).
..
.
CHAPTER 7
136 Using t hese results, we can compute
Ttl
X
= ( 2u/ a2 , 2v/ b2 ,1)
Tv
e,
Now, we calculate
N · ()uu N · ()uv
= =
n
= liT" X Tvl1 2 = 1 +4u 2/a 4 +4v2/b 4 .
m and n:
e = m
W
and
= [(Tu x T )/VW] · (0 ,0 , 2/a 2 ) = 2/a 2 /W. = N . (0 , 0,0) = 0; and lI
N · 4'1111 = [(T " x T,, )/Vw] . (0 , 0, _ 2/b 2 )
;:::
2/b 2 J'{V.
Finally, t he Gauss curvatUIe is
2
2 ) ( 22) JW  °2]
){ = en W m = [( a2 J'{V 4 a 2 b2 W 2  (a 4 b4
b
.W
'7"
 4a6 b6 + 4b4 u 2 + 4a4 v 2 )2 .
Notice th at we did not have to compu te E, F and G t o find the G auss curvature.
8. T he torus T is defined by ~ ( ¢, () )
so
= (x ,y,z ) = (( R + cos ¢ ) cos B, (R +cos¢ )sin(),sin cP)'
~ ¢ = T ¢ = ( sin ¢ cos f),  sin rf; sin () , cos ¢ ) ;
4' e = T e = ( (R + cos cP ) sin If, (R + cos .p) cos () , 0).
T hroughout this exercise, we use the identity
C05
2
A + sin 2 A = 1, so we get the cross product
 (R + cos ¢ ) cos ¢ cos ei  (R + cos ¢ ) cos ¢ sin Ifj ( R + cos ¢) sin ¢k.
T tl> x T o
Nex t , we calculate
W
= li T X Tol12 (R + COS ¢ ) 2 cos 2 1; cos 2 e + (R + cos cP )2C05 2 ¢sin 2 e + (R + cos ¢)2 sin 2 ¢ (R + cos ¢)2 cos 2 ¢ + (R + cos ¢ )2 sin 2 if; (R + cosif; f .
Then, we calculate N = (T ¢ x T 8)/ J'{V =  cos 1> cos (lj  cos ¢ sin €Ij  sin .pk.
Now, we com p ute
e, m and
£ = N · ()¢¢
n:
N . ( cos ¢ cos If,  cos ¢ sin If,  sin rP ) cos 2 ¢ cos 2 B + cos 2 ¢ sin 2 () + sin 2 .p cos 2 if; + sin 2 ¢
m
= N · ~¢ o
= 1;
N . (sin ¢ sin () ,  sin ¢ cos 0, 0)
e
 sin ¢ cos ¢ sin () cos (I + si n if; cos 1> sin () cos + 0
= 0;
N . (  (R + cos ¢ ) cos () ,  (R + cos ¢ ) sin (), 0)
n = N ·4' OB
(R + cos if; ) cos if; cos 2 B + (R + cos if; ) cos ¢ sin 2 If + 0
(R + cos ¢) cos ¢ . F inally, the Gauss curvatUIe is
K
Cn  m 2 HI
=
(1)( R + cos ¢ ) cos ¢  0 2
(R + cos¢ )2
cos if; R + cos¢
137
INTEGRALS OVER PATHS AND SURFACES
For the torus, which has genus 1, the GaussBonnet theorem tells us that we should get (1/ 271") ffs K dA = 0. In this case, we have
= ~ 271" 1
= 271"
{27r {27r
io 10
127r127r 0
0
os¢l
R + cos¢l R
cos ¢l + cos ¢ (R + cos ¢) de d¢l
= {2r. {27r cos ¢ de d¢ =
io 10
=
sin¢ C
II Tt,6 x T ell dBd¢l
271" {2rc cos ¢l d¢l 271"
10
= o.
9. We begin by computing all of the first and second partial derivatives of (I(u , v): (I" (I ~
(I"u
c)u ~ (I""
Using these result
= (l,h'(u)cosv,h' (u)sin v); T~ = (0 ,  h(u) sin v, h(u) cos v);
= = = = =
Tu
(0, h"(u) cos v, hI! ('U ) sin v) j (0,  h'(u) sin v, h' (u) cos v); and (0,  h(u) cos v,  h(u) sin v) .
we can compute Tu
X
T il
= (h'(u)h(u) ,h(u)cos v,  h(u) sin v).
Furthermore, we calculate
w
=
IIT u
=
(h(u»2[(h'(u)2
Next, we calculate
e= N
· c)uu
X
T~1I 2
t. m and
=
= (h'(u ») 2(h(u» 2 + (h(U ))2 cos 2 V + (h(U» )2sin 2 v + 1].
n:
[(Tu x TtI) / Vw]· (0, h"(u) cos v, h"(u) sin v)
= ( h(u)h// (u) cos 2 V m = N . (lU ti
n = N·(ln
Hence, the Gau
K

h(u)h"(u) sin 2 v)/Vw
= [h(u )hl!(u)]/Vw;
=
[(Tu x Tu )/Vwj · (0,  h'(u) sin v, h' (u) cos v)
=
(h(u)h' (u)sin vco v  h(u)h'(u) sin v cos v)/Vw = 0;
=
[(TuxTu) / Vw]·(O,  h(u ) cosv,h(u)sin v)
=
[(h(U»2 cos 2 v + (h(U))2 sin 2 v]/Vw
= (h(u))2/Vw.
curvature is
= en  m 2 = en W
=
 h3(h" )
W
°
= (h(h
_
ll
VW
) )
( ~) ~
VW
'
W
= h3 (h") W2
 hI!
[( 1 + (h' )2)h2J2  [(1 + (h' )2)2h] '
SOLUTIONS TO SELECTED REVIEW EXERCISES FOR CHAPTER 7 l. (b) First, we calculate c'(t ) = ( sint,cost, 1), so lI e'(tl ll
{{2'1r
1c fds = i o
= .;2. Then the path integral is
{27r xyz ll e' (t) ll dt = io
V2 (cost )(sint)t dt .
CHAPTE R 7
138
Use integration by parts with becomes V2
1.1
= t,
dv
t 2 t 127r _ 127f ~sin2tdt ) sin ( 2 0 0 2
2. (b) As in exercise 1(b), we have ds =
1f c
= cos t sin t dt, du =
{27f
In = _~
( 2 7rl cos2fdt =7rV2 2 io 2 2 .
II c' (t) II dt =
V2 dt . Then, the path integral is
2
i o (sin t + cos t)V2 dt =
ds
dt and v = ~ sin 2 t. T he integral
{27f (
V2 i o
sin t +
1 + cos 2
2t ) dt
SI: 2t) 1027r = V27r.
t'
V2 (  cost + "2 +
3. (a) Compute t he line integral over each segment of C and add them together. Be careful
with the orientation. Recall from chapter 1 that the equation of the line I(t) = (x, y , z) =
(1 , 0,0) + t(  1,1,0), 0 :5 t :5 1 satisfies the condition 1(0) = (1 , 0,0) and 1(1) = (0, 1,0). On this segment , x = 1  t, Y = t and z = O. Also dx =  dt, dy = dt and dz = O. Substitute these values into the given line integral to get
11 o
11
[sin 7r(1  t) dt  (cos 7rt)(O) ] =
0
sin 7r(1  t) dt = 1 cos 7r(1  t) 7r
1 0
= 1 (1  ( 1)) = 2 . 7r 7r
T he next segment can be parametrized by I(t) = (x , y, z) = (0,1,0) + t(O, 1 , 1) = (0,1 t, t ) for 0 :5 t :5 1. Also, dx = 0, dy =  dt, and dz = dt , so we get
11
1 11
[sin 7r(0)( dt )  cos 7r( 1  t ) dt ] =  sin 7r(1  t) = O. o 7r 0
Finally, the last segment can be parametrized by the equation
I(t ) = (x, y, z) for 0 :5 t :5 1. Also, dx
= (0, 0, 1) + t(l, 0, 1) = (t , 0,1 
= dt, dy = 0 and dz =
t)
 dt , so we get
11
[sin 7rt (O)  cos 7r(0)(  dt )] =
T herefore, the line integral around the triangle is 2/7r
11
1 dt = l.
+ 0 + 1 = 2/7r + 1.
5. We start at (0, 0) and compute t he line integral over each side of the square. For t he line
segment from (0,0 ) to (a, 0), we have y = 0, dy = 0 and :5 x :5 a. Thus
J
F . ds
r
=i o
[( x
2 
°
2
0 ) dx
+ 2x (0)
For the line segment from (a, 0) to (a, a), we have x
J
F· ds
=
l
. (0)]
2 
2
3
x dx
a = a
= a, dx = 0 and 0 :5 y:5 a.
a
[( a
r
=i o
y2)(0 ) + 2(a) ydy] =
l
a
2ay dy
=
ay2 1:
Thus,
= a3 .
°and x goes from a to O. Thus,
=  (a a  = 32a '
For t he line segment from (a, a) to (0 , a), we have y = a, dy =
J
F · ds
= iar (x 2 
2
a ) dx
=
l
0
a
(x
2

2
a ) dx
3
3 a )
3
Finally, for t he line segment from (0, a) to (0, 0) , x = 0, dx = 0 and y goes from a to O. T hus
J
F · ds
= O.
By addition, we find t hat the line integral around the square is a3 /3 + a3
+ 2a3 / 3 + 0 =
2a 3 .
...
139
INTEGRALS OVER PATHS AND SURFACES
7. (b) First complete the square: (2x2  8x + 8) + y2 + z2 = 1 + 8 or 2(x  2)2 + y2 + Z2 = 32. This has the form X 2 + y2 + z2 = p2, so use spherical coordinates: For the x parametrizat ion, v'2(x  2) 3cos Bsin<,h , so x = 2 + (3 cos 11 sin <,h )/ v'2. Thus, the parametrization is let X
=
=
2 + (3 cos hin <,h)/v2 3 sin 11 sin
x y z
with 0 ~ B ~ 271' and 0 ~
lID
10. The surface area of the graph lying over D is II Tx x Ty II dx dy and the area of D is The' parametrization" of t he graph is x = x, y = y, z = / (x, V) . Thus,
Tr = i + (8// 8x)k
and
lID dx dy.
T y = j + (8!1 8y) k.
T herefore, Tx x Ty = (8/18x)i + (8//8y)j + k
and
II Tr x Ty ll = [( 8/18x )2 + (8 //8y )2 + l f/2.
Since (8/18z)2 + (8//8y)2 = c, II Tx x Ty ll = JI+C. Returning to t he original formula, we have lID IITr x Tylldxdy = lIDJI+Cdxdy. Since c is a constant , we factor the constant from the integral to get VI+C lIDdx dy = v'I+C . (area of D) . 12. (b) Use cylindrical coordinates . Let x = rcosB, y = r sinl1. In addition, Z = x = "cos l1 , and the intervals are 0 ~ l' ~ 1, 0 ~ 11 ~ 27l' since we want to be inside the cylinder x 2 + y2 = 1. We calcuJate
(cosBi + sin OJ + cos 11k ) x ( rsinOi + l' cos OJ (ri + rk)=r(i + k )
so
l'
sin Ok)
IIT r xTell= v2r .
Therefore,
lIs x
15. We want to compute dS , where S is the trian gle with vertices (I, I, 1), (2, 1,1) and (2 ,0,3). First , we need to find the normal to the triangle: Two vec tors on the triangle are (1 , 0,0) and (0 , I,  2) (found by subtracting the coordinates of the vertices). Take t heir cross product and normalize it. We get the unit normal (0 ,2, l)/v'5, so cosB n · k = 1/ v'5. Next, the n projection of S onto the X!t"plane can be described by  y + 2 ~ x ~ 2, 0 ~ y ~ I , as shown. Thus,
=
Jis
(1,1,1)
=
xdS
=
v'sJ'rJD
v's1 1 1
x dx dy =
0
2
y+2
1
v'2 5 1 4  (2  y)2 dy v'5 [4+ (2 y)311] = 5V5.

0
2
3
o
I
, (1,1,0)
6
X
dx dy
x
(2,1,0)
y
CHAPTER 7
140
19. Here , we have F
= e'i + tj + t 2k and ds = etj + j + 2tk.
1
1
= 10
F . ds
(et . et + t . 1 + t 2 . 2t ) dt e
t2 t4) 11 = 21 (e
2t
2
(
Thus
2
+ "2 +"2
0
= 10
1
(e2t + t + 2t 3 ) dt
+ 1) .
=
24. Our surface is z2 1  x 2  y2 . We want the part above the xy plane , so we want z ~ 0, or 2 f (x, y ) z (1  x y2 )1 / 2. For the surface area of a graph over D, the formul a is
= =
fL V
1 + f ; + f;dxdy .
A= In this case,
_  1
2
x
2  1/ 2 _
fx   2 . 2x (1  x  y )
 (
2
2)1 / 2
I  x  y
_
and fy  (
y ( ) 2 2 )1/2 by sym m etry . I  x  y
Hence , the integrand is
+
x2
1+ 1_
y2 x 2 _ y2 
f afa
The area we want is
(1 _
x 2 _ y2 )1/ 2 '
1 _
2 _
2) 1/ 2
dydx.
Y 2 We can eval uate one of the integrals. Let u = (1  x 2)1/ 2. Since  a a ( 1
X
tI
is independent of y, the
integral in ybecomes
f a(u  a
Substituting v
f
= y/u
a ti /
 a / ti
with dy
dV 1
v'f=tj2 = V
1
2 _
y
1
2) 1/ 2
dy
=~
fa a
1 '" dy.
I. "
= tI dv, we get .
sm
1
(v)1
a/ ti  a/ti
a = 2sm. 1 () U
=
.
2sm
 1 (
~. 1  ;z; a
)
P lug this back into the original integral and get t h e form ul a for t he surface area: 2
I:
sin
1
(b)
dx.
=
=
=
27. (c) Use cylindrical coordinates: x 2cosO, y 2sinO , z z, where 0 :5 (J :5 21r and o :5 z :5 x + 3 = 2 cos (J + 3. (Note the difference in parametrizations, as compared t o exercise 112cos Bi + 2sinBj il 2 . T hen , 12). We calcula te IIT8 x Tzil = IIC 2sinOi + 2 cos OJ) x kll the surface integral beco mes
=
211'
1
o
12
cos 8+ 3
0
2z2 dzd()
=
f211' 2
lo
a (2cos O+ 3)3 dB
2 f211'
lo (8 cos3 () + 36 cos 2 () + 54 cos B + 27) dO
a
2 f21r [ . 8( 1 S10
al
2
o
(J)cosO+ 36
3
(1 + cos2 2B ) + 54cosB+ 27J dO 2
. . O+ 270J/0 11' a2 [ 8 ( smO  Sin3  (J ) +18 ( ()+ Sin20) 2+ 54sm
2
a [18 . 21r + 27 . 21r] = 601r.
141
IN TEGRALS OVER PATHS AND SUR FACES
TEST FOR CHAPTER 7 1. True or false . If false, explain why.
°
(a) Let c(t ) = (e t + 1, sin 2t ,  2t + 4), ~ t ~ 1 and let b (u) = (eu / 2 ~ u ~ 2. Then the line integrals F . ds and F . ds are equal.
°
Ie
(b) On a given domain, jf g(x , z) ~ surface y = g(x, z) is negative.
+ 1, sin u, u + 4),
Ib
°and f is a positive scalar, t hen the integral of f over the
(c) Over a triangle with vertices at (1 , 1), (2, 3) and (5 ,  3), the surface area of the graph of I( x, y) 2xy+y+5 is the same as t he surface area of the graph of g(x , y) x 2 + y2 + x +3.
=
=
(d) Consider t he cone z = J x2 + y2. Cl t( u , v) = parametrization if u ~ 0 and v ~ 0, but tl2 (r , 0) r ~ 0 is not onetoone.
(u , v,";U2
+v2) yields a onetoone for 0 ~ (I < 211',
= (r cos (I , r sin (I , Jr)
(e) If the line integral of F along one closed curve is 0, then F is a gradient. 2. An ellipsoid centered at the origin wi th intercepts at x = ± 1, Y = ±2 , z = ± 3 can be paramet rized by tI (B,
°
=
°
3. Find the work done by a force F = xi + x yj along the path beginning at (0, 0), going to (1,1 ), to (2,0) and t hen back to (0,0) all along straight line segments. 4. Let F(x, y , z) = xi + yj + (x + z) k. Compute t he surface integral of F over the elliptical cylinder x 2 /4 + y2 = 1, 0 ~ z ~ l. 5. At a point (x , y, z) on a spherical shell of radius 2, the density is Iz I. (a) Find a paramet rization for t he sphere using the spherical coordinates
(I
and ¢.
(b) Write the mass of the sh II as a surface integral. (c) Compute the mass. 6. (a) Compute the gradient of I (x ,y,z) = eZ + 2x z sin y. (b) Let p be the path from the origin to t he point. (1r/ 2, 1, 0) along the curve y = sinx . From there, p continues along a st raight line to (5, e,  3). Then p cont.inues on a straight line fro m (5, e,  3) and ends at (2, 37r/ 2, 1). Compute the line integral of F (x , y,z) 2z sin yi + 2xz cos yj + (eZ + 2;1; sin y)k over p. 7. Let C) (u v)
= (u2 , V 
2u , 3uv 2 ) be a parame trized surface.
(a) Find the equat ion of the tangent plane when (u v)
= (1, 1) and (x, y , z) = (1, 1, 3) .
(b) Write an expression for the surface area of tI over the region (u, v) E [0, 2] x [1,1]. (Do not evaluate the integral. ) 8. Find the fl ux of F(x , y , z) = ;ri  3yj
+ 2zk
across the surface z = x 3y + y2 on [ 1,1] x [0, 2].
9. Let r (t) = (t 2 , et , t + 3, 2t ), 0 ~ t ~ 1 b . a path in 1R4 and let F(w, x , y, z) Compute the line integral of F over the path r .
= (z , 3, w + y, 2x ).
10. A new amusement park is being built in town . Its perimeter has an elliptical shape and can be described by x 2 /9 + y2 /4 = 1. A roller coaster ride is to be built on top of the fe nce enclosing the park . Its height is IxYI. Use the path integral to compute the surface area of the park's fencing.
143
8
THE INTEGRAL THEOREMS OF VECTOR ANALYSIS
8. 1: GREEN'S THEOREM GOALS 1. Be able to state Green's t heorem .
2. Be able to use Green's theorem to compute a line integral or a double integral. 3. Be able to use Green 's theorem to fi nd an area. STUDY HINTS 1. Notati on. Recall that
a,
t he same symbol used for part ial derivatives, means boundary. Thus, aD is the boundary of D.
2. Green's the07'em. Under c rtain conditions, a line integral is converted into a double integral:
iL (~~ :)
L (PdX+QdY) =
dxdy .
Of en, one ide of the equation will be much easier to evaluat than the other side. 3. Required conditions .
(a) C must be a closed curve wit h an ou tside counterclockwise orientation. Interior "holes" have a clockwise orientation . In both cases, the region is on your left if you walk around C with the correct orientation. (b) P and Q must have continuou first derivatives. If Green's theorem does not apply di rec ly to a region, the region can usually be subdivided so th at the theorem can be applied. See figure 8.1.5 of the text . 4. Green's theorem and area. T he are of D m ay be computed by the formu la
A = ihD (xdyydX). Again, aD is oriented counterclockwise. This for ula is most useful if the boundary has a simple parametrization ; otherwise , double integration is usually si mpler. 5. Vector form. If F = Pi + Qj , then \7 x F G reen's theorem becomes
hD F . ds where ds
=
= (aQ / ax  ap/ ay)k,
so another formulation for
iL
(\7 x F ) . kdA,
= dx i + dyj.
6. Divergence theorem in the plane . If n is an outward unit normal to aD , then
hD F · nds
=
iL
div FdA.
Compare this with Gauss' divergence theorem in space studied later in Section 8.4.
144
CHAPTER 8
SOLU T IO N S TO SELECTED EXERCISES 2. According to Green's theorem, the area of a region is
A=
~laD(X dy  ydx) .
Since D is a d isk centered at (0,0) with radius R, the boundary aD can be parametrized by x
= R cos e
and
y = R sin B,
0
~
e ~ 27r .
T hen, dx =  R sin Bde, dy = R cos e de , and so the area is
A
1 f2 Tr
= 2" lo
1 f 2Tr [(Rc os B) (R cos e)  (Rsin B)(  Rsin e)] dB = 2" l o R2 dB
= 7r R2 ,
which is indeed the area that is calculated by using elementary geometry. 3. (b) We use the same parametrization for aD as in exercise 2. T he lefthand side ofthe identity in Green 's t heorem is
f (P dx + Qdy)
JaD
f (x + y) dx + f y dy Ja D JaD f2Tr f 2Tr Jo (R cose+ Rsin e)(RsinB)dB+ Jo (R sinB)(R cose) de f 2Tr Jo ( R2 cos Bsin B R2 sin 2 e + R2 sin e cos B) dB _ R2 1
2Tr
2Tr
sin 2 BdB = _ R2 1
0
= _ R2 7r.
1  cos 2B dB
0
2
By Green's theorem, the same integral should be equal to
OP) oX oy J'f (OQ
dx dy.
JD
Again, we use polar coordinates. D can be described by 0 ~ r ~ Rand 0 ~ B ~ 27r. Calculating = 1 and remembering that the J acobian for polar coordinates is r, the righthand side of Green 's t heorem becomes
oQ/ Ox = 0 and fJ P/ fJy
f 2Tr f R Jo Jo (01) rdr dB=
1 2
0
"
fR 2 Jo rdr dB=7r R .
Therefore, Green 's theorem is verified for this case, 5. Green 's theorem t ells us that the area is
A=
~ f
2 laD
(x dy  y dx).
From the given parametrization of the cycloid , we compute dx = a(1  cos B) dB and dy = a sin B dB. Remember th at t he x axis fo rms part of the boundary. This part of the boundary can be described by x = 27ra(1  u) and y = 0 fo r 0 ~ u ~ 1. Notice that our cycloid is traversed clockwise, t hat is, from left to right, starting at the origin along the cycloid, and t hen back along the 3.; axis from right to left. Since y = dy = 0, the desired area is
1 f 2Tr
"2 J
o
[a(B  sin (1) . asin BdB  a(1  cos B) . a(1 cos e) dB]
1
t
+ 2" J (x· 0 
o 2 2 2 2 2 ~ (a2B sin B a sin B a + 2a cos B a cos 2 B) dB
1 12Tr (a2 f) sin e 2a 2 
2 0
+ 2a 2 cos e) dB ,
0 . dx)
THE IN TEGRAL TH EOREMS OF VECTOR ANALYSIS
145
Using integration by parts, we get
To get t he correct answer, recall that t he direction should be counterclockwise. By simply changing the sign , we get the area spanned by one are of the cyeliod, 3a 2 '1T". 8. Let D be t he union of sim ple regions, Di , i = 1,2, 3, ... ,n, where each boundary, aD;, is oriented counterclockwise. Suppose P and Q are continuously differentiable on D. We want to show th at the boundary of D is the same as the boundary of all of the Di. Each intersecting boundary lies on the bou nd ary oftwo regions. W hen orientations are considered, contributions to the integral cancel ou t as in figure 8.1.5. Hence
{ (P dx
J& D
t.=1 J& {D, + = til .(~~  ~P) Jl (~~ ~:)
+ Q dy)
(P dx
Qdy )
.= 1
D,
Y
dx dy
dx dy.
11. (a) Using polar coordinates , we let x = r cos () and y = r sin (). Then D is described by 0 ~ r ~ 1 and 0 ~ () ~ 2'1T". We com pute that div F 1 + 1 2. (If necessary, you should review how to compu te a divergence in section 4.4.) Since the J acobian for polar coordinates is 1', we get
=
Jl
div FdA
=
= 127<11 (2)1' dr d() = 211" .
On the other hand , we know t hat the outward unit normal to t he unit circle is n::o: (cos (), sin ()), so
1
F . n ds
=
&D
27< (cos () , sin ()) . (cos () , sin ()) ds = 127< ds = 2'1T". 1 a a
(b) We want to compu te
{ F · nds ,
J&D
where F
= 2xyi 
y2j . By the divergence theorem, this integral is equal to
Jl
div F dA
=
Jl
(2y  2y ) dx dy
= O.
15. Since cos 2 () + sin 2 () = I , let x/a = cos() and y/ b = sin (), 0 ~ () ~ 211". Then x = acos(), dx =  asin () dO, y = bsin() and dy = bcos ()d(). By Green's t heorem , t he area of the ellipse is A =
~ lD (x dyY dx) "2112 a 7< [( a cos ())( b cos ())  (b sin ())( a sin ()) 1d()
! (21r ab d() = ab'lT". 2 Jo
=
16. In polar coordinates, x = l' cos B and y rsin (). Since r is a function of (), we get dx = (1" cos ()  l' sin ()) d() and dy = (1" sin () + r cos ()) d() (here 1" denotes the derivative of l' with
CHAPTER 8
146 respect t o e). Substituting into Green 's theorem , we get
~ [ (xdy ydx)
A
2Jc
~ f b(1' cos e)(r' sin e + 1'COS e) de  (r sin B)(r' cos e  rsine) de 2 Ja 1 fb ') . ') 1 fb 2 J ,.2 (cos~ e + sm~ e) de == 2 Ja 1'2 de . a 19. To fo rm one loop of the rose, we go from e = 0 to e == 7r /2. Using the result of exercise 16 , the a rea
IS
1
r/
2 Jo
r/ (9 sin 2 Jo
~
2
(3 sin 2e) 2 dB
~
2
Bu/an
f oou ds = jf
J8B
n
8B
2B)dB =
r/ (1+ cOS2 4e ) 2 Jo
~
2
de
(B+ Sin4e) 11f/2 = 97r
4 22 . Use the given definition of B = Bp:
2
4
0
8
and the divergence theorem in the pl ane for the regIOn
\7u . n ds
= J' f
\7 . (\71/,) dA
JB
= J'{ \7 2 u dA.
JB
26. (a) Suppose u (p ) is a maximum point on D. From exercise 25 , u (p) is the average value of u on a disk of rad ius R centered at p. T his is possible only if u (p ) = u( q) for all q in D . Indeed, if u (q) < u(p ) for some q in D , there must be a u(r ) > u(p ) to m aintain the average. Therefore, tI must be constant on some disk centered at p .
8. 2: STOKES ' TH EOREM GOALS 1. Be a ble to state and use Stokes ' theorem. 2. Be able to use Stokes ' theorem t o calculate a line integral on a closed curve or a surface integral over a surface wi th a closed curve as its boundary.
STUDY HINTS 1. R eview. Surface integrals are used in t his section . You should review section 7.6 if you have
forgott en how to calculate a surface integral.
2, R elation to Green's theorem. Like Green 's theorem , Stokes ' t heorem converts an integral in one dimension to an integral in two dimensions. Stokes ' theorem is a generalization of Green 's theorem . 3. Stokes ' th eorem for graphs. If z
= f(x , y) , then
Jis
curl F · dS
=
lasF .
ds.
This is a formula you should memorize. As with Green 's theorem , oS is oriented so that the surface is on your left as you walk upright around as, which must be a closed curve . 4. Genemlized surfaces. If the surface S is not the graph of a functio n, then Stokes ' theorem is still true if S can be descri bed using a onetoone parametriz ation. T he boundary of D, oD, gets m apped to as, so oD should h ave t he correct orientation. For an example when the orient ation becomes imp ortant , see the solution of exercise 5 of section 8. 1.
THE IN TEGRAL TH EOREMS OF VECTOR ANALYSIS
147
5 , Application, According to Stokes' theorem, to evaluate ffs curl F ' dS , we can change the surface 5 to any other surface with the same boundary 85, In most cases, we will change to
a planar surface or another simple surface, Imagine a loop of wire with an elastic sheet S on it , The surface integral of the curl of a vector field over any surface formed by a deformation of the elastic sheet will be equ al to t he line integral over the wire (assu ming the wire itself is not deformed), 6, Circulation, You should know t hat curl V, n is the circulation of V per unit area of surface
perpendicular to n ,
SOLUTIONS TO SELEC TED EXERCISES 1. By Stokes' theorem , we only need to evaluate
r F . ds,
Jas
where 8S is the circle x 2 + y2 = 1, Z = 0, the boundary of OUI surface . Parame trize the circle using polar coordinates, i.e., x = cos B, y = sinB, Z = 0, 0 ~ B ~ 27r. T hen the integral becomes
1 1 F · ds =
&S
2~
[sin 8( sin 8) cos 8(cos B)] d8= 
0
12~
d8 = 27r .
0
4. We want to show that the derivative of he magnetic flu x with respect to time is O. We begin
wit h
~
J'{
J'r
J' {
H . dS = 8H . dS = 8H . dS .
8t Js Js at Js at Th first step is justified because S is not a function of time. By Faraday's law ,
Jis
('v x E ) . dS .
Then by Stokes' theorem and the fact t.hat E· ds = 0, since E is perpendicular to t he boundary of 5, we get
 { E· ds = O.
Jas
5. The boundary of th surface is a closed curve, so we m ay take advan age of Stokes' theorem:
Jis T he bou ndary is the circle x 2
+ y2
(V' x F ) . dS:;:: la s F . ds .
= 1,
Z
{ F · ds
Jas
= O. The righthand side becomes
={
Jas
(x, y) . (dx, dy).
Now use polar coordin ates: Let x = cos 8, dx =  sin 8 d8; y = sin 8, dy = cos 8 d8; 0 ~ 8 ~ 27r . Substitution into the last integral gives us
r ( cos 8 sin 8 + sin Bcos 8) dB = Jo 21r
O.
9. Use Stokes' theorem . The boundary of S is the circle y2 + z2 = I , x = O. On the boundary, F becomes  i'j . Use polar coordinates : y = sinO, Z = cosB , 0 ~ f) ~ 27r. By Stokes ' theorem , we get
(HAPTER 8
148
12. The flow rate is JJs curl I) ·dS and by Stokes ' theorem, this is Ja s I) . ds. We have the boundary on the xy plane where z ~ O. Parametrize the boundary by
x = (R/4) cos O, Then the fl ux is
f2Tr R2
1
y
= (R/4) sinO . 11"R2
asl)·ds = io 16(sin B+cos B) dO=8·
14. By Stokes' theorem,
J1s
2
(V' x F) . dS
2
~ las F . ds.
J
Since F is perpendicular to the tangent to the boundary S, F·ds ~ 0. Hence, Js ('V' x F ) ·dS ~ O. If F is an electric field, this means the rate of change of magnetic flux is zero by Faraday's
law. See example 5 in the text . 18. The solu tion relies on the basic identities of vector analysis (see table in section 4.4) . Since C is a closed curve which is the boundary of a surface S, we have C = as in Stokes' theorem. T hus, Stokes' theorem tells us JJs (V' x F) . dS = Je F . ds . (a) Here, we use Stokes' theorem where F = IV' g. We need to show that \l x fV' 9 ~ V'I x V' g, or curl(JV'g ) = \llx\lg . Bybasicidentity10,we havecurl (lV'g) =/ curl(V'g )+\llx V'g, but by basic identity 11, curl( V' g) = 0 , and 80 we have curl (IV' g) = 1 (0) + V'I x V' 9 ~ V'I x V'g , as required. Al tern ati vely, we could have computed the curl of IV'g = I (ag/o x )i + I(og/oy)j + I (og/ oz) k and t hen used the equality of mixed partials since I and 9 are C 2 func tions. (b) Here, we use Stokes' theorem with F = I V' g+ gV' I. By basic identity 6, curl(fV' g+ gV' f) = curl(fV' g) + curl (g\l I ). Applying basic identity 10, we have
f cUrl(V'g) + V' I x \lg + I curl( V'g) + 9 curl (\lf)
curl (IV'9 + 9 V'f)
9 curl (V' f) + V'g x V'I
since V'I x V' 9 = (V' 9 x V' f) by properties of the cross product. According to basic identity 11, curl(\ll) = cu rl (V'g) = 0, and so curl(l\lg + gV'f) = O. By Stokes' theorem ,
J JIs o·
[(IV'9 + gV' f) . ds
is V' x (IV' 9 + 9V'/) . dS dS
~ O.
2l. First , we wili calcuiate JJs (V' xF ) .dS . Weco m pute \lxF ~ (1,1, 1). Alsol)r = (cosO , sin O,O) and 1) 0 = ( r sin 0, "cos 0, 1), so I)r x I) e = (sin e, cos 0, r ). Therefore , we get 1
Jis (1,1,1) . {Sin 8,  cOs8, r ) d7>dO = 1 1"/2 (sino coso+ r ) dBdr =
11
Cir)
dr = ~.
On the other hand , the boundary, aS, is composed of four parts. First, when r = 1, we have 1)(1, B) = (cos e, sin fJ, B), so F = (e, cos B, sin e) and ds = dl) (1, 8) = ( sin e, cos B, 1) dB . T herefore, Tr / 2 F· ds = (0, cos 8,8in B) . ( sin 0, cos B, 1) dO.
1
l
BS,
0
e sin B and the halfangle formula to integrate cos 2 8,
Using integration by parts to integrate we get
[(Bcos B When
e=
sine) +
. 20)  cosO]1"'0/2 ( "20+ SI:
11" 4
11"/2, orientation is maintained by letting r go from 1 to O. T hus, we have
f
iaS2
F . ds =
fO(~ ,o, r). (O,1, O)dr=O.
il
THE INTEGRAL THEOREMS OF VECTOR AN ALYSIS
149
W hen r = 0, B goes from 7r/2 to 0, so we get
[0 (B , 0,0)
[ F · ds =
las.
o.
. (0 , 0,1 ) dB =
lrr /2
Si milarly, when B = 0, we get [ F · ds = [l(O,,·, O).(l , O, O) dr =O .
las.
lo
Adding all the parts together, we get verified.
Ias F . ds
= 7r/4 + 0 + 0+0 = 7r/4, so theorem 6 is
25. For a direct computation , parametrize the sUl'face as follows: Le x = T' cos B and y = rsinB, so z = t(x2 + y2 ) = r 2 /2. Also , we want ~ z ~ 2, so 0 ~ ,. 2 / 2 ~ 2, or ~ r ~ 2. In addition, we have 0 ~ B ~ 210 . W calculate T o = ( 1' sin B,rcosB, O) and Tr = (cosB , in B, r), so the outward normal is T 8 X T r = (1'2cos e, r2 sin B, 1'). Also, we calcu la te
°
i \7 x F =
Finally,
j
°
k
a/ax a/ay a/az 3y  xz _ yz 2
(z 2 + x , 0 , z  3) 
(1  r
4
4
1 )
+ 1'cosB ' 0 1'2 3 . , 2
IIs(\7 x F) . dS becomes
2 r2rr
1lo o
(\7 x F) . (T o x T r ) dBd1'
12
(2 7r1· 3 + 67rr ) dr = ( 7r;4 + 37r7'2)
I1s (\7
I: = 207r .
Ias
On the other hand, by Stokes' theorem, x F ) . dS = F . ds. T he boundary is as, which is the circle ofradi us 2 in the plane z = 2. It can be parametrized by (2 cos t,  2sin t , 2) for ~ t ~ 27r. We use this orientation b cause the surface lies below the boundary, so we should traverse it in a clockwise orientation. We com pute ds = ( 2 sin t ,  2 cos t , 0) dt, so
°
[ F · ds
las
=
1
211'
( 6 sint,  4cost,8sint )· ( 2sint, 2 cost ,O) dt
r rr (12 sin t + 8 cos t ) dt 2
lo
[12
2
2
C
(~  Si: 2t ) + 8 (~ + Si:2t )]
= 207r.
If one choo es the other orientation, the answer should be  201T.
8.3: CONSERVATIVE FIELDS GOALS 1. Understand that the line integral of a gradient field is path independent. 2. Be able to determine whether a field is conservative.
CHAPTER 8
150
3. Given a conservative vector field , be able to fi nd a scalar function whose gradient is equal to that vector field .
STUDY HINTS 1. Theorem 7. T his is a very important theorem . To summarize, it states that if F is a gradient, then \l x F = 0, the line integral depends only on the endpoints, and all lin e integrals around closed curves are O. Also, if one these con ditions hold, then F is a gradient. If the conditions of the t heorem hold , then we say that t he line integrals are "independent of path." Note that if a single line integral is 0, then F may not necessarily be a gradient. 2 . Example 1. In method 2, part (a), notice how after integrating in x, we add a "constant" h 1 (y, z). It is a "constant" because it only involves the other variables. Study example 1
carefully. You should know how to use at least one of the two methods .
3. Gra di ents in 1R 2. The corollary precedi ng example 3 tells you that if the integrand in Green's theorem, oQ/ Bx  BP/By , is 0, then F = Pi + Qj is a gradient . Be careful!! F must be C 1 on all of ]R. 2, unlike theorem 7, which allows some exceptional points in 1R3. Exercise 12 el aborates this point . If the origin is an exceptional point in IR 2, then the integral x of F over t.he path on the left (a closed path) is 0 because the path does not include the origin, while the integral of F over the path on the right is not.
= O.
4. Is F a curl? F is a curl of some vector field if div F for fin ding a G such t hat F = curl G.
Exercise 16 explains the procedure
SOLU TIONS TO SELECTED EXERCISES 2. (a) Si nce y = 2X2, we have dy = 4x dx, and so
[
t
[1
1
8
19
lc F· ds = 10 (x · 2X2 ) dx + (2X 2 ) 2. 4xdx = 10 (2x 3 + 16x 6 )dx = 2" + 3' = 6' (b) The answer is yes, since
\7 x F
=I
j
k
B/ox a/ ay o/Bz xy
y2
0
1= (0 , 0, x) f O.
Alternatively, one can pick a different path and show t hat the line integral is not equal to T his would mean that t he line integral of F is path dependent. 3. If F
= \l f , then t he x component of F
must be
li.
of /ax, i. e.,
of/ox = 2xyz + sin x.
(1)
Similarly, the y component of F must be oj/ oy and the z component of F must be oj/ oz, % .e. ,
of/ay Bj/oz
x2z 2
x y.
(2) (3)
Integrating (1) with respect to x, we get f = f( 2xyz + sin x) dx = z 2yz  cos x + h(y, z), where h is a fu nction of y and z only. When we integrate with respect to x we should restore all the terms with x. The terms without x are treated as a constant when differentiated , and
THE INTEGRAL TH EOREMS OF VECTOR ANALYSIS
151
integration with respect to x cannot restore them. Similarly, integrate (2) with respect to y to get 1= x 2z dy = x 2z y + g( X, z),
J
where g(x, z) is a function of x and z only. Integrating (3) with respect to z gives us I = x 2y dz = x 2yz + k (x, V), where k(x , y) is a fu nction of x and y only. Compare the three results: I(x, y, z) = x 2yz  cos x + h(y, z ) = x 2yz + g(x, z) = x 2yz + k(x, V) .
f
We conclude (by insp ction) t hat g(x , z) = k( x, y) =  cos x + C and h(y, z) a constant. Thus, I(x , y, z) = x 2 yz  cos X + C.
= C,
where Cis
6. (a) This fact should already be familiar (see Example 6 in Section 2.6) . Here are t he details. We use the chain rule to compute (alox )(1 /J x2 + y2 + z2 ) = xl (x 2 + y2 + z2)3/2. Then, by sym metry, we get "V
(~)
1
= '\7 (
J x2
l'
+ y2 + Z2
)
=
= ~ = r
 (xi + yj + zk) (x 2 + y2 + z2)3/2
IIrl13
r3 .
(b) Since F is the gradient of a function j, fcF . ds = I (c(a))  I (c (b)) for all paths c(t) beginning at a and ending at b, unless the path passes t hrough the origin , where 1/r is not continuous. Thus , t he work done by F in moving a particle from ro "to 00" is 1 J x 2 + y2
+ z2
( ~)
_ lim
r
r+ oo
_ 1
..;x 2 + y2
+ z2 .
Notice th at this tells you it is more difficult to move a particle from a position near the origin. 9. T he reader should verify that '\7 x F = O. Hence, F is a gradient of some function I(x, y, z). Integrate the i component with respect to x, the j component with respect to y, and the k component wi h respect to z. Comparing the results, we see t hat I(x, y, z) = eX sin y + z 3 /3. Also, we compute c(O) = (0,0 , 1) and c(l ) = (1 , 1, e). Since F is a gradient, we have
1
c F· ds
= l(c (l))
e3
 l(c(O))
1
= esin 1 + 3  3'
=
13. (b) F is not the gradient of a scalar function f. If such j were to exist, then a I I ax xy and 2 2 of I 8y xv· By t h equality of mixed partials, we would expect that 8 I I 8x8y 8 I I 8y8x. But in this case, (818y)(8f!8x ) = x and (818x )(8118y) y.
=
=
=
15. (b) To show that F is conservative, we can show t hat it is a gradient . (We can also show that anyone of the four conditions in theorem 7 is met) . We will use the same method as in exercise 13. The partial derivatives are:
8 ( 2X ) 8y y2+1
=
(2 x )(  2y) 4xy (y2 +1)2 = (y2 +1)2 ;
1))
8 (2y(x 2 + 8x \ (y2 + 1)2
=
 4x y
(y2 + 1)2 '
Since these part ials are equal, F is conservative. Finding I so that F = "V I makes the evaluation of the path integral easy. You should verify t hat if I(x, y) = (x 2 + 1) /(y2 + 1) , then F = '\7 f. Then substitute x = t 3  1 and y = t 6  t to get I (t) = [(t 3  1)2 + 1]/[(t 6  t)2 + 1], and so
[
lc F· ds =
[(t3 _1)2+ 1]1 (t 6 _ t)2
+1
1
o::::;;l.
CH APTER 8
152 18 . First, we compute "V . F = (8/8x)xz + (8/oy)( yz) + (%z)y = z  z + 0 exists a G such that F = "V x G . To find G , use the result of exercise 16:
G1
=
Hence, G answer:
l
z
o
yt dt 
lY
t dt
_
y2 = yz2  , 2
0
2
= (_~)[(yz2 + y2)i + xz 2jJ.
"Vx G
21 I
G2
J a/oy yz2 + y2 xz 2 xzi  yzj + yk = F.
a/ax
=
lz 0
= O.
Thus, there
2
xt dt
 xz = , 2
and G 3
= O.
It is a good idea to compute curl G to verify your
k 8/8z
= ~l ( 2xzi + 2yzj + (z2 
z2  2y) k)
0
Note that G is not unique; for example, arbitrary constants may be added to each component of G, or any gradient may be added to G, and "V x G would still be equal to F . 23. (a) Recall from chapter 4 that F is not irrotational when curl F # O. Thus, we com pute curl F : J k "V x F = I 0/ ox 0/ 8y 8/ OZ I = 2k # O.
x
 y
0
(b) Let c(t) = (x(t), y(t), z(t)) be the trajectory of the cork. By the definition of flow lines, c/(t ) = F(c(t)), and so c/(t) = (y(t),x(t),O) . Equi valently, we have the following system of differential equations :
Xl (t) yl (t) Zl
(t)
(1) (2) (3)
y(t) x(t)
o.
Equati on (3) can be solved easily. Its solution is z(t) = constant. T aking one more deri vative of (1) with respect to t yields X"(t) _yl(t), but (2) implies that x"(t)  x (t), or x"(t) + x (t ) = O. T his equation represents a harmonic oscillator. T he famous solution to the differential equation x "(t) + x(t) = 0 is
=
=
x(t) =Asint+B cos t,
where A and B are cons t ants. If t he reader is un familiar with the techniques of solving ord in ary differential equations , the reader may verify this is indeed the solution by substitution. Similarly, y(t) = Csin t + Dcos t , where C and D are constants. Since x"(t) = y'(t), we differentiate x and y and compare terms to find that C = B and D = A. T herefore, y(t) = B sin t  A cos t. Squaring x and y and addi ng them gives us
x2
+ y2
A 2 sin 2 t + 2AB sin t cos t + B 2 cos 2 t + B2 sin 2 t 2A B sin t cos t + A2 cos 2 t = A2 + B2.
Since A2 + B2 is a constant, we recognize t he equation x 2 + y2 = A 2 + B 2 as the equation of a circle of radius VA2 + B2 centered at (0,0) . Thus, the cork has a ci rcu lar trajectory about the z axis in a plane parallel to the ~'y pl ane. (c) As y increases, x decreases, since xl(t) = yo We also kn ow t hat the cork is goin g in a circle. T hus t he cork is revolving counterclock wise. 24. (c) T he property of being rotational is a local property, t.h at is, the field is rotational at a poin t. In exercise 23, the cork whi rls while going around in a circle, but in exercise 24 , the cork does not. Single trajectories have little to do with the r otationalness of the fluid.
THE INTEGRAL TH EOR EMS OF VECTOR ANALYSIS
153
8.4: GAUSS' THEOREM GOALS 1. Be able to s ate and use Gauss' theorem.
2. Be able t o use Gauss ' t heorem to compute a double integral over a closed surface or a triple int egral over a volume enclosed by a surface.
STUDY HINTS 1. D efi nition. A clos ed surf ace is a surface wh ich, rough ly speaking, must be p unctured in order
to get into the region it encloses . The enclosed region is denoted Wand the closed surface is denoted oW. 2. Gauss} dw ergence theorem. If oW is a closed surface, then
fffw (diV F)
dV
= fhw(F.n ) dS .
T hus, a triple integral is reduced to a double integral , or vice versa. Compare th is to the divergence theorem in the plane (section 8.1) .
3. Physlcal interpretation. div F(P) is the net outward flow at the poi nt P per unit volume. If div F(P) > 0 material flo ws out from P, and if div F (P) < 0, material flows in towards P. If div F(P ) = 0, t he vector field is divergence free, that is, what goes in must come out .
SOLUTIONS TO SELECTED EXERCISES 4. (a) For the faces parallel to the yz plane, the outward pointing unit normal vectors are j and
 i, respectively. For those two faces , we have
fiaw F'dS
1
1
= 1 1 \ i + j + k) .idY dZ + 1 1 \i + j +k).(  i)dy dz
111\
i
+ j + k) . (i 
i) dy dz
= o.
T he outward pointing unit normal vectors n for any two parallel faces are the exact opposite, so the integrals over any two parallel faces of the cube cancel. Therefore, the int gral is O. Next , we calculate that div F = 0, so by the divergence theorem, t he desired integral is
fffwdiVFdV =O. 6. (b) First , we see th at div F
= 1 + 1 + 1 = 3, so
flaw
F . dS =
by the di vergence theorem ,
1ffw div F
dV = 3
fffw
dV,
which is t hree times the vol ume of W. T he problem now is to find th volume of the region of interest. Use "cylindrical" coordinates. We will let f) range fro m 1r / 2 t o 7r / 2 since x ~ O. In addition , we have :S r :S 1 and x 2 + y2 = 1'2 :S Z :S 1. Therefore,
°
fli
W
dV
=
J,"/2 1111r dz dr de = 1f' /
2 0
r 2
7r
11
r (l  1'2) dr
0
Thus,
flaw F· dS = 3 (~) = 3:.
= 7r
(1 1) 
2

4
7r
4
CHAPTER 8
154
= (x 2 + y2)2, so
10 . Use the divergence theorem . We have div F
fhsF . dS = f fis (x2 + y2)2 dV. Using cylindrical coordinates, the cylinder S can be described by 0 ::;: 1. Since the Jacobian is r, we get
T ::;:
1, 0 ::;: B ::;: 21!', and
o::;: z ::;:
fhsF . dS = 121<1110
1 T'
4
r drdz dB
= 21!'
11 5
r dr
=
i'
16. From section 8. 3, we know that curl F = 0 implies that F = \7 f, so t aking the divergence of both sides gives us div F = \7 . (\7 J) \7 2 f. But we are given th at div F = O. Hence, 2 \7 f = O.
=
18. By the divergence theorem , fffw div F dV = f fow F . dS . G iven th at F is t angent to the surface, we know t hat F is perpendicular to t he unit normal of the surface S, and so F · dS = O. Hence, di v F dV = O.
IIIw
8. 5: S OM E DIFFERENTIAL EQ UATIO NS O F MAT H E MATICAL P HYSI CS GO ALS 1. Be able to use vector an alysis techniques in applications such as conservation laws , heat
t ransfer and electrom agnetism .
STUDY HINTS 1. N otation. There 's a lot of new notation , so you m ay find it useful to make a list of new symbols and t heir definitions. 2. Old concepts. No new vector calculus material is presented in this section . Only the term inol ogy has changed . You should be able to justify each step with the knowledge t hat you have acquired in the previous ch apters.
S OL U T I ONS T O SELE C TED EXERCI SES 3. Using t he tr ansport equation for p, we have
:t f Jfw
p(x, y , z, t ) dx dy dz
=
fJfw (:
+ pdiv
Jf /w ( ~: + Jf fw(~: +
F)
\7 P . F
dx dy dz
+ pdiv
F) dx dy dz
\7 . (PF )) dx dy dz
o for any HI in R3. Thus, the integrand must be identically 0, i.e.,
at + \7 . (pF) = 0,
ap
which is t he law of conservation of mass. 8. Given \7 2 rjJ = 0 and V = \7rjJ , we want to show that p(aV lOt + V· \7V ) =  \7p if av lot = O. Using exercise 7, V . 'VV = ! 'V (II V I1 2) + ('V x V ) x V. Since V is a gradient , 'V x V = O. It suffices to show that p [ ~\7( I IV I 1 2) l =  'Vp. Choose p = (pI2) IIV W.
l55
THE INTEGRAL THEOREMS OF VECTOR ANALYSIS
9. Start with Ampere's law: \7 x H  fJE / fJt = J . Take the divergence of both sides: div J = \7 . (\7 x H )  \7 . (8j&t )E = 0  (fJ / fJt )(\7 . E ) (the change of order of t he two derivatives is j ustified since \7 takes the derivatives of spatial variables only). Now , by Gauss' law, \7 . E = p. Combining the two results, we get div J =  op/ at , or div J + fJp/ fJt = 0, which is the equation of continuity. 10. (a) By definition , the integral of the Poyn ting vector field is
By Gauss' theorem, we may rewrite the righthand side as
JJ1
\7 . (E x H) dV.
Equation 8 from the table of basic identities of vector analysis in section 4.4 tells us that div(F x G ) = G . curl F  F . curl G . Thus, the integrand above is div(E x H ) = H . curl E  E· curl H H . (\7 x E )  E . (\7 x H )
V · (E x H )
H · O  E· J = E· J since
x E = 0 and V x H = J. Therefore,
Jis
p . dS =
JJ1
\7. (E x H ) dV = 
JJ1
E . J dV.
8.6: D IFFERENTIAL FORMS GOALS 1. Be able to add and wedge multiply kforms.
2. Be able to integrate and differentiate kforms. 3. Be able to tate Green s, Stokes' and Gauss' theorems in terms of kforms.
STUDY HI TS 1. Notation . In this textbook, w is used for Iforms,
7}
is used for 2forms and
3forms. (This is generally the case, but not always .)
2. Definition . (a) Oforms are scalar functions. (b) lforms contain dx, dy, or dz . (c) 2forms contain dx dy , dy dz, or dz dx in the order specified . (d) 3forms contain t he expression dx dy dz .
These definitions will differ slightly in higher dimensions.
IJ
is used fo r
CHAPTER 8
156 3. Integra tion.
(a) Iforms are integrated like line integrals. (b) 2forms are integrated like surface integrals. Notice that each term of the double integral is equ a.tion (2 ) contains a Jacobian. The variables in the J acobian's "numerator" follow the cyclic nature of the differentials. (c) 3forms are integrated li ke ordinary triple integral s. 4. W edge produ cts. Different forms may be multiplied. In our discussion , you can only multiply
a kform and an Iform if 0 ~ k + I :::; 3. Otherwise, products are zero. Be aware that w /\ J1 = (1 )101 (I' /\ w) . This property is called anticommutativity. Notice that dx /\ dx dy /\ dy = dz /\ dz = O. The other properties are similar to t hose of real number multiplication.
=
5. Differen tiation. If f is a Oform, df is like the ordinary derivative. The "sum" rule (linearity) remains the same as in onevariable calculus. The "product" rule differs slightly and d(dw) = 0; tbis last ident ity captures the two identities 'V x 'V f = 0 and 'V . ('V x F ) = 0 in one formula.
SOLUTIONS T O SELECTED EXERCISES l. (d) Since this is a 2form wedged with a I form , i. e., a 3form, the product is a (2
+ I)form.
By the distribut ive property, we get w/\ry
=
=
(xy dy dz + x 2 dxdy )/\(dx + dz ) xy dy dz /\ dx + x 2 dx dy /\ dx + x y dy dz /\ dz + x 2 dx dy /\ dz x y(dy dz /\ dx ) + x 2 (dx dy /\ dx ) + xy(dy dz /\ dz) + x 2(dx dy /\ dz) xy dx dy dz + x 2 dx dydz = (xy + x 2) dx dydz .
We used the anticommutativity fact that dydz /\ dx = (I)2( dx /\ dydz) . Also, we used the associativity property along with the identities dx /\ dx = dz /\ dz = O. 3. (b) When we differentiate this Iform, we get a 2form: dw =
= =
[d(y2 cos x) /\ dy + (_I)0(y2 cos x ) /\ d( dy )] + [d(x y) /\ dx + (l)o(xy) /\ d(dx)] + d(dz) [(_ y2 sin x dx + 2y cos xdy) /\ dy + 0] + [(y dx + xdy) /\ dx + OJ + 0 _ y2 sin x dx dy  x dx dy = (_ y2 sin x  x ) dx dy
since dx /\ dx = dy /\ dy = 0 and dy /\ dx =  dx dy. (e) When we differentiate this 2form, we get a 3form : dw
= d[ ( X2 + y2 ) dyJ/\ dz +(1)1( x2 + ~ ) dy /\d( dz )
= [d( x 2 + ~ ) /\ dy + (_1)0 (x 2 + y2) /\ d( dy)]/\ dz + 0
= [( 2x dx + 2y dy ) /\ dy]/\ dz = [2x dx /\ dy + 2y dy /\ dy]/\ dz = 2x dx dy dz
= d(dy) = dy /\ dy = O. 4. Notice t hat dx /\ dx dy = dy /\ dx dy = dy /\ dy dz = dz /\ dy dz = dz /\ dz dx = dx /\ dz dx = O. since d(dz)
The derivative of F dx dy is
oF of dy + 0OFdz ] /\ dx dy + (1) ° F /\ d( dx dy) [8 x dx + ;:;ay z of of of of ox dx /\ dx dy + oy dy /\ dx dy + OZ dz /\ dx dy == o z dz /\ dx dy of of oz dz /\ (d:c /\ dy ) = (dz/\dx )/\ dy
az of of  7h (d:c /\ dz ) /\ dy =  7h dx /\ (dz /\ dy ) of
of
8z dx /\ (dy /\ dz ) = fu dx dydz .
THE INTEGRAL TH EO REMS OF VECTOR ANALYSIS
157
Si milarly the derivative of G dy dz is (oG j ax) dx dy dz and the derivative of H dz dx (oH j oy) dx dydz . Therefore,
OG
dl1= ( ox
8H OF ) + ay + Tz
IS
.
dx dydz= (d1v V)dx dy dz .
8. By a direct calculation, w note that as is the unit circle in the xy plane. The parametrization is (x, y, z) = (cos t, sin t , 0), 0 S; t S; 27r. T herefore,
is
= 12Tr [(cos t + sin t)O + (sin t + 0)( 
w
r = inr ( sin 2 t + cos2 t) dt = io 2Tr
sin t dt ) + (cos t
2Tr
cos 2t dt
+ O)(cos t dt)]
= S1;' 2tl2Tr 0 = O.
By Stokes theorem, the above calculation is equal to ffs dw . We compute dw:
=
d
= because d(dx)
[d(x + y) !\ dz + (x + y) !\ d(dz)] + [d(y + z) !\ dx + (y + z) [d(x + z ) !\ dy + (x + z ) !\ d( dy)] [(dx + dy ) !\ dz ) + [(dy + dz) !\ dx ) + [(dx + dz) !\ dy)
= d(dy) = d( dz ) = O.
dx A dz
= 0,
d(dx)]
This simplifies to
+ dy !\ dz + dy !\ dx + dz !\ dx + dx !\ dy + dz !\ dy = dz dx + dy dz  dx dy + dz dx + dx dy 
Therefore, we have ffs dw
!\
dy dz = O.
aLso.
l! . By Stokes theorem, we have fJaR T}
= f f f R d1J.
We will look at the righthand side:
d(x dy dz + y dz dx + z dx dy) = d[(x dy Adz) + (y dz !\ dx ) + (z dx A dy)) = [d(x dy)!\ dz + (l) l (X dy A d{dz ))] + [d(ydz ) !\ dx + (1/ (y dz!\ d(dx))) +[d(zdx)!\ dy + (1)1( z dx !\ d(dy))] = [dx A dy + x!\ d(dy ))!\ dz + [dy !\ dz + y!\ d(dz )) !\ dx + [dz A dx + z!\ d(dx))!\ dy. Since d(dx)
= d(dy) = d(dz ) = 0, we get (dx A dy)
!\
dz
+ (d y !\ dz ) !\ dx + (dz !\ dx)!\ dy =
=
=
dx dy dz + dx !\ (dy !\ dz ) + (dx !\ dz ) !\ dy dx dyd z + dxdydz  dx !\ (dz Ady) 3 dx dydz.
Thus, the righthand side is ffJR3dxdyd z. We recognize f ffRdx dy dz as the volume of R. Dividing by 3 gives us the desi red result. SOLUTIONS TO SELECTED REVIEW EXERCISES FOR CHAPTER 8 2. By Gauss' t heorem, we have ffaw H· dS = fffw C'v· H) dV = fffw( div H) dV, so we need to how that div[F x (\7 x G )] = (\7 x F ) . (\7 x G)  F · (\7 x \7 x G). By formula 8 in the table of basic identities of vector analysis in section 4.4, we have div(A x B ) = B . curl A  A . curl B = B· (\7 x A)  A · (\7 x B). Now let A
=F
and B
= \7 x G . This gives the desired result since a . b = b· a.
CHAPTER 8
158
4. According to Green's theorem, Ie P dx + Q dy = IID( 8Q/o x  {)P/fJy) dx dy, where C is the boun dary of D and C has a counterclockwise orientation. W ith P(x, y) = x 2 y and Q (x, y) = y, the rightband side is easily computed as
fL(0 
x
2 )
111:(
=
dx dy
_x
2
)
dy dx =
11 (  X2yl:=x3)
1\x + x ) dx= (_;4+~6)1: 3
1
11
t 2 (t 3 )(dt)
1
5
For the lefthand side , first param etrize the curve C 1 T hen x 2 y dx+ y dy=
dx
:
= x3
Y
+ (t 3)(3t 2 dt ) =
12
= t, y = t 3 ,
as x
11
6
4t 5 dt =4t 11
0 ~ t ~ l.
2 3
6 0 For the curve C 2 : Y = x, a counterclockwise parametrization is x = 1  t, Y = 1  t , 0 Then C, O D
[ x 2y dx
le.
~
t
~
l.
+y dy = ~O( [(1  t)2(1  t)( dt) +(1 t)( dt)l = lot [(1  t)3 +(1  t)l( dt).
Substitu ting u = 1  t yields
(u4+ '2u 4
3
[ 0
I I (u +u) du =
Thus, the lefth and side is ~  ~
=  112 '
0
2
)
1
1
3
4'
and G reen 's theorem is verified for this case.
7. (a) To show that F is conservative, the easiest. thing to do is to show that Y' x F
a/ax
Y'x F
6x y(cos z) ( 3x 2 sin
= 0:
k
J %y 3x 2 (cosz)
%z  3x 2 y(sin z)
z + 3x 2 sin z, 5xysin z 
6x y sin z , 5xycos z  6xycos z)
O.
(b) If F is the gradient of some f (x, y, z), then /(x, y, z) must sat isfy
o//ox of/oy of/oz.
6xycosz 3x 2 cosz 3x 2 ysin z
(1) (2) (3)
Integration of (3) with respect to z gives f( x, y, z) = 3x 2 ycos Z + g(x, V); integration of (2) with respect to y gi ves f (x,y , z ) = 3x 2 y cos z + h(x,z); integration of (1) with respect to x gives f( x , y,z) = 3x 2 ycosz + k (y,z). Comparing these three f's, we can see that f( x , y ,z ) = 3x 2 y cos z + C . (c) Since F is a gradient, we only need to evaluate f at the endpoints to calcula.te the line integral . We get
1
F · ds
= f (c(1r/2))
 f(c (O))
=
3(cos 3 tI)2(sin 3 tI)
c
9. We want to com pute
f ffw
IIIw V' . F dV Y' . F dV
n/2
I0
= (J.
where W is the unit cube. By Gauss' theorem,
t t ~o( = lolo
(5z
1111 (~ +
+ x 2 + y) dx dy dz 5z
+ y)
1 (1+ +1) 1
o
3
6z
2
dy dz
dz =
(1 ++ 1 ) 23 3
2
3
=. 6
159
THE IN TEGRAL THEOREMS OF VECTOR ANALYSIS 13. (a) We compute
'\7 1 = (al/ax)i + (al/ ay)j + (af/az)k = 3y exp(z2 )i + 3x exp (z2)j + 6xyz exp (z2 )k.
(b) '\71 i a gradient, so the integral is l (c(11"))  l (c( O)). Since I (x , y, z) = 3xyexp(z2 ), we have I (c(t)) 3(3cos3 t)( in 2 t ) exp (e2t ) ; thus l (c(11")) 0 and l(c(O)) O. Th refore,
=
Ie '\71 . ds = O.
=
=
(c) Let A be the region whose boundary is c(t). Then, by Stokes' theorem ,
1
'\71 · ds =
1
['\7 x ('\7 f)] . dS.
This is 0 becau_ the uri of a gradient is always OJ the oth r part of Stokes' theorem has been illustrated in part (b).
=
IID(3x 2 + 3y2 ) dxdy, where D is the unit disc. Now 14. By Green' heorem, Ie x 3 dy  ~ dx use polar coordinates : x = r cos 8, y r sin 8. The unit circle is described by 0 ~ r ~ 1 and o ~ 8 ~ 2 . ince the Jacobian is r, the integral b omes
=
12"1 o
17. By Green' Area
1
3r 2
r
.
dr d8 = 211"
0
11
3r 3 dr = 211" . 3
4
0
3 . = 11" 2
eo em .
Jl
=
=
~
dz dy
1·
=~
l
(x dy  y dx)
_ = 1 a2 fo ( 2sin 2 8cos 2 8  sm 2 ()cos 2 ()
Jo
= ~Q2 f 2 Jo ;~
=
Q
?
2
~
(a sin 8 cos 8)(2asin 8 cos 8) d8 
in 2 O(cos 2
()
+ sin 2 () ) d8 =
1 C~os 2())
1'fr
(a sm 2 8)[a (cos 2 8  sin 2 8)] de
+ sin 4 8) de 2
a
2
r sin 2 8 d8
Jo
d()
I:
(~  Si~2()) = 11":2.
20. (c) If F ' a gradient, then curl F
curl F
=
= O.
, I
. J
We compute
a/ax a/ ay 2x~ x 2 z 3
Excep a the origin, curl F is not 0 , so F is not a gradient.
TEST FOR CHAPTER 8 1. True or false. If false, explain why. (a) For any region to which Gre n's theorem applies and if C is traversed counterclo kwise, the line integral Ie (4x  sin x + eX + 2y) dx + (4x + cos y + 2Y ) dy has positive value. (b) The vector field F (z. y z) = (y 2 e· cos z)i + (2xye Z cosz is a conservative vector field.
+ 2ycos(y2))j + (xy2 ez cos z  xy 2 ez sin z)k
CH APT ER 8
160 (c) If F is a gradient, then curl F (d) The line integral
= O.
Ie P dx + Q dy is independent
of path if a P / o y
= oQ / ax on all of lR 2.
(e) If y = l (x ) is a posi tive function on [0 , 1] , then th e area under the curve is
~
11
[t (J' (t) dt )  I(t) dt].
2. Let C be the perimeter of a trapezoid with vertices at (0 , 0), (4, 0) , (4, 1) and (2,1). IT C is traversed in t he counterclockwise direction, starting and ending at the origin, calculate
fc
(2x
+ x3 
Y + sin x ) dx
+ (eY 
x  tan y + Vii) dy.
'il l for I (x , y, z ) = 2X 2 Z 2 sin ycos (e (b) Let F = 'il l as in part (a). Com pute Ic F. ds where C is t he portion of t he pl ane X
3. (a) Compute 2x
+ y + Z = 3 that
lies inside the sphere x 2
) .
+ y 2 + Z2 = 16.
4. Let F (x , y, z ) = (3x + ysin z, x 2 + y + ze x , xe:ll y + z ). Let P be t he parallelepiped spanned by the vectors (1, 2, 1), (3, 0,1) and ( 1,3, 2) and having one vertex at t he origin . Compute the i.ntegral of F over oP. 5. (a) Compute curl F for F( x , y, z) =  2yi (b) For F as in part (a), compute
+ 2xj + xyz 3 k.
IIs curl F· dS where S is the semiellipsoid x2
2)1/ 2
z= ( l    !L 4 9
.
6. A curve is described parametrically by x = sin t , y = sin 2t , 0 ~ t ~ 1r. F ind the area enclosed by t he curve. (Hint: sin(m x) cos(n x) = ~ [sin (m  n)x + sin(m + n)x].) 7. A solid W is described by 1 ~ x ~ 1, 0 ~ y ~ 4  x , 0 ~ z ~ 10  x 2 of F( x , y, z ) = (y + z2 )i + (x y  sin z )j + (5xe!l )k across oW . 8. Suppose t hat
Iis curl F · dS = 5 for a given F
and S is the cone y

y.
Compute the fl ux
= 4  x2 
Z2 ,
0 ~ Y ~ 4.
T is the semiellipsoid y = 5(1  x 2 /4 z2 /4)1/2 , 0 ~ Y ~ 5.
(a) Compu te
f IT curl F ·dS , where
(b) Com pu te
Ilu curl F . dS , where
U is t he paraboloid (y
+ 2)2 = x2 + z2, 
2 ~ y ~ O.
9. For a closed curve C that is traversed in a counterclockwise di rection, it has been determined that Ie y dx  x dy is 6 . C alculate Ie(x + 2y) dx + (5x  ye!l) dy. 10. (a) Show that F(x , y , z ) = {2xyz3 + z)i + (X2Z3 + l)j + (3x 2yz2 + x )k is a conservative vector field . (b) Find a fUDction I such t hat
'ill = F .
(c) Suppose the same force field F as in (a) causes a ba by to m ake a random crawl fro m (0, 0, 0) to (5, e, 0) where she knocks over a priceless obj ect. How much work is done by the force field on the baby? (d) T he force field F then causes t he baby to crawl up some cabinet drawers to (I, 1, 1). How much work was done by the force field in taking the baby from (0,0 , 0) to (1 , 1, I) ?
161
9
SAMPLE EXAMS
9.1: COMP REHENSIVE TEST FOR CHAPTERS 1  4 1.
(a) Find the relative extrema of f(x, y)
= x2 
2x + y2  1.
(b) For the same t, find the relative extrema on the curve x 2 + y2 = 1. 2. (a) Fido is on a strange mo untain which has a shape described by M (x,y) = eY xsiny. If Fido is at (1,1':/ 2, e rr / 2 ) , what is the d irection Fido must take to get down the mountain as fast as possible? (b) Fido has to follow a certain trail of tree trunks (which he has previously m arked) along e(i) = (2t , eexp ( t 2 ), 2tsin(t 2 )) . How fast is Fido's alt it ude changing at t = I? 3. (a) Given the vector field F(x, y) taking a good guess.
= (y, x) , find
an expression for a flow line c(t) for F by
(b) Do the same fo r F (x, y) = (2x ,  y). 4.
(a) Find the arc length ofe (t)
= (2t,t2), O:'S t:'S l.
(b) Find the arc I ngth of e(t)
= (2sin t,sin 2 t ), 0 :'S t:'S 1': / 2.
(c) Explain your answers. 5. A wooden box is to be m ade with $120.00 worth of wood . The lid is to be made from wood that costs $2 .00 per square inch, and the rest of the box is to be made from wood that costs $2.50 per square inch . What is t he biggest box that could be made? 6.
(a) Find an equation of the line of intersection ofthe planes 3x + 2y z
= 7 and x 4y+ 2z =
O.
(b) Find the equation of the plane wh ich contains the points (1 ,3, 2) and (0,  2, 1) and is perpend icular to the plane 3x  y  2z = 5. 7. Let t: B C ~ 2 t ~3 be defined by (u,v) t (uv 2 ,v 3 defined by (x , y, z) t (yz e"', y3 cosx z).

u,v sinu), and g: B C IR3 t IR 2 be
(a) Calculate D(J 0 g)(O , 1, 0). (b) Calculate D (g 0 f )(O, 1) . 8. An aquarium m anufacturer bas advertised that its tanks hold exactly 0.49 m 3 of water . Due to production error , t he tanks have di mension 1.01 m x 0.72 m x 0.67 m rather than the specified 1.00 m x 0.70 m x 0.70 m. Use the linear approximation to estim ate t he error in the advertised capacity of t he tank . 9. A budding bioengineering st udent who is an amateur ornithologist wanted to fly sou th for Christmas vacation, but couldn 't afford an airline ticket. He knew that his pet swallows also wan ed to fly south for t he winter, so his plan was to tie lis feathered friends to a hang glider and have them fly him home. His welltrained swallows had a veloci ty vector of 3i + 4j + k (km/ hr) until they r ached their cruising altitude at (0, 0, ~ ). At that point, they continued along the same path, but with no change in altitude, (Assume the Earth is flat. ) (a) What was the original starting point at (x, y, O)? (b) How long did it take to reach cruising altitude?
CHAPTER 9
162
(c) When they reached (0, 0, ~ ), a strong wind with veloci ty 2i + j affected the flight. What was the swallow's veloci ty vector to maintain a total velocity of 3i + 4j? (d) After 3 hours, the bioengineer calculates that they will pass over his worst enemy's horne, where some of his pets will give a 21bird dropping salute. What is the position of the enemy's house? (e) Assuming that the swallows were unable to compensate for the wind, how far from the target point would they be after 3 hours? 10. Tightwad Terry, the miser , has performed a computer analysis of his earning power. He has determined that his earning fun ction is $(t, u , v, w) = t 2 e ti + v sin w .
(a) Compute $t, $u , $t1 and $w · (b) Assuming T ightwad Terry wants to in crease his earnings as fast as possible and he was at (1 , 0,2, 0) on his $ graph, how should he change t, u, v and w? (c) If he has exactly one unit of each resource (represented by t, u, v and w), what. is the largest increase in earning th at he can expect? Note that he can use fractional amounts of resources, such as of t and ~ of w.
t
9.2: COMPREHENSIVE TEST FOR CHAPTERS 5  8 1. (a) Evaluate
Jin
3(x
+ y)e(xv)
over the region bounded by the lines x + y = 1, x (b) Evaluate
dx dy
+Y=
1, x  y = 1 and x  y =
~ l.
1111
siny2 dydx .
2. A block has a slanted top described by x + y + z = 2. Its edges are perpendicular to the xy plane, and the botto m of t he block is formed by the triangle with vertices (1,0,0), (0, 1,0) and (0 , 1,0). What is the volume of this block?
3. Find t he m ass of a wall described by o(x, y) = 21yl + 3. 4. Let F (x , y)
a$
y $ _x 2

2x
+ 3,
 3 $ x $ 3, having cJiensity
= (eX cos 3y,  3ex sin 3y).
(a) Find an f( x,y) such that (b) Evaluate
Vf =
F for aJi (x,y).
IeF . ds for the path c(t ) = (cos t, sin t), °$ t $
211'.
(c) Compute V x F. (You should be able to do parts (b) and (c) in your head.) 5. (a) Suppose information flows through an idiot's skull (a surface descri bed by x 2 + y2 + 2z2 = 1) at the velocity 2xi + 3yj + zk. After brai n surgery, his mW1bed skull, which remained the same shape, receives infor mation at the velocity 5xi + yj + 3zk. Thinkrng of this as a fl ux of information through a surface, did the surgery help? Explain . (b) A particle moves in a path c(t ) = (2t, 3t , t ) in a force field F work done in the ti me interval 0 $ t $ 5?
= (2x, 2y, 4z).
What is the
6. (a) A contact lens can be described as a cap of a sphere of radius R cut ou t by a cone of angle 1T / 4. Find the surface area of the lens. (Hint: Set up the lens in a correct and convenient coordinate system. )
163
SAM PLE EXAMS (b) Slice a sphere anywhere with 2 parallel plane which are separated by a fixed di tance d. The bands obtained always have the sa me surface area. Prove this. What is the area? (Hint : Use spherical coordinates.)
7. (a) Find the volume enclosed by the surface
a;2
+ 9y2 + z2 
2x
+ 54y 
10z
+ 106 = O.
(b) F ind an expression for the surface area of the surface above. 8. Which of the following vector fields are conservative?
(a) F(x, y, z ) = (x 2
+ l/ x , In(y + 1) , z)  3x
(b) F (x, y, z) = ( (x 2
+ y2 + z2)3/2'
3 y
(x2
+ y2 + z2)3/2'
3Z) + z2)3/2
(x 2 + y2
(c) F( x, y, z) = (3yz , 2xz , 5xy) 9. Complete t he following st atement: A vector field F is conservative if _ _.
(i) T here is a vector field G such that F
= curl G.
(ii) T here is a scalar function fsu ch that \7 f = F. (iii) div F = O. (a) (i) only
(b) (i) and (iii)
(c) (ii) only
(d) (ii) and (iii).
10 . Consider the follo wing argument: given a vector field F,
by Stokes' theorem but
JJ.}wf
\7. (\7 x F ) . dS = {
}s=&W
(\7 x F ) . dS
by Gauss' t heorem, so everyt hing is 0 because the divergence of the curl of any vector field is O. W hat's wrong?
9.3: C OMPREHENSIVE T EST A FOR C HAPTERS 1  8 l. If true, state true. If false, expl ain why.
Ie
(a) The path integral 27l' ds equals the surface area of a cylinder of radius 1 and height 27l' ifc (t) = (cost,sin t,t ), 0 ~ t ~ 27l'. (b) A continuous fu n tion is a differentiable fu nction .
(c) The line integral of a mass density function along a curve is the mass of the curve.
= \7 . F . f /8 x 8z = 8 2f/8 zax for all continuous functions
(d) curl F (e) 8
2
f( x,y,z ).
CHAPTER 9
164 2. M ultiple choice. Choose the correct a nswer.
(1) Which pair of vectors have the smallest angle between them? (a) i  j , j + 2k (b) 3i + 2k,  2j (c) 2i  j + k, i  j
+k
(d) T here is insufficient informat ion.
(2) Green 's theorem requires (i) continuous fi rst partial deri vat ives (ii) continuous functions (iii) any closed curve which is the boundary of a region (a) (i) only
(c) (ii) and (iii)
(b) (ii) onl
(d) (i) and (iii) .
(3) Let !(x , y, z) = z2 xe x cos(yz) . A particle t ravels along a path c from (0,  2/ 5, 1T/4) to (3,5, 1T /2) . If the force acting on t he particle is 'V I, then t he work done on the particle IS
(a) negat ive (b) zero (c) positive (d) unknown since c is unknown
(4) For any vectors u and v , U· (u x v) =
(i) v· (u x v )
°
(ii) (iii) (v x u ) . u (a) (i) only 3. Let F = xi
+ yk.
(b) (ii) only Compute
(d) (i) , (ii) and (iii ).
(c) (i) and (ii)
IIs curl F . dS for the following surfaces S:
+ y2 + (z  3)2 = 1, z ~ 3 2 (b) 9x + 9y2 + z2 = 9, z ~
(a) x 2
°
4. Consider the system 2
F1(u, v, x, y)
u 2u
F2(U, v, x, y) (a) Show that one cannot solve for (b) Show th at
tl
2
v + 2x + x y = + 3v  5x 2 = 0 
°
and v in terms of x and yat (u, v) = (0 , 0).
au/ax exists at (u , v, x , y) = (3, 2, 1 , 0). Compute it .
5. Suppose g(u,v) = (uv 2 ,u+2v,v) . At (x, y, z ) = (0,0,0), (u ,v) = (1,2) and
D! = Let h = 9 0
f.
[11 0] 2
3
1
.
°~
u ~ 2,
Wh at is Dh (O, 0, o)?
6. A surface is parametrized by x = u
2
,
Y= v
2
,
Z
= uv ,
°~
v ~ 2.
165
SAMPLE EXAMS
= (1, 1) as a function of x and
(a) Find the tangent plane at (u, v)
(b) When u = 1, find the arc length of the curve in space for 0 7. Let
~
~
y.
v ~ 2.
be the surface parametrized by U
x:;:;;: e cosv,
y
=e
(a) F ind the area of the surface
U
sin v,
z
= v,
0 ~ u ~ 1, 0 ~ v ~ 1r/2.
~.
(b) Compute the average of z over
~.
8. Letf(x,y)=x 2 3x+yy2+5.
(a) At all critical points (xo, Yo), express (xo, Yo, f( Xo, Yo» in spherical coordinates. (b) Find the extrema of f on the circle x 2 9. A surface is described by the equation z =
+ y2 = ~. 2y + cos 1rX  ft.
(a) Find the directional derivative in the direction of i
Consider the point (4, 1, 1).
+ 2j.
(b) F ind the equation of the tangent plane. 10. A region in space lies between the graphs of z also lies inside the cylinder x 2 + y2 4.
=
= 16 + x 2 + y2 and z = J x 2 + y2.
The region
(a) Express the volume of the region as a triple integral with cartesian coordinates. (b) Rewrite your answer in (a) with cylindrical coordinates. (c) Find t he volu me of the region .
9.4: COMPREHENS IVE T ES T B FOR CHAPTERS 1  8 1. P hysical and geometric interpretations.
= O? x v = O?
(a) What do you know about u and v if u . v (b) What do you know about u and v if u
(c) What physical interpretation is associated with a negative divergence? (d) Give a phy ical interpretation of curl V. 2. Consider the point (1, 1,1 ,1) of the function f(x, y, z, t) = x 2 y + zt  z. (a) Compute the directional derivative in the direction of (2,0,6,4) . (b) Is fincreasing faster in the direction of (2, 0, 6, 4) or in the direction of (1, 1,2, I)? Explain your answer.
3. (a) The cyli nder x 2 + y2 = 4 is cut by the plane x the intersecting curve is
J8127r VI
+y+z
= 1.
Show that the arc length of
cos 0 sin 0 dO.
(b) For the intersecting curve, find an equation for the tangent line at (1,)3, V3). 4. Find t he m inimum and maximum of x {( x,y) I (x,y) E [3, 5] x [0,2]}. 5. Let S be the boundary of a box B G = x3 i + 3zj + 2yk.
+ yz
= [2,2] x [ 1,1] x
(a) Com pute the integral of \l . F over B. (b) Compute
ffs
G . dS .
subject to the constraints x
[3,3], F
2xi
+z =
2y and
+ 3zj + 2yk
and
CHAPTER 9
166
(c) Suppose t.he origin at the center of B is shifted to (8, 15, 20) and then rotated 30 0 around the yaxis. Compute IIs F . dS . 6. A hole of radius ~ is drilled through the axis of symmetry of the hemisphere x 2 + y2 z ~ O.
+ z2 =
1,
(a) Write the volume of the remaining piece in Cartesian coordinates. (b) Write the volume of the remaining piece in cylindrical coordinates. (c) Compute the volume. 7. A painter is scared of heights, so he charges z2 dollars per square
meter to paint objects located at height z. His latest job is to paint
the silo 5 shown here. The height of the cylindrical part is 3 meters
and the radius of the hemispherical part is 2 meters.
(a) Compute IIs dS and interpret geometrically. 3
(b) Compute how much the painter charges. 8. (a) ComputeI;Ix1cos(y2+3)dydx.
roo Joroo Joroo e ('+'+ (b) Compute Jo x Y Z ')'/' dx dy d z. 9. (a) Verify Stokes' theorem for F z ~ O.
= z3 j + (x 3 
y3)j
+ y3k over
the hemisphere x 2 + y2
(b) For the same F as in (a), evaluate IIs('V' x F) . dS for the surface x 2 + y2 10.
+ 5z 2 , z
+ z2, ~ O.
(a) Find a vectorvalued function /(x, y, z) such that CD8
D/(x, y, z) = [  yz sin(xy)e y2 sin z
(x
y
)
 x z sin(xy)eCDs(x y ) 2xysin z
(b) For the region D shown at the right, let V be
the volume of the solid lying between /(x, y) =
x 3 sin y and the xy plane and lying over D. Write
Vin the form III g(x,y,z) dzdydx.
eeDS(XY)
xy2 cos z
]
.
y
1 x
(c) Rewrite your answer to part (b) in the form III g(x,y,z)dzdxdy.
167
Appen dix
ANSWERS TO CHAPTER TESTS AND SAMPLE EXAMS
ANSWERS TO T E ST FO R C H APTER 1 1.
(a) False; dot products need to be defined for vectors in the same space, so u · v is undefined. (b) True .
(c) True. (d) False; le t t he vectors be i, j and k. (e) False; v x a v
2.  2x  2y + z
= 0 for a
being any real number .
=1
3. (a) a = 5i+ j + k;h = 4i  2k
= [0,1 v'I4
(b) s
(c)
and t
= [0, s)
4. } 11/15 5.
±6V5
6. (a) l(t) = (1,3,  2, 0) + t(  1, 2, 1,1) (b) (1 ,  2,1, 1)· (2 , 0, 3, 1)= 2 #0
7. (a) [!3
~]
(b) 1; area of th pa rallelogr am spanned by (1,0) and (3, 1). (c) O' volume of th paralJelpiped spa.nned by (2, 0,  3), (8, 1, 5) a nd (2, 0,3) is 0; all three vectors lie in a single plane .
8. (a) (/3,311"/4, 1)
(b) ( .../6/2, .../6/2 , 1)
9.
(a) lu ,vi = 2 ~ 2V15 = IIull II vll (b) (2, 1 , 0, 1)/3
10 . (a) x +y  3z =3 (b)
5IIT
(c)
10/ IIT
ANSWERS TO T E ST F O R C H APTER 2 1.
(a) False ; let f (x, y , z ) = 1 if x, yor z = (b) False; should b e perpendicular.
°
and f (x , y, z ) = 0 otherwise.
APPEN DIX
168 (c) True. (d) False; consider z
= Ix!
(e) True.
3
16 ]
2. [ ; 3. 8.969
4. Positive x direction . 5. z
= 12x + 8y 
9
6. 8g / 8x does not exist ; 8g/8y == 0. 7. 8u/8s = 0; 8v/8t = 1 8. (a) 12 (b) lim(x ,y)+(O,O)!(x,y) does not exist because limx+o!(x , O) therefore, !( x, y) cannot be made continuous .
= 1 and
iimy+o!(O , y)
= 0;
9. ~~ (1, 1) = 2 + e > ~~ (1, 1) = 2 10. (a) (3.6, 36,  0.02) (b) J)ecreases by 3.6 (c ) No; the directional derivative is  ~;~
«
1.
ANSWER S TO TEST FOR CH APTER 3 1. (a) False; a minus sign is missing. (b) True . (c) False; the fourthorder derivatives must be continuous to guarantee equality. (d) True. (e) False; let !(x , y) = x 2 + y2 on the whole plane. It has a minimum but no maximum. 2. 4. 28; 4.2665
3. Since 8 2 z/8x 2 = 30xy, concave up if x > 0; concave down if x
< 0.
4. (0,1,2, 2), (0, 1,2, 2) , (0 , 1,  2,  2), (0, 1, 2, 2) 5. (a) and (c) since 8 2 f/ 8 y8x = fj2 f/8 x8y. 6. Minimum =
 v'6/2; maximum = v'6/2.
= ~ ; maximum = 0. Min imum = ~ ; no maximum.
7. (a) Minimu m (b)
8. The only critical point is at ( ~, ~). Also, D 0, so ( ~ , ~) is a saddle point. 9. (a) 8x/8v
=2
(b) Can't be done. 10. (a) (g , c) = (10,5) (b) (10 ,5) is a m aximum point.
= (8 2 ! /8x 2 )(82 ! /8y2) 
(8 2 f/8x8y) 2
=  12 <
169
ANSWERS TO CHAPTER TESTS AN D SAMPLE EXAMS
A N SWERS T O TEST F OR CH A PTER 4 1.
(a) True . (b) False; limits of integration should be and 11'. (c) Fa.lse; the last term should be b x dc/dt. (d) False; v(t) = (2,0,0) and w (t) = (0,1,0) both have 0 acceleration .
°
(e) True. 2. Both satisfy the desired conditions. 3. (a) All of ~ 3.
(b) Nowhere. 4. (a)
ftr/ 2 J4 + 5 cos
2
t dt
(b) (4, 0)
5. (a) 6g l/(t)i + 6[(g'(t))2 + g(t)gl/(t)Jj (b) g(t) must be a constant. 6.
z > 1
7.
fa4 J~ + 4y2 + 4y2 exp(2y2  12) dy
=
8. (a) and (b) since div F 0. 9. (a) (3:c 2  2:c + 3)[( 2sin(x3  x 2 + 3x  4) cos(x 3  x 2 + 3x  4))i 3(x 3  x 2 + 3x  4))j]
(b) pl/(t)c(t)
+ (4(x 3 
+ 2p' (t)c'(t) + p(t)cl/(t)
10. (a) T hey are bot h running counterclockwise on the ellipse 4x 2 + y2 = 4. (b) 12500i (c) l (t) (d)
= (cos( 4rr /5),2 sin( 4rr /5)) + (t VI + cos 2 4t dt
rr /5)( 4 sin( 4rr /5) , 8 cos( 4rr /5))
2 4 f 2 11:11:/54
A N SWER S TO T EST FOR CHAPTER 5 1.
(a) False; shouJd be (b) True. (c) True .
2 4 12 fa 
x~
dydx.
(d) False; I( x, y) should be nonnegative at all points of D. (e) False; the righthand side should be
[1
2
(x
2
+ 4x + v'x) dX]
[[11 (y3 
y2
+ 4y) dY] .
2.
i11VYl a a
1 Y  dzdx
a
l 1 lVY = l 1 iVY = 1 1i~l 1
dy
=
a a 1
a a
1
Z

1 Y 
a
a
dx dydz dxdzdy
1 z 
a a
=
dydx dz
x'
i 111X'11Y dydz dx : a a
x'
4
15
x 2 + 3x  4)3_
APPENDIX
170
3.
181 12
4. (1  sin(1))/2 by changing the order of integration. S. (a) 1811"
3 f'; 9,,2 ( 2 2
(b ) f 3 _';9,, 2 9  x  y ) dydx (c) 2711"/2
6.
i
7. (a) mv(W) :S fffw f( x,y , z) dxdydz:S MV(W). (b) The minimum value of f on W is V2/2e 2 at (11"/ 4, 1, 1) and the maximum value is 1 at (0 , 0, 0). The vol ume of W is 11"/8. Apply these values to the statement in part (a). 8 . (e 8

1) /3 by changing the order of integration of x and z.
9. ~ 3 10.
(a) 5~2
(b) 8 
if¥
A NS W ERS TO TEST FOR CHAPTER 6 1. (a) True. (b) False;
fooo cos xdx doesn't converge , so Fubini's theorem doesn't
apply.
(c) False; the integral in polar coordinates is only being integrated over the region between x = ±2y and x = 4  y2 .
J
(d) True.
(e) True. 2. 11"2 3. 8/311" 4. 16011"/3 5 . ~
6. (a) 2 J~1 f01 J~l(ll+v)dlldv dw
(b) 2 7. 4811" 8. 211" 9. The integral is improper near z = 0, so the integral should be calculated as f01 f~lI(l/x) dx dy+
f o1 f~ (l/x) dx dy, which diverges. 10. (0 , 10/(4 + 311") )
ANSWERS TO T EST F OR CHAPTER 1 1. (a) True. (b) False; (c) True.
liT
ti
x T Ii II is positive and
f
is positive, so the scalar surface integral is positive.
171
5\ ERS TO CHAPTER TESTS AND SAMPLE EXAMS
(d) True. (e) False' Ie F(x , y) == xi + xyj and choose any closed curve in the plane x ==
o.
2. 4A 3. ~ 4. 41T
5. (a) C)(O,r/» == (2cosOsinr/>,2sinOsinr/>,2cosr/», 0
(b) f fs Iz l dS == fa" f02"
12 cos r/>I (4 sin r/»
S 0 S 21T, 0 S r/> S 1T.
dO dr/>
(c) 321T 6. (a) V/== (2zsiny,2xzcosy,e Z +2xsiny)
(b) e  5 7. (a) 15x12y+2z==3
(b) f02 f~l (144u 2v 2 + 72uv 3 + 9v 4
+ 144u 4 v 2 + 4U 2)1/2 dv du
8. 1~8 9. 7e
+ 361
10. 8(3V3  2V2)/fJ
ANSWERS TO TEST FOR CHAPTER 8 1. (a) True.
(b) True. (c) Tru e. (d) True. (e) False; the formula A == ~ f aD x dy  y dx must also be applied to the lines x == 0, x == 1 and y == o.
2. 6 3. (a) (4xz 2 sin y cos(eX)  2e x x 2z2 sin y sine eX))i
(b) 0 4. 70 5. (a) xz 3i  yz3j
+ 4k
(b) 241T 6. ~ 7.
_ 158
8. (a) 5
(b)  5 9. 9 10 . (a) V x F == 0 (b) I (x y , z ) == x 2 yz 3
(e) e
+ y + xz
+ 2x2 z2 cos Y cos(eX)j + 4x 2 z sin y cos( eX)k
172
APPENDIX (d) 3
ANS W ERS TO COMPREHENSIVE T E ST F OR C HAPTERS 1
~
4
(a) (1 , 0) is a minimum
1.
(b) (1 , 0), (1,0)
(a) (0, _e 7r / 2 , 1)
2.
(b) (3e 1 cos 1  2e i sin 1)/(e 2 sin 2 1 8e 2 sin 1 cos 1 + 1 + 4e  2 )1/2
= (cos t ,sint) c(t) = (exp(2t) ,exp(t))
3. (a) c(t)
(b)
4. (a) (b)
V2 + In(1 + V2) V2 + In(1 + V2)
(c) The two parametrizations actually trace out the same path . 5. 32V5/3 cubic inches 6. (a) (2 , 1,1)+t(0,:,2)
(b) l:3x
7. (a)
+ 7y+ 16z = 2
OIl
[
00 9 1 o 0 1
(b) [1 3 0] 9 8. 0.0021 m 3
9. (a) (~,2) (b) ~ hour
(c) i + 3j (d) (125,10 , 0) (e) 5V5/2 km (a) $t = 2te", $u = t 2 eu , $v = sinw, $w = vcosw
10.
(b) Go in the direction of (2, 1, 0, 2)
(c) (2 e1 / 3 /3,4e 1 / 3 /9,0,0) A NSWERS TO COMPREH ENSIVE T E ST FOR C H A P TERS 5  8 1.
(a) 0 (0) (lcosl)/2
2. ~
3. 4.
1848 5
(a) J(x , y)=excos3y+C (b) 0 (c) 0
173
ANSWERS TO CHAP TER TESTS AND SAMPLE EXAMS
ar e. helps; the flux is 6V before and it is 9V after , where V = 41211"/3 is the volume e· ull
5. (a)
0"
(b) 3,5 6. (a) '!  R 1 (b)
ea
:/ 2) 2 Rd.
7.
9. Ie) 10. In
he urf ce should no be closed (so it would have a real boundary), wher m. the su rface has to be closed (to enclose a volume).
ANSWERS TO CO .
HE . SIVE TEST A FOR CHAP T ERS 1  8
1.
area of such a cylinder is 211", while the value of the integral is 21211" . pie. in::3 let z =
(e) (d) (e)
a
Ixl; it 's not
differentiable at x =
product , not a dot product .
eed continuously diffe rentiable second partials.
2. (1) (c
(3) (b)
(4) (d)
3. (a)
(b) 0 Fl / F./8
4.
(b) 5.
6.
[¥ a z
 ..IU)l
(b) 7.
0 0 1 1 2 3 1 0


(a)
 1)
(b) ~
8. 9.
10.
a) '4 (b (a) (b)
(e) 4


T
 /3
z'! 
y2 16
r2
J x 2 + y2 dy dx
 2. f 2 Jo r dr dz
rr ~ + 211" Jr4 o Jo r dz dr
o.
APPENDI'X
174
A NSWERS TO COM P R EHENSIVE TES T B FOR CHAPTERS 1  8 1. (a ) The two vectors are orthogonal, or one of them is O.
(b) The two vectors are parallel, or one of t hem is O. (c) There's more material going in tha.n out; i.e., we have a sink. (d) curl V . n tells you about the circulation of Von the surface with normal n.
2. (a) 8 (b) In t he direction of (1, 1, 2, 1) because the directional deri vati ve using normalized vectors is larger in the second case.
3. (a) T he intersecting curves can be parametrized by (x, y, z)
= (2 cos B, 2 sin B, 1 
(b) (x , y,z ) = (1,V3,v'3) +t(V3,1,V31) 4. Maximum
= 6, minimum = 3
5. (a) 96 (b) 192 (c) 96 l Iv'1'=? n/1  y' dz dydx ) f 1/2 Vl/4x 2J O (b) f~1r f01/2 fo"lr rdzdrdB x 2
6. (a 4
2
(1 _ ( ~ ) 3/2)
(c) %
7. (a) 161!', which is the surface area of the silo. (b) (5001!'/3) dollars 8. (a) [sin(4)  sin(3)J/2 (b) 1!' 9. (a) 31!'/4 (b) 31!'/4 10.
(a) f (x ,y,z) (b)
= (ze cos (xY),xy 2 sin z)
tl fOX'+l fox'sin y dzdy dx
(c) f01 r!';~':yfOx3siny dzd:r: dy
2 cos B2 sin B).