Download - Chapter 15 Solutions Calculus
-
7/30/2019 Chapter 15 Solutions Calculus
1/37
620
CHAPTER 15
Multiple Integrals
EXERCISE SET 15.1
1.
10
20
(x + 3)dydx =
10
(2x + 6)dx = 7 2.
31
11
(2x 4y)dydx =31
4x dx = 16
3.
42
10
x2ydxdy =
42
1
3y dy = 2 4.
02
21
(x2 + y2)dxdy =
02
(3 + 3y2)dy = 14
5.
ln 30
ln 20
ex+ydydx =
ln 30
exdx = 2
6.
20
10
y sin xdydx =
20
1
2sin x dx = (1 cos2)/2
7.
01
52
dxdy =
01
3 dy = 3 8.
64
73
dydx =
64
10dx = 20
9.
10
10
x
(xy + 1)2dydx =
10
1 1
x + 1
dx = 1 ln 2
10.
/2
21
x cos xydydx =
/2
(sin2x sin x)dx = 2
11.
ln 20
10
xy ey2xdydx =
ln 20
1
2(ex 1)dx = (1 ln2)/2
12.
43
21
1
(x + y)2dydx =
43
1
x + 1 1
x + 2
dx = ln(25/24)
13.
11
22
4xy3dydx =
11
0 dx = 0
14.
10
10
xyx2 + y2 + 1
dydx =
10
[x(x2 + 2)1/2 x(x2 + 1)1/2]dx = (3
3 4
2 + 1)/3
15.
10
32
x
1 x2 dydx =10
x(1 x2)1/2dx = 1/3
16.
/20
/30
(x sin y y sin x)dydx =/20
x
2
2
18sin x
dx = 2/144
17. (a) xk = k/2 1/4, k = 1, 2, 3, 4; yl = l/2 1/4, l = 1, 2, 3, 4, R
f(x, y) dxdy 4
k=1
4l=1
f(xk, yl )Akl =
4k=1
4l=1
[(k/21/4)2+(l/21/4)](1/2)2 = 37/4
(b)
20
20
(x2 + y) dxdy = 28/3; the error is |37/4 28/3| = 1/12
-
7/30/2019 Chapter 15 Solutions Calculus
2/37
Exercise Set 15.1 621
18. (a) xk = k/2 1/4, k = 1, 2, 3, 4; yl = l/2 1/4, l = 1, 2, 3, 4, R
f(x, y) dxdy 4
k=1
4l=1
f(xk, yl )Akl =
4k=1
4l=1
[(k/2 1/4) 2(l/2 1/4)](1/2)2 = 4
(b)
20
20
(x 2y) dxdy = 4; the error is zero
19. V =
5
3
2
1
(2x + y)dydx =
5
3
(2x + 3/2)dx = 19
20. V =
31
20
(3x3 + 3x2y)dydx =
31
(6x3 + 6x2)dx = 172
21. V =
20
30
x2dydx =
20
3x2dx = 8
22. V = 3
0
4
0
5(1
x/3)dydx =
3
0
5(4
4x/3)dx = 30
23. (a)
(1, 0, 4)
(2, 5, 0)
x
y
z (b)(0, 0, 5)
(3, 4, 0)
(0, 4, 3)
z
x
y
24. (a)
(1, 1, 0)
(0, 0, 2)
x
y
z (b)
(2, 2, 0)
(2, 2, 8)
x
y
z
25.
1/20
0
x cos(xy)cos2 xdydx =
1/20
cos2 x sin(xy)0
dx
=
1/20
cos2 x sin xdx = 13
cos3 x1/20
=1
3
-
7/30/2019 Chapter 15 Solutions Calculus
3/37
622 Chapter 15
26. (a)
y
x
z
5
3
(0, 2, 2)
(5, 3, 0)
(b) V =
50
20
y d y d x +
50
32
(2y + 6) dydx
= 10 + 5 = 15
27. fave =2
/20
10
y sin xydxdy =2
/20
cos xy
x=1x=0
dy =
2
/20
(1 cos y) dy = 1 2
28. average =1
3
30
10
x(x2 + y)1/2dxdy =
30
1
9[(1 + y)3/2 y3/2]dy = 2(31 9
3)/45
29. Tave =1
2 1
0 2
0 10 8x2 2y2
dydx =
1
2 1
0
44
3 16x2
dx =
14
3
30. fave =1
A(R)
ba
dc
k dydx =1
A(R)(b a)(d c)k = k
31. 1.381737122 32. 2.230985141
33.
R
f(x, y)dA =
ba
dc
g(x)h(y)dy
dx =
ba
g(x)
dc
h(y)dy
dx
=
ba
g(x)dx
dc
h(y)dy
34. The integral of tan x (an odd function) over the interval [1, 1] is zero.
35. The first integral equals 1/2, the second equals 1/2. No, because the integrand is not continuous.
EXERCISE SET 15.2
1.
10
xx2
xy2dydx =
10
1
3(x4 x7)dx = 1/40
2. 3/2
1
3y
y
ydxdy = 3/2
1
(3y
2y2)dy = 7/24
3.
30
9y20
ydxdy =
30
y
9 y2 dy = 9
4.
11/4
xx2
x/y dydx =
11/4
xx2
x1/2y1/2dydx =11/4
2(x x3/2)dx = 13/80
-
7/30/2019 Chapter 15 Solutions Calculus
4/37
Exercise Set 15.2 623
5.
2
x30
sin(y/x)dydx =
2
[x cos(x2) + x]dx = /2
6.
11
x2x2
(x2 y)dydx =11
2x4dx = 4/5 7.
/2
x20
1
xcos(y/x)dydx =
/2
sin x dx = 1
8. 1
0
x
0
ex2
dydx = 1
0
xex2
dx = (e
1)/2 9.
1
0
x
0
yx2 y2 dydx =
1
0
1
3
x3dx = 1/12
10.
21
y20
ex/y2
dxdy =
21
(e 1)y2dy = 7(e 1)/3
11. (a)
20
x20
xydydx =
20
1
2x5dx =
16
3
(b)
31
(y+7)/2(y5)/2
xydxdy =
31
(3y2 + 3y)dy = 38
12. (a) 1
0x
x2
(x + y)dydx = 1
0
(x3/2 + x/2 x3 x4/2)dx = 3/10
(b)
11
1x21x2
xdydx +
11
1x21x2
y d y d x =
11
2x
1 x2 dx + 0 = 0
13. (a)
84
x16/x
x2dydx =
84
(x3 16x)dx = 576
(b)
42
816/y
x2 dxdy +
84
8y
x2 dxdy =
84
512
3 4096
3y3
dy +
84
512 y33
dy
=640
3+
1088
3= 576
14. (a)21
y0
xy2dxdy =21
12
y4dy = 31/10
(b)
10
21
xy2 dydx +
21
2x
xy2 dydx =
10
7x/3 dx +
21
8x x43
dx = 7/6 + 29/15 = 31/10
15. (a)
11
1x21x2
(3x 2y)dydx =11
6x
1 x2 dx = 0
(b)
11
1y2
1y2(3x 2y) dxdy =
114y
1 y2 dy = 0
16. (a)5025x2
5x y d y d x =50 (5x x
2
)dx = 125/6
(b)
50
25y25y
y dxdy =
50
y
25 y2 5 + y
dy = 125/6
17.
40
y0
x(1 + y2)1/2dxdy =40
1
2y(1 + y2)1/2dy = (
17 1)/2
-
7/30/2019 Chapter 15 Solutions Calculus
5/37
624 Chapter 15
18.
0
x0
x cos y d y d x =
0
x sin x dx =
19.
20
6yy2
xydxdy =
20
1
2(36y 12y2 + y3 y5)dy = 50/3
20. /4
0
1/2
sin y
xdxdy = /4
0
1
4cos2y dy = 1/8
21.
10
xx3
(x 1)dy dx =10
(x4 + x3 + x2 x)dx = 7/60
22.
1/20
2xx
x2dydx +
11/2
1/xx
x2dydx =
1/20
x3dx +
11/2
(x x3)dx = 1/8
23. (a)
-1.5 -1-2 -0.5 0.5 1 1.5
x
y
1
2
3
4
(b) x = (1.8414, 0.1586), (1.1462, 3.1462)
(c)
R
x dA 1.14621.8414
x+2ex
xdydx =
1.14621.8414
x(x + 2 ex) dx 0.4044
(d)
R
x dA 3.14620.1586
ln yy2
xdxdy =
3.14620.1586
ln2 y
2 (y 2)
2
2
dy 0.4044
24. (a)
1 2 3
5
15
25
x
y
R
(b) (1, 3), (3, 27)
(c) 3
1
4x3x4
34x+4x2xdydx =
3
1
x[(4x3
x4)
(3
4x + 4x2)] dx =
224
15
25. A =
/40
cos xsinx
dydx =
/40
(cos x sin x)dx =
2 1
26. A =
14
y23y4
dxdy =
14
(y2 3y + 4)dy = 125/6
-
7/30/2019 Chapter 15 Solutions Calculus
6/37
Exercise Set 15.2 625
27. A =
33
9y21y2/9
dxdy =
33
8(1 y2/9)dy = 32
28. A =
10
cosh xsinhx
dydx =
10
(cosh x sinh x)dx = 1 e1
29. 4063x/20
(3
3x/4
y/2) dydx = 40
[(3
3x/4)(6
3x/2)
(6
3x/2)2/4] dx = 12
30.
20
4x20
4 x2 dydx =
20
(4 x2) dx = 16/3
31. V =
33
9x29x2
(3 x)dydx =33
(6
9 x2 2x
9 x2)dx = 27
32. V =
10
xx2
(x2 + 3y2)dydx =
10
(2x3 x4 x6)dx = 11/70
33. V =30
20
(9x2 + y2)dydx =30
(18x2 + 8/3)dx = 170
34. V =
11
1y2
(1 x)dxdy =11
(1/2 y2 + y4/2)dy = 8/15
35. V =
3/23/2
94x294x2
(y + 3)dydx =
3/23/2
6
9 4x2 dx = 27/2
36. V =
30
3y2/3
(9 x2)dxdy =30
(18 3y2 + y6/81)dy = 216/7
37. V = 8
50
25x20
25 x2dydx = 8
50
(25 x2)dx = 2000/3
38. V = 2
20
1(y1)20
(x2 + y2)dxdy = 2
20
1
3[1 (y 1)2]3/2 + y2[1 (y 1)2]1/2
dy,
let y 1 = sin to get V = 2/2/2
1
3cos3 + (1 + sin )2 cos
cos d which eventually yields
V = 3/2
39. V = 4 1
01x2
0
(1
x2
y2)dydx =
8
3
1
0
(1
x2)3/2dx = /2
40. V =
20
4x20
(x2 + y2)dydx =
20
x2
4 x2 + 13
(4 x2)3/2
dx = 2
41.
20
2y2
f(x, y)dxdy 42.
80
x/20
f(x, y)dydx 43.
e21
2lnx
f(x, y)dydx
-
7/30/2019 Chapter 15 Solutions Calculus
7/37
626 Chapter 15
44.
10
eey
f(x, y)dxdy 45.
/20
sinx0
f(x, y)dydx 46.
10
xx2
f(x, y)dydx
47.
40
y/40
ey2
dxdy =
40
1
4yey
2
dy = (1 e16)/8
48. 1
0
2x
0
cos(x2)dydx = 1
0
2x cos(x2)dx = sin1
49.
20
x20
ex3
dydx =
20
x2ex3
dx = (e8 1)/3
50.
ln 30
3ey
xdxdy =1
2
ln 30
(9 e2y)dy = 12
(9ln3 4)
51.
20
y20
sin(y3)dxdy =
20
y2 sin(y3)dy = (1 cos8)/3
52. 1
0 e
e
x
xdydx = 1
0
(ex xex)dx = e/2 1
53. (a)
40
2x
sin y3 dydx; the inner integral is non-elementary.
20
y20
sin
y3
dxdy =
20
y2 sin
y3
dy = 13
cos
y3 2
0
= 0
(b)
10
/2sin1 y
sec2(cos x)dxdy ; the inner integral is non-elementary.
/20
sin x0
sec2(cos x)dydx =
/20
sec2(cos x)sin x dx = tan 1
54. V = 4
2
0
4x20
(x2 + y2) dy dx = 4
2
0
x2
4 x2 + 13
(4 x2)3/2
dx (x = 2sin )
=
/20
64
3+
64
3sin2 128
3sin4
d =
64
3
2+
64
3
4 128
3
2
1 32 4 = 8
55. The region is symmetric with respect to the y-axis, and the integrand is an odd function of x,hence the answer is zero.
56. This is the volume in the first octant under the surface z =
1 x2 y2, so 1/8 of the volume ofthe sphere of radius 1, thus
6.
57. Area of triangle is 1/2, so f = 210
1x
1
1 + x2 dydx = 210
11 + x2
x
1 + x2
dx =
2 ln 2
58. Area =
20
(3x x2 x) dx = 4/3, so
f =3
4
20
3xx2x
(x2 xy)dydx = 34
20
(2x3 + 2x4 x5/2)dx = 34
8
15= 2
5
-
7/30/2019 Chapter 15 Solutions Calculus
8/37
Exercise Set 15.3 627
59. Tave =1
A(R)
R
(5xy + x2) dA. The diamond has corners (2, 0), (0,4) and thus has area
A(R) = 41
22(4) = 16m2. Since 5xy is an odd function of x (as well as y),
R
5xydA = 0. Since
x2 is an even function of both x and y,
Tave =4
16 R
x,y>0
x2 dA =1
4
2
0
42x
0
x2 dydx =1
4
2
0
(4
2x)x2 dx =1
44
3x3
1
2x4
2
0
=2
3
C
60. The area of the lens is R2 = 4 and the average thickness Tave is
Tave =4
4
20
4x20
1 (x2 + y2)/4 dydx = 1
20
1
6(4 x2)3/2dx (x = 2cos )
=8
3
0
sin4 d =8
3
1 32 4
2=
1
2in
61. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so
V = a
0 sinx
x/2 1 + x + y dy dx = 0.676089
EXERCISE SET 15.3
1.
/20
sin 0
r cos dr d =
/20
1
2sin2 cos d = 1/6
2.
0
1+cos 0
rdrd =
0
1
2(1 + cos )2d = 3/4
3. /2
0
a sin
0
r2drd = /2
0
a3
3
sin3 d =2
9
a3
4.
/60
cos30
rdrd =
/60
1
2cos2 3 d = /24
5.
0
1sin 0
r2 cos drd =
0
1
3(1 sin )3 cos d = 0
6.
/20
cos 0
r3drd =
/20
1
4cos4 d = 3/64
7. A = 2
0
1cos
0
rdrd = 2
0
1
2
(1
cos )2d = 3/2
8. A = 4
/20
sin 20
rdrd = 2
/20
sin2 2 d = /2
9. A =
/2/4
1sin2
rdrd =
/2/4
1
2(1 sin2 2)d = /16
-
7/30/2019 Chapter 15 Solutions Calculus
9/37
628 Chapter 15
10. A = 2
/30
2sec
r d r d =
/30
(4 sec2 )d = 4/3
3
11. A = 2
/2/6
4sin 2
rdrd =
/2/6
(16 sin2 4)d = 4/3 + 2
3
12. A = 2
/2
1
1+cos
rdrd =
/2
(
2cos
cos2 )d = 2
/4
13. V = 8
/20
31
r
9 r2 drd = 1283
2
/20
d =64
3
2
14. V = 2
/20
2sin 0
r2drd =16
3
/20
sin3 d = 32/9
15. V = 2
/20
cos 0
(1 r2)rdrd = 12
/20
(2cos2 cos4 )d = 5/32
16. V = 4 /2
0
3
1
drd = 8 /2
0
d = 4
17. V =
/20
3 sin 0
r2 sin drd = 9
/20
sin4 d = 27/16
18. V = 4
/20
22cos
r
4 r2 drd +/2
20
r
4 r2 drd
=32
3
/20
sin3 d +32
3
/2
d =64
9+
16
3
19. 2
0
1
0
er2
rdrd =1
2
(1
e1)
2
0
d = (1
e1)
20.
/20
30
r
9 r2 drd = 9/20
d = 9/2
21.
/40
20
1
1 + r2rdrd =
1
2ln 5
/40
d =
8ln 5
22.
/2/4
2cos 0
2r2 sin drd =16
3
/2/4
cos3 sin d = 1/3
23. /2
0
1
0
r3drd =1
4
/2
0
d = /8
24.
20
20
er2
rdrd =1
2(1 e4)
20
d = (1 e4)
25.
/20
2cos 0
r2drd =8
3
/20
cos3 d = 16/9
-
7/30/2019 Chapter 15 Solutions Calculus
10/37
Exercise Set 15.3 629
26.
/20
10
cos(r2)rdrd =1
2sin1
/20
d =
4sin1
27.
/20
a0
r
(1 + r2)3/2drd =
2
1 1/
1 + a2
28. /4
0
sec tan
0
r2drd =1
3
/4
0
sec3 tan3 d = 2(
2 + 1)/45
29.
/40
20
r1 + r2
drd =
4(
5 1)
30.
/2tan1(3/4)
53csc
r d r d =1
2
/2tan1(3/4)
(25 9csc2 )d
=25
2
2 tan1(3/4)
6 = 25
2tan1(4/3) 6
31. V =
20
a0
hrdrd =
20
ha2
2d = a2h
32. (a) V = 8
/20
a0
c
a(a2 r2)1/2 rdrd = 4c
3a(a2 r2)3/2
a0
=4
3a2c
(b) V 43
(6378.1370)26356.5231 1,083,168,200,000 km3
33. V = 2
/20
a sin 0
c
a(a2 r2)1/2r d r d = 2
3a2c
/20
(1 cos3 )d = (3 4)a2c/9
34. A = 4
/40
a2cos20
rdrd = 4a2/40
cos2 d = 2a2
35. A =
/4
/6
4sin
8cos2
r d r d +
/2
/4
4sin
0
rdrd
=
/4/6
(8sin2 4cos2)d +/2/4
8sin2 d = 4/3 + 2
3 2
36. A =
0
2a sin 0
rdrd = 2a20
sin2 d = a2 12
a2 sin2
37. (a) I2 =
+0
ex2
dx
+0
ey2
dy
=
+0
+0
ex2
dx
ey
2
dy
=+0
+0
ex2
ey2
dxdy =+0
+0
e(x2
+y2
)dxdy
(b) I2 =
/20
+0
er2
rdrd =1
2
/20
d = /4 (c) I =
/2
38. (a) 1.173108605 (b)
0
10
rer4
drd =
10
rer4
dr 1.173108605
-
7/30/2019 Chapter 15 Solutions Calculus
11/37
630 Chapter 15
39. V =
20
R0
D(r)rdrd =
20
R0
kerrdrd = 2k(1 + r)erR0
= 2k[1 (R + 1)eR]
40.
tan1(2)tan1(1/3)
20
r3 cos2 drd = 4
tan1(2)tan1(1/3)
cos2 d =1
5+ 2[tan1(2) tan1(1/3)] = 1
5+
2
EXERCISE SET 15.4
1. (a) z
xy
(b)
x
y
z
(c)
x
y
z
2. (a)
y
x
z (b) z
y
x
(c) z
yx
-
7/30/2019 Chapter 15 Solutions Calculus
12/37
Exercise Set 15.4 631
3. (a) x = u, y = v, z =5
2+
3
2u 2v (b) x = u, y = v, z = u2
4. (a) x = u, y = v, z =v
1 + u2(b) x = u, y = v, z =
1
3v2 5
3
5. (a) x = 5cos u, y = 5sin u, z = v; 0 u 2, 0 v 1(b) x = 2 cos u, y = v, z = 2sin u; 0 u 2, 1 v 3
6. (a) x = u, y = 1 u, z = v;1 v 1 (b) x = u, y = 5 + 2v, z = v; 0 u 3
7. x = u, y = sin u cos v, z = sin u sin v 8. x = u, y = eu cos v, z = eu sin v
9. x = r cos , y = r sin , z =1
1 + r210. x = r cos , y = r sin , z = er
2
11. x = r cos , y = r sin , z = 2r2 cos sin
12. x = r cos , y = r sin , z = r2(cos2 sin2 )
13. x = r cos , y = r sin , z =
9
r2; r
5
14. x = r cos , y = r sin , z = r; r 3 15. x = 12
cos , y =1
2 sin , z =
3
2
16. x = 3cos , y = 3sin , z = 3cot 17. z = x 2y; a plane
18. y = x2 + z2, 0 y 4; part of a circular paraboloid
19. (x/3)2 + (y/2)2 = 1; 2 z 4; part of an elliptic cylinder
20. z = x2 + y2; 0 z 4; part of a circular paraboloid
21. (x/3)
2
+ (y/4)
2
= z2
; 0 z 1; part of an elliptic cone22. x2 + (y/2)2 + (z/3)2 = 1; an ellipsoid
23. (a) x = r cos , y = r sin , z = r, 0 r 2; x = u, y = v, z = u2 + v2; 0 u2 + v2 4
24. (a) I: x = r cos , y = r sin , z = r2, 0 r 2; II: x = u, y = v, z = u2 + v2; u2 + v2 2
25. (a) 0 u 3, 0 v (b) 0 u 4,/2 v /2
26. (a) 0 u 6, v 0 (b) 0 u 5, /2 v 3/2
27. (a) 0 /2, 0 2 (b) 0 , 0
28. (a) /2 , 0 2 (b) 0 /2, 0 /2
29. u = 1, v = 2, ru rv = 2i 4j + k; 2x + 4y z = 5
30. u = 1, v = 2, ru rv = 4i 2j + 8k; 2x + y 4z = 6
31. u = 0, v = 1, ru rv = 6k; z = 0 32. ru rv = 2i j 3k; 2x y 3z = 4
-
7/30/2019 Chapter 15 Solutions Calculus
13/37
632 Chapter 15
33. ru rv = (
2/2)i (2/2)j + (1/2)k; x y +
2
2z =
2
8
34. ru rv = 2i ln 2k; 2x (ln2)z = 0
35. z =
9 y2, zx = 0, zy = y/
9 y2, z2x + z2y + 1 = 9/(9 y2),
S =2033
39 y2 dydx =
2
0 3 dx = 6
36. z = 8 2x 2y, z2x + z2y + 1 = 4 + 4 + 1 = 9, S =40
4x0
3 dydx =
40
3(4 x)dx = 24
37. z2 = 4x2 + 4y2, 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus
z2x + z2y + 1 = (16x
2 + 16y2)/z2 + 1 = 5, S =
10
xx2
5 dydx =
5
10
(x x2)dx =
5/6
38. z2 = x2 + y2, zx = x/z, zy = y/z, z2x + z
2y + 1 = (z
2 + y2)/z2 + 1 = 2,
S =R
2 dA = 2 /2
0
2cos 0 2 rdrd = 42
/20 cos
2
d = 2
39. zx = 2x, zy = 2y, z2x + z2y + 1 = 4x2 + 4y2 + 1,
S =
R
4x2 + 4y2 + 1 dA =
20
10
r
4r2 + 1 drd
=1
12(5
5 1)20
d = (5
5 1)/6
40. zx = 2, zy = 2y, z2x + z2y + 1 = 5 + 4y
2,
S =10
y0
5 + 4y
2
dxdy =10 y
5 + 4y
2
dy = (27 55)/12
41. r/u = cos vi + sin vj + 2uk, r/v = u sin vi + u cos vj,
r/u r/v = u4u2 + 1; S =20
21
u
4u2 + 1 du dv = (17
17 5
5)/6
42. r/u = cos vi + sin vj + k, r/v = u sin vi + u cos vj,
r/u r/v = 2u; S =/20
2v0
2 u du dv =
2
123
43. zx = y, zy = x, z2x + z
2y + 1 = x
2 + y2 + 1,
S =
R
x2 + y2 + 1 dA =
/60
30
r
r2 + 1 drd =1
3(10
101)
/60
d = (10
101)/18
-
7/30/2019 Chapter 15 Solutions Calculus
14/37
-
7/30/2019 Chapter 15 Solutions Calculus
15/37
634 Chapter 15
54. (a) x = v cos u, y = v sin u, z = f(v), for example (b) x = v cos u, y = v sin u, z = 1/v2
(c) z
yx
55. (x/a)2 + (y/b)2 + (z/c)2 = cos2 v(cos2 u + sin2 u) + sin2 v = 1, ellipsoid
56. (x/a)2 + (y/b)2 (z/c)2 = cos2 u cosh2 v + sin2 u cosh2 v sinh2 v = 1, hyperboloid of one sheet
57. (x/a)2 + (y/b)2 (z/c)2 = sinh2 v + cosh2 v(sinh2 u cosh2 u) = 1, hyperboloid of two sheets
EXERCISE SET 15.5
1.
11
20
10
(x2 + y2 + z2)dxdydz =
11
20
(1/3 + y2 + z2)dydz =
11
(10/3 + 2z2)dz = 8
2.
1/21/3
0
10
zx sin xy dz dy dx =
1/21/3
0
1
2x sin xydydx =
1/21/3
1
2(1 cos x)dx = 1
12+
3 24
3.
20
y21
z1
yz dx dz dy =
20
y21
(yz2 + yz)dz dy =
20
1
3y7 +
1
2y5 1
6y
dy =
47
3
4.
/4
0
1
0
x2
0
x cos y dz dxdy =
/4
0
1
0
x3 cos ydxdy =
/4
0
14
cos y dy = 2/8
5.
30
9z20
x0
xydydxdz =
30
9z20
1
2x3dxdz =
30
1
8(81 18z2 + z4)dz = 81/5
6.
31
x2x
ln z0
xey dy dz dx =
31
x2x
(xz x)dz dx =31
1
2x5 3
2x3 + x2
dx = 118/3
7.
20
4x20
3x2y25+x2+y2
x dz dydx =
20
4x20
[2x(4 x2) 2xy2]dydx
=
2
0
43
x(4 x2)3/2dx = 128/15
8.
21
2z
3y0
y
x2 + y2dxdydz =
21
2z
3dydz =
21
3(2 z)dz = /6
-
7/30/2019 Chapter 15 Solutions Calculus
16/37
Exercise Set 15.5 635
9.
0
10
/60
xy sin yz dz dy dx =
0
10
x[1 cos(y/6)]dydx =0
(1 3/)x dx = ( 3)/2
10.
11
1x20
y0
y dz dydx =
11
1x20
y2dydx =
11
1
3(1 x2)3dx = 32/105
11.20
x0
2x20
xyz dz dy dx =20
x0
1
2 xy(2 x2)2dydx =20
1
4 x3(2 x2)2dx = 1/6
12.
/2/6
/2y
xy0
cos(z/y)dz dx dy =
/2/6
/2y
y sin xdxdy =
/2/6
y cos y dy = (5 6
3)/12
13.
30
21
12
x + z2
ydzdydx 9.425
14. 8
10
1x20
1x2y20
ex2y2z2 dz dy dx 2.381
15. V =
40
(4x)/20
(123x6y)/40
dz dy dx =
40
(4x)/20
1
4(12 3x 6y)dydx
=
40
3
16(4 x)2dx = 4
16. V =
10
1x0
y0
dz dy dx =
10
1x0
y d y d x =
10
2
3(1 x)3/2dx = 4/15
17. V = 2 2
0 4
x2 4y
0
dz dy dx = 2 2
0 4
x2(4 y)dydx = 2
2
0 8 4x2 +
1
2x4 dx = 256/15
18. V =
10
y0
1y20
dz dx dy =
10
y0
1 y2 dxdy =
10
y
1 y2 dy = 1/3
19. The projection of the curve of intersection onto the xy-plane is x2 + y2 = 1,
V = 4
10
1x20
43y24x2+y2
dz dy dx
20. The projection of the curve of intersection onto the xy-plane is 2x2 + y2 = 4,
V = 420
42x20
8x2y23x2+y2
dz dy dx
21. V = 2
33
9x2/30
x+30
dz dy dx 22. V = 8
10
1x20
1x20
dz dy dx
-
7/30/2019 Chapter 15 Solutions Calculus
17/37
636 Chapter 15
23. (a)
(0, 0, 1)
(1, 0, 0)(0, 1, 0)
z
y
x
(b)
(0, 9, 9)
(3, 9, 0)
z
x
y
(c)
(0, 0, 1)
(1, 2, 0)
x
y
z
24. (a)
(3, 9, 0)
(0, 0, 2)
x
y
z
(b)(0, 0, 2)
(0, 2, 0)
(2, 0, 0)
x
y
z
(c)
(2, 2, 0)
(0, 0, 4)
x
y
z
25. V = 1
0 1x
0 1xy
0
dz dy dx = 1/6, fave = 6 1
0 1x
0 1xy
0
(x + y + z) dz dy dx =3
4
26. The integrand is an odd function of each of x, y, and z, so the answer is zero.
27. The volume V =3
2, and thus
rave =
2
3
G
x2 + y2 + z2 dV =
2
3
1/21/
2
12x212x2
67x2y25x2+5y2
x2 + y2 + z2dzdydx 3.291
-
7/30/2019 Chapter 15 Solutions Calculus
18/37
Exercise Set 15.5 637
28. V = 1, dave =1
V
10
10
10
(x z)2 + (y z)2 + z2 dxdydz 0.771
29. (a)
a0
b(1x/a)0
c(1x/ay/b)0
dz dy dx,
b0
a(1y/b)0
c(1x/ay/b)0
dz dx dy,
c0
a(1z/c)0
b(1x/az/c)0
dydxdz,
a0
c(1x/a)0
b(1x/az/c)0
dy dz dx,
c
0
b(1z/c)
0
a(1y/bz/c)
0
dxdy dz,
b
0
c(1y/b)
0
a(1y/bz/c)
0
dx dz dy
(b) Use the first integral in Part (a) to geta0
b(1x/a)0
c
1 xa y
b
dydx =
a0
1
2bc
1 xa
2dx =
1
6abc
30. V = 8
a0
b1x2/a20
c1x2/a2y2/b20
dz dy dx
31. (a) 2
04x2
0
5
0
f(x,y,z) dzdydx
(b)
90
3x0
3xy
f(x, y, z) dzdydx (c)
20
4x20
8yy
f(x,y,z) dzdydx
32. (a)
30
9x20
9x2y20
f(x, y, z)dz dy dx
(b)
40
x/20
20
f(x, y, z)dz dy dx (c)
20
4x20
4yx2
f(x,y,z)dz dy dx
33. (a) At any point outside the closed sphere {x2 + y2 + z2 1} the integrand is negative, so tomaximize the integral it suffices to include all points inside the sphere; hence the maximumvalue is taken on the region G = {x2 + y2 + z2 1}.
(b) 4.934802202
(c)
20
0
10
(1 2)ddd = 2
2
34.
ba
dc
k
f(x)g(y)h(z)dz dy dx =
ba
dc
f(x)g(y)
k
h(z)dz
dydx
=
ba
f(x)
dc
g(y)dy
dx
k
h(z)dz
= b
a
f(x)dxd
c
g(y)dy
k
h(z)dz35. (a)
11
x dx
10
y2dy
/20
sin z dz
= (0)(1/3)(1) = 0
(b)
10
e2xdx
ln 30
eydy
ln 20
ezdz
= [(e2 1)/2](2)(1/2) = (e2 1)/2
-
7/30/2019 Chapter 15 Solutions Calculus
19/37
638 Chapter 15
EXERCISE SET 15.6
1. Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum is5(0 a) + 10(5 a) + 20(10 a) = 0 for equilibrium, so 250 35a = 0, a = 50/7. The fulcrumshould be placed 50/7 units to the right of m1.
2. At equilibrium, 10(0 4) + 3(2 4) + 4(3 4) + m(6 4) = 0, m = 25
3. A = 1, x =
1
0
1
0
xdydx =1
2, y =
1
0
1
0
y d y d x =1
2
4. A = 2, x =1
2
G
xdydx, and the region of integration is symmetric with respect to the x-axes
and the integrand is an odd function of x, so x = 0. Likewise, y = 0.
5. A = 1/2,
R
x dA =
10
x0
xdydx = 1/3,
R
y dA =
10
x0
y d y d x = 1/6;
centroid (2/3, 1/3)
6. A =
10
x20
dydx = 1/3,
R
x dA =
10
x20
xdydx = 1/4,
R
y dA =
10
x20
y d y d x = 1/10; centroid (3/4, 3/10)
7. A =
10
2x2x
dydx = 7/6,
R
x dA =
10
2x2x
xdydx = 5/12,
R
y dA =
10
2x2x
y d y d x = 19/15; centroid (5/14, 38/35)
8. A =
4,
R
x dA =
10
1x20
xdydx =1
3, x =
4
3, y =
4
3by symmetry
9. x = 0 from the symmetry of the region,
A =1
2(b2 a2),
R
y dA =
0
ba
r2 sin drd =2
3(b3 a3); centroid x = 0, y = 4(b
3 a3)3(b2 a2) .
10. y = 0 from the symmetry of the region, A = a2/2,
R
x dA = /2
/2 a
0
r2 cos drd = 2a3/3; centroid 4a3
, 0
11. M=
R
(x, y)dA =
10
10
|x + y 1| dxdy
=
10
1x0
(1 x y) dy +11x
(x + y 1) dy
dx =1
3
-
7/30/2019 Chapter 15 Solutions Calculus
20/37
Exercise Set 15.6 639
x = 3
10
10
x(x, y) dydx = 3
10
1x0
x(1 x y) dy +11x
x(x + y 1) dy
dx =1
2
By symmetry, y =1
2as well; center of gravity (1/2, 1/2)
12. x =1
M
G
x(x, y) dA, and the integrand is an odd function of x while the region is symmetric
with respect to the y-axis, thus x = 0; likewise y = 0.
13. M =
10
x0
(x + y)dydx = 13/20, Mx =
10
x0
(x + y)y d y d x = 3/10,
My =
10
x0
(x + y)xdydx = 19/42, x = My/M = 190/273, y = Mx/M = 6/13;
the mass is 13/20 and the center of gravity is at (190/273, 6/13).
14. M =
0
sin x0
y d y d x = /4, x = /2 from the symmetry of the density and the region,
Mx =0sinx0 y
2
dydx = 4/9, y = Mx/M =
16
9 ; mass /4, center of gravity
2 ,
16
9
.
15. M =
/20
a0
r3 sin cos drd = a4/8, x = y from the symmetry of the density and the
region, My =
/20
a0
r4 sin cos2 drd = a5/15, x = 8a/15; mass a4/8, center of gravity
(8a/15, 8a/15).
16. M =
0
10
r3drd = /4, x = 0 from the symmetry of density and region,
Mx =
0
1
0
r4 sin drd = 2/5, y =8
5
; mass /4, center of gravity 0,8
5.
17. V = 1, x =
10
10
10
x dz dydx =1
2, similarly y = z =
1
2; centroid
1
2,
1
2,
1
2
18. symmetry,
G
z dz dy dx =
20
20
10
rz dr d dz = 2, centroid = (0, 0, 1)
19. x = y = z from the symmetry of the region, V = 1/6,
x =1
V
10
1x0
1xy0
x dz dydx = (6)(1/24) = 1/4; centroid (1/4, 1/4, 1/4)
20. The solid is described by 1 y 1, 0 z 1 y2, 0 x 1 z;
V =
11
1y20
1z0
dx dz dy =4
5, x =
1
V
11
1y20
1z0
x dxdz dy =5
14, y = 0 by symmetry,
z =1
V
11
1y20
1z0
z dx dz dy =2
7; the centroid is
5
14, 0,
2
7
.
-
7/30/2019 Chapter 15 Solutions Calculus
21/37
640 Chapter 15
21. x = 1/2 and y = 0 from the symmetry of the region,
V =
10
11
1y2
dz dy dx = 4/3, z =1
V
G
z dV = (3/4)(4/5) = 3/5; centroid (1/2, 0, 3/5)
22. x = y from the symmetry of the region,
V = 2
0
2
0
xy
0
dz dy dx = 4, x =1
VG
x dV = (1/4)(16/3) = 4/3,
z =1
V
G
z dV = (1/4)(32/9) = 8/9; centroid (4/3, 4/3, 8/9)
23. x = y = z from the symmetry of the region, V = a3/6,
x =1
V
a0
a2x20
a2x2y20
x dz dydx =1
V
a0
a2x20
x
a2 x2 y2 dydx
=1
V
/20
a0
r2
a2 r2 cos drd = 6a3
(a4/16) = 3a/8; centroid (3a/8, 3a/8, 3a/8)
24. x = y = 0 from the symmetry of the region, V = 2a3/3
z =1
V
aa
a2x2a2x2
a2x2y20
z dz dy dx =1
V
aa
a2x2a2x2
1
2(a2 x2 y2)dydx
=1
V
20
a0
1
2(a2 r2)rdrd = 3
2a3(a4/4) = 3a/8; centroid (0, 0, 3a/8)
25. M =
a0
a0
a0
(a x)dz dy dx = a4/2, y = z = a/2 from the symmetry of density and
region, x =1
M a
0
a
0
a
0
x(a x)dz dy dx = (2/a4)(a5/6) = a/3;
mass a4/2, center of gravity (a/3, a/2, a/2)
26. M =
aa
a2x2a2x2
h0
(h z)dz dy dx = 12
a2h2, x = y = 0 from the symmetry of density
and region, z =1
M
G
z(h z)dV = 2a2h2
(a2h3/6) = h/3;
mass a2h2/2, center of gravity (0, 0, h/3)
27. M = 1
1 1
0
1y2
0
yz dz dy dx = 1/6, x = 0 by the symmetry of density and region,
y =1
M
G
y2z dV = (6)(8/105) = 16/35, z =1
M
G
yz2dV = (6)(1/12) = 1/2;
mass 1/6, center of gravity (0, 16/35, 1/2)
-
7/30/2019 Chapter 15 Solutions Calculus
22/37
Exercise Set 15.6 641
28. M =
30
9x20
10
xz dz dy dx = 81/8, x =1
M
G
x2z dV = (8/81)(81/5) = 8/5,
y =1
M
G
xyz dV = (8/81)(243/8) = 3, z =1
M
G
xz2dV = (8/81)(27/4) = 2/3;
mass 81/8, center of gravity (8/5, 3, 2/3)
29. (a) M =10
10
k(x2 + y2)dy dx = 2k/3, x = y from the symmetry of density and region,
x =1
M
R
kx(x2 + y2)dA =3
2k(5k/12) = 5/8; center of gravity (5/8, 5/8)
(b) y = 1/2 from the symmetry of density and region,
M =
10
10
kxdydx = k/2, x =1
M
R
kx2dA = (2/k)(k/3) = 2/3,
center of gravity (2/3, 1/2)
30. (a) x = y = z from the symmetry of density and region,
M =10
10
10
k(x2
+ y2
+ z2
)dz dy dx = k,
x =1
M
G
kx(x2 + y2 + z2)dV = (1/k)(7k/12) = 7/12; center of gravity (7/12, 7/12, 7/12)
(b) x = y = z from the symmetry of density and region,
M =
10
10
10
k(x + y + z)dz dy dx = 3k/2,
x =1
M
G
kx(x + y + z)dV =2
3k(5k/6) = 5/9; center of gravity (5/9, 5/9, 5/9)
31. V =G
dV =0
sinx0
1/(1+x2+y2)0
dz dy dx = 0.666633,
x =1
V
G
xdV = 1.177406, y =1
V
G
ydV = 0.353554, z =1
V
G
zdV = 0.231557
32. (b) Use polar coordinates for x and y to get
V =
G
dV =
20
a0
1/(1+r2)0
r dz dr d = ln(1 + a2),
z =1
V G
zdV =a2
2(1 + a2) ln(1 + a2)
Thus lima0+
z =1
2; lima+
z = 0.
lima0+
z =1
2; lima+
z = 0
(c) Solve z = 1/4 for a to obtain a 1.980291.
-
7/30/2019 Chapter 15 Solutions Calculus
23/37
642 Chapter 15
33. Let x = r cos , y = r sin , and dA = rdrd in formulas (11) and (12).
34. x = 0 from the symmetry of the region, A =
20
a(1+sin )0
rdrd = 3a2/2,
y =1
A
20
a(1+sin )0
r2 sin drd =2
3a2(5a3/4) = 5a/6; centroid (0, 5a/6)
35. x = y from the symmetry of the region, A =/20
sin 20
rdrd = /8,
x =1
A
/20
sin20
r2 cos drd = (8/)(16/105) =128
105; centroid
128
105,
128
105
36. x = 3/2 and y = 1 from the symmetry of the region,R
x dA = xA = (3/2)(6) = 9,
R
y dA = yA = (1)(6) = 6
37. x = 0 from the symmetry of the region, a2/2 is the area of the semicircle, 2y is the distancetraveled by the centroid to generate the sphere so 4a3/3 = (a2/2)(2y), y = 4a/(3)
38. (a) V =
1
2a2
2
a +
4a
3
=
1
3(3 + 4)a3
(b) the distance between the centroid and the line is
2
2
a +
4a
3
so
V =
1
2a2
2
2
2
a +
4a
3
=
1
6
2(3 + 4)a3
39. x = k so V = (ab)(2k) = 22abk
40. y = 4 from the symmetry of the region,
A =
22
8x2x2
dydx = 64/3 so V = (64/3)[2(4)] = 512/3
41. The region generates a cone of volume1
3ab2 when it is revolved about the x-axis, the area of the
region is1
2ab so
1
3ab2 =
1
2ab
(2y), y = b/3. A cone of volume
1
3a2b is generated when the
region is revolved about the y-axis so1
3a2b =
1
2ab
(2x), x = a/3. The centroid is (a/3, b/3).
42. Ix = a
0 b
0
y2 d y d x =1
3ab3, Iy =
a
0 b
0
x2 d y d x =1
3a3b,
Iz =
a0
b0
(x2 + y2) d y d x =1
3ab(a2 + b2)
43. Ix =
20
a0
r3 sin2 d r d = a4/4; Iy =
20
a0
r3 cos2 d r d = a4/4 = Ix;
Iz = Ix + Iy = a4/2
-
7/30/2019 Chapter 15 Solutions Calculus
24/37
Exercise Set 15.7 643
EXERCISE SET 15.7
1.
20
10
1r20
zr dz drd =
20
10
1
2(1 r2)rdrd =
20
1
8d = /4
2.
/20
cos 0
r20
r sin dz drd =
/20
cos 0
r3 sin drd =
/20
1
4cos4 sin d = 1/20
3.
/2
0
/2
0
1
0
3 sin cos ddd =
/2
0
/2
0
14
sin cos dd =
/2
0
18
d = /16
4.
20
/40
a sec 0
2 sin ddd =
20
/40
1
3a3 sec3 sin dd =
20
1
6a3d = a3/3
5. V =
20
30
9r2
r dz drd =
20
30
r(9 r2)drd =20
81
4d = 81/2
6. V = 2
20
20
9r20
r dz drd = 2
20
20
r
9 r2drd
=
2
3 (27 55) 2
0 d = 4(27 55)/37. r2 + z2 = 20 intersects z = r2 in a circle of radius 2; the volume consists of two portions, one inside
the cylinder r =
20 and one outside that cylinder:
V =
20
20
r220r2
r dz drd +
20
202
20r220r2
r dz drd
=
20
20
r
r2 +
20 r2
drd +
20
202
2r
20 r2 drd
=4
3(10
5 13)
20
d +128
3
20
d =152
3 +
80
3
5
8. z = hr/a intersects z = h in a circle of radius a,
V =
20
a0
hhr/a
r dz drd =
20
a0
h
a(ar r2)drd =
20
1
6a2h d = a2h/3
9. V =
20
/30
40
2 sin ddd =
20
/30
64
3sin dd =
32
3
20
d = 64/3
10. V =
20
/40
21
2 sin ddd =
20
/40
7
3sin dd =
7
6(2
2)
20
d = 7(2
2)/3
11. In spherical coordinates the sphere and the plane z = a are = 2a and = a sec , respectively.They intersect at = /3,
V =
2
0
/3
0
a sec
0
2 sin ddd +
2
0
/2
/3
2a
0
2 sin ddd
=
20
/30
1
3a3 sec3 sin dd +
20
/2/3
8
3a3 sin dd
=1
2a320
d +4
3a320
d = 11a3/3
-
7/30/2019 Chapter 15 Solutions Calculus
25/37
644 Chapter 15
12. V =
20
/2/4
30
2 sin ddd =
20
/2/4
9sin dd =9
2
2
20
d = 9
2
13.
/20
a0
a2r20
r3 cos2 dz drd =
/20
a0
(a2r3 r5)cos2 drd
=1
12a6
/2
0
cos2 d = a6/48
14.
0
/20
10
e3
2 sin ddd =1
3(1 e1)
0
/20
sin dd = (1 e1)/3
15.
/20
/40
80
4 cos2 sin ddd = 32(2
2 1)/15
16.
20
0
30
3 sin ddd = 81
17. (a) /2
/3
4
1
2
2
r tan3 1 + z
2dz dr d =
/3
/6
tan3 d4
1
r dr2
2
11 + z
2dz
=
4
3 1
2ln 3
15
2
2ln(
5 2)
=
5
2(8 + 3ln 3)ln(
5 2)
(b)
/2/3
41
22
y tan3 z1 + x2
dxdy dz; the region is a rectangular solid with sides /6, 3, 4.
18.
/20
/40
1
18cos37 cos dd =
2
36
/20
cos37 d =4,294,967,296
755,505,013,725
2 0.008040
19. (a) V = 2
20
a0
a2r20
r dz drd = 4a3/3
(b) V =
20
0
a0
2 sin ddd = 4a3/3
20. (a)
20
4x20
4x2y20
xyz dz dy dx
=
20
4x20
1
2xy(4 x2 y2)dydx = 1
8
20
x(4 x2)2dx = 4/3
(b)
/20
20
4r20
r3z sin cos dz drd
=/20
20
1
2 (4r3
r5
)sin cos drd =
8
3/20 sin cos d = 4/3
(c)
/20
/20
20
5 sin3 cos sin cos ddd
=
/20
/20
32
3sin3 cos sin cos dd =
8
3
/20
sin cos d = 4/3
-
7/30/2019 Chapter 15 Solutions Calculus
26/37
Exercise Set 15.7 645
21. M =
20
30
3r
(3 z)r dz drd =20
30
1
2r(3 r)2drd = 27
8
20
d = 27/4
22. M =
20
a0
h0
k zrdz dr d =
20
a0
1
2kh2rdrd =
1
4ka2h2
20
d = ka2h2/2
23. M = 2
0
0
a
0
k3 sin ddd = 2
0
0
1
4ka4 sin dd =
1
2ka4
2
0
d = ka4
24. M =
20
0
21
sin ddd =
20
0
3
2sin dd = 3
20
d = 6
25. x = y = 0 from the symmetry of the region,
V =
20
10
2r2r2
r dz drd =
20
10
(r
2 r2 r3)drd = (8
2 7)/6,
z =1
V
20
10
2r2r2
zr dz drd =6
(8
2 7) (7/12) = 7/(16
2 14);
centroid 0, 0, 7162 14
26. x = y = 0 from the symmetry of the region, V = 8/3,
z =1
V
20
20
2r
zr dz drd =3
8(4) = 3/2; centroid (0, 0, 3/2)
27. x = y = z from the symmetry of the region, V = a3/6,
z =1
V
/20
/20
a0
3 cos sin ddd =6
a3(a4/16) = 3a/8;
centroid (3a/8, 3a/8, 3a/8)
28. x = y = 0 from the symmetry of the region, V =20
/30
40
2 sin ddd = 64/3,
z =1
V
20
/30
40
3 cos sin ddd =3
64(48) = 9/4; centroid (0, 0, 9/4)
29. y = 0 from the symmetry of the region, V = 2
/20
2 cos 0
r20
r dz drd = 3/2,
x =2
V
/20
2 cos 0
r20
r2 cos dz drd =4
3() = 4/3,
z =2
V /2
0 2cos
0 r2
0
rz dz dr d =4
3(5/6) = 10/9; centroid (4/3, 0, 10/9)
30. M=
/20
2 cos 0
4r20
zr dz drd =
/20
2cos 0
1
2r(4 r2)2drd
=16
3
/20
(1 sin6 )d = (16/3)(11/32) = 11/6
-
7/30/2019 Chapter 15 Solutions Calculus
27/37
646 Chapter 15
31. V =
/20
/3/6
20
2 sin ddd =
/20
/3/6
8
3sin dd =
4
3(
3 1)/20
d
= 2(
3 1)/3
32. M =
20
/40
10
3 sin ddd =
20
/40
1
4sin dd =
1
8(2
2)
20
d = (2
2)/4
33. x = y = 0 from the symmetry of density and region,
M =
20
10
1r20
(r2 + z2)r dz drd = /4,
z =1
M
20
10
1r20
z(r2+z2)r dz drd = (4/)(11/120) = 11/30; center of gravity (0, 0, 11/30)
34. x = y = 0 from the symmetry of density and region, M =
20
10
r0
zr dz drd = /4,
z =1
M 2
0 1
0 r
0
z2r dz drd = (4/)(2/15) = 8/15; center of gravity (0, 0, 8/15)
35. x = y = 0 from the symmetry of density and region,
M =
20
/20
a0
k3 sin ddd = ka4/2,
z =1
M
20
/20
a0
k4 sin cos ddd =2
ka4(ka5/5) = 2a/5; center of gravity (0, 0, 2a/5)
36. x = z = 0 from the symmetry of the region, V = 54/3 16/3 = 38/3,
y =1
V
0
0
32
3 sin2 sin ddd =1
V
0
0
65
4sin2 sin dd
=1V
0
658
sin d =3
38(65/4) = 195/152; centroid (0, 195/152, 0)
37. M =
20
0
R0
0e(/R)32 sin ddd =
20
0
1
3(1 e1)R30 sin dd
=4
3(1 e1)0R3
38. (a) The sphere and cone intersect in a circle of radius 0 sin 0,
V = 2
1 0 sin0
0 20r2
r cot 0
r dz drd = 2
1 0 sin0
0 r20 r2 r2 cot 0 drd
=
21
1
330(1 cos3 0 sin3 0 cot 0)d =
1
330(1 cos3 0 sin2 0 cos 0)(2 1)
=1
330(1 cos 0)(2 1).
-
7/30/2019 Chapter 15 Solutions Calculus
28/37
Exercise Set 15.8 647
(b) From Part (a), the volume of the solid bounded by = 1, = 2, = 1, = 2, and
= 0 is1
330(1 cos 2)(2 1)
1
330(1 cos 1)(2 1) =
1
330(cos 1 cos 2)(2 1)
so the volume of the spherical wedge between = 1 and = 2 is
V =1
332(cos 1 cos 2)(2 1)
1
331(cos 1 cos 2)(2 1)
=
1
3 (
3
2 3
1)(cos 1 cos 2)(2 1)(c)
d
dcos = sin so from the Mean-Value Theorem cos 2cos 1 = (21)sin where
is between 1 and 2. Similarlyd
d3 = 32 so 3231 = 32(21) where is between
1 and 2. Thus cos 1cos 2 = sin and 3231 = 32 so V = 2 sin .
39. Iz =
20
a0
h0
r2 r dz dr d =
20
a0
h0
r3dz dr d =1
2a4h
40. Iy = 2
0
a
0
h
0
(r2 cos2 + z2)r dz drd = 2
0
a
0
(hr3 cos2 +1
3
h3r)drd
=
20
1
4a4h cos2 +
1
6a2h3
d =
4
a4h +
3a2h3
41. Iz =
20
a2a1
h0
r2 r dz dr d =
20
a2a1
h0
r3dz dr d =1
2h(a42 a41)
42. Iz =
20
0
a0
(2 sin2 ) 2 sin ddd =
20
0
a0
4 sin3 ddd =8
15a5
EXERCISE SET 15.8
1.(x, y)
(u, v)=
1 43 5 = 17 2. (x, y)(u, v) =
1 4v4u 1 = 1 16uv
3.(x, y)
(u, v)=
cos u sin vsin u cos v = cos u cos v + sin u sin v = cos(u v)
4.(x, y)
(u, v)=
2(v2 u2)(u2 + v2)2
4uv(u2 + v2)2
4uv
(u2 + v2)22(v2 u2)(u2 + v2)2
= 4/(u2 + v2)2
5. x =2
9u +
5
9v, y = 1
9u +
2
9v;
(x, y)
(u, v)=
2/9 5/91/9 2/9 = 19
6. x = ln u, y = uv;(x, y)
(u, v)=
1/u 0v u = 1
-
7/30/2019 Chapter 15 Solutions Calculus
29/37
648 Chapter 15
7. x =
u + v/
2, y =
v u/2; (x, y)(u, v)
=
1
2
2
u + v
1
2
2
u + v
12
2
v u1
2
2
v u
=
1
4
v2 u2
8. x = u3/2/v1/2, y = v1/2/u1/2;(x, y)
(u, v)
=
3u1/2
2v1/2 u
3/2
2v3/2
v1/22u3/2
12u1/2v1/2
=1
2v
9.(x, y, z)
(u, v, w)=
3 1 01 0 20 1 1
= 5
10.(x, y, z)
(u, v, w)=
1 v u 0
v vw u uw uvvw uw uv
= u2v
11. y = v, x = u/y = u/v,z = w
x = w
u/v;(x, y, z)
(u,v,w)=
1/v u/v2 00 1 0
1/v u/v2 1 = 1/v
12. x = (v + w)/2, y = (u w)/2, z = (u v)/2, (x, y, z)(u, v, w)
=
0 1/2 1/2
1/2 0 1/21/2 1/2 0
= 1
4
13.
x
y
(0, 2)
(1, 0) (1, 0)
(0, 0)
14.
1 2 3
1
2
3
4
x
y
(0, 0) (4, 0)
(3, 4)
15.
-3 3
-3
3
x
y
(2, 0)
(0, 3) 16.
1 2
1
2
x
y
17. x =1
5u +
2
5v, y = 2
5u +
1
5v,
(x, y)
(u, v)=
1
5;
1
5
S
u
vdAuv =
1
5
31
41
u
vdu dv =
3
2ln 3
-
7/30/2019 Chapter 15 Solutions Calculus
30/37
Exercise Set 15.8 649
18. x =1
2u +
1
2v, y =
1
2u 1
2v,
(x, y)
(u, v)= 1
2;
1
2
S
veuvdAuv =1
2
41
10
veuvdu dv =1
2(e4 e 3)
19. x = u + v, y = u v, (x, y)(u, v)
= 2; the boundary curves of the region S in the uv-plane are
v = 0, v = u, and u = 1 so 2 S
sin u cos vdAuv = 2 1
0
u
0
sin u cos v dv du = 1
1
2
sin2
20. x =
v/u, y =
uv so, from Example 3,(x, y)
(u, v)= 1
2u; the boundary curves of the region S in
the uv-plane are u = 1, u = 3, v = 1, and v = 4 so
S
uv2
1
2u
dAuv =
1
2
41
31
v2du dv = 21
21. x = 3u, y = 4v,(x, y)
(u, v)= 12; S is the region in the uv-plane enclosed by the circle u2 + v2 = 1.
Use polar coordinates to obtainS
12
u2
+ v2
(12) dAuv = 14420
10 r
2
dr d = 96
22. x = 2u, y = v,(x, y)
(u, v)= 2; S is the region in the uv-plane enclosed by the circle u2 + v2 = 1. Use
polar coordinates to obtain
S
e(4u2+4v2)(2) dAuv = 2
20
10
re4r2
dr d = (1 e4)/2
23. Let S be the region in the uv-plane bounded by u2 + v2 = 1, so u = 2x, v = 3y,
x = u/2, y = v/3,(x, y)
(u, v)
= 1/2 00 1/3
= 1/6, use polar coordinates to get1
6
S
sin(u2 + v2)dudv =1
6
/20
10
r sin r2 drd =
24( cos r2)
10
=
24(1 cos1)
24. u = x/a,v = y/b,x = au,y = bv;(x, y)
(u, v)= ab; A = ab
20
10
r d r d = ab
25. x = u/3, y = v/2, z = w,(x, y, z)
(u,v,w)= 1/6; S is the region in uvw-space enclosed by the sphere
u2 + v2 + w2 = 36 so
S
u2
9
1
6dVuvw =
1
54
20
0
60
( sin cos )22 sin d d d
=1
54
20
0
60
4 sin3 cos2 d d d =192
5
-
7/30/2019 Chapter 15 Solutions Calculus
31/37
650 Chapter 15
26. Let G1 be the region u2 + v2 + w2 1, with x = au,y = bv,z = cw, (x, y, z)
(u, v, w)= abc; then use
spherical coordinates in uvw-space:
Ix =
G
(y2 + z2)dxdydz = abc
G1
(b2v2 + c2w2) dudv dw
= 2
0
0
1
0
abc(b2 sin2 sin2 + c2 cos2 )4 sin ddd
=
20
abc
15(4b2 sin2 + 2c2)d =
4
15abc(b2 + c2)
27. u = = cot1(x/y), v = r =
x2 + y2
28. u = r =
x2 + y2, v = ( + /2)/ = (1/)tan1(y/x) + 1/2
29. u =3
7x 2
7y, v = 1
7x +
3
7y 30. u = x + 4
3y, v = y
31. Let u = y
4x, v = y + 4x, then x =
1
8
(v
u), y =
1
2
(v + u) so(x, y)
(u, v)
=
1
8
;
1
8
S
u
vdAuv =
1
8
52
20
u
vdu dv =
1
4ln
5
2
32. Let u = y + x, v = y x, then x = 12
(u v), y = 12
(u + v) so(x, y)
(u, v)=
1
2;
12
S
uv dAuv = 12
20
10
uv du dv = 12
33. Let u = x
y, v = x + y, then x =
1
2
(v + u), y =1
2
(v
u) so
(x, y)
(u, v)
=1
2
; the boundary curves of
the region S in the uv-plane are u = 0, v = u, and v = /4; thus
1
2
S
sin u
cos vdAuv =
1
2
/40
v0
sin u
cos vdu dv =
1
2[ln(
2 + 1) /4]
34. Let u = y x, v = y + x, then x = 12
(v u), y = 12
(u + v) so(x, y)
(u, v)= 1
2; the boundary
curves of the region S in the uv-plane are v = u, v = u, v = 1, and v = 4; thus1
2
S
eu/vdAuv =1
2
41
vv
eu/vdu dv =15
4(e e1)
35. Let u = y/x,v = x/y2, then x = 1/(u2v), y = 1/(uv) so(x, y)
(u, v)=
1
u4v3;
S
1
u4v3dAuv =
41
21
1
u4v3du dv = 35/256
-
7/30/2019 Chapter 15 Solutions Calculus
32/37
Exercise Set 15.8 651
36. Let x = 3u, y = 2v,(x, y)
(u, v)= 6; S is the region in the uv-plane enclosed by the circle u2 + v2 = 1
so
R
(9 x y)dA =S
6(9 3u 2v)dAuv = 620
10
(9 3r cos 2r sin )rdrd = 54
37. x = u, y = w/u,z = v + w/u,(x, y, z)
(u,v,w)
=
1
u
;
S
v2w
udVuvw =
42
10
31
v2w
udu dv dw = 2ln 3
38. u = xy,v = yz,w = xz, 1 u 2, 1 v 3, 1 w 4,
x =
uw/v,y =
uv/w,z =
vw/u,(x, y, z)
(u,v,w)=
1
2
uvw
V =
G
dV =
21
31
41
1
2
uvwdw dv du = 4(
2 1)(
3 1)
39. (b) If x = x(u, v), y = y(u, v) where u = u(x, y), v = v(x, y), then by the chain rule
x
u
u
x+
x
v
v
x=
x
x= 1,
x
u
u
y+
x
v
v
y=
x
y= 0
y
u
u
x+
y
v
v
x=
y
x= 0,
y
u
u
y+
y
v
v
y=
y
y= 1
40. (a)(x, y)
(u, v)=
1 v uv u = u; u = x + y, v = yx + y ,
(u, v)
(x, y)=
1 1
y/(x + y)2 x/(x + y)2
=x
(x + y)2+
y
(x + y)2=
1
x + y=
1
u;
(u, v)
(x, y)
(x, y)
(u, v)= 1
(b)(x, y)
(u, v)=
v u0 2v = 2v2; u = x/y, v = y,
(u, v)
(x, y)=
1/
y x/(2y3/2)0 1/(2
y)
= 12y = 12v2 ; (u, v)(x, y) (x, y)(u, v) = 1(c)
(x, y)
(u, v)=
u vu v
= 2uv; u = x + y, v = x y,
(u, v)(x, y) = 1/(2
x + y) 1/(2
x + y)
1/(2x y) 1/(2x y) = 12x2 y2 = 12uv ; (u, v)(x, y) (x, y)(u, v) = 1
41.(u, v)
(x, y)= 3xy4 = 3v so
(x, y)
(u, v)=
1
3v;
1
3
S
sin u
vdAuv =
1
3
21
2
sin u
vdu dv = 2
3ln 2
-
7/30/2019 Chapter 15 Solutions Calculus
33/37
652 Chapter 15
42.(u, v)
(x, y)= 8xy so
(x, y)
(u, v)=
1
8xy; xy
(x, y)(u, v) = xy
1
8xy
=
1
8so
1
8
S
dAuv =1
8
169
41
du dv = 21/8
43.(u, v)
(x, y)= 2(x2 + y2) so (x, y)
(u, v)= 1
2(x2 + y2);
(x4 y4)exy(x, y)(u, v)
= x4 y42(x2 + y2) exy = 12 (x2 y2)exy = 12 veu so1
2
S
veudAuv =1
2
43
31
veudu dv =7
4(e3 e)
44. Set u = x + y + 2z, v = x 2y + z, w = 4x + y + z, then (u, v, w)(x, y, z)
=
1 1 21 2 14 1 1
= 18, and
V =R
dxdy dz =66
22
33
(x, y, z)
(u, v, w) dudv dw = 6(4)(12)
1
18 = 16
45. (a) Let u = x + y, v = y, then the triangle R with vertices (0, 0), (1, 0) and (0, 1) becomes thetriangle in the uv-plane with vertices (0, 0), (1, 0), (1, 1), andR
f(x + y)dA =
10
u0
f(u)(x, y)
(u, v)dvdu =
10
uf(u) du
(b)
10
ueu du = (u 1)eu10
= 1
46. (a) (x,y,z)(r,,z)
=
cos
r sin 0
sin r cos 00 0 1
= r,(x, y, z)(r,,z) = r
(b)(x, y, z)
(,,)=
sin cos cos cos sin sin sin sin cos sin sin cos
cos sin 0
= 2 sin ;(x, y, z)(,,)
= 2 sin
CHAPTER 15 SUPPLEMENTARY EXERCISES
3. (a) R
dA (b) G
dV (c) R
1 +
z
x2
+ z
y 2
dA
4. (a) x = a sin cos , y = a sin sin , z = a cos , 0 2, 0 (b) x = a cos , y = a sin , z = z, 0 2, 0 z h
7.
10
1+1y21
1y2f(x, y) dxdy 8.
20
2xx
f(x, y) dydx +
32
6xx
f(x, y) dy dx
-
7/30/2019 Chapter 15 Solutions Calculus
34/37
Chapter 15 Supplementary Exercises 653
9. (a) (1, 2) = (b, d), (2, 1) = (a, c), so a = 2, b = 1, c = 1, d = 2
(b)
R
dA =
10
10
(x, y)
(u, v)dudv =
10
10
3dudv = 3
10. If 0 < x,y < then 0 < sin
xy 1, with equality only on the hyperbola xy = 2/4, so
0 =
0
0
0 dydx <
0
0
sin
xy dy dx <
0
0
1 dydx = 2
11.
11/2
2x cos(x2) dx =1
sin(x2)
11/2
= 1/(
2)
12.
20
x2
2ey
3
x=2yx=y
dy =3
2
20
y2ey3
dy =1
2ey
3
20
=1
2
e8 1
13.
10
22y
exey dxdy 14.
0
x0
sin x
xdydx
15.
6
1
x
y
y = sinx
y = tan (x/2)
16.
0
p/2
p/6
r = a
r= a(1 + cos u)
17. 2
80
y1/30
x2 sin y2dxdy =2
3
80
y sin y2 dy = 13
cos y280
=1
3(1 cos 64) 0.20271
18.
/2
0
2
0
(4 r2)rdrd = 2
19. sin2 = 2sin cos =2xy
x2 + y2, and r = 2a sin is the circle x2 + (y a)2 = a2, so
a0
a+a2x2aa2x2
2xy
x2 + y2dydx =
a0
x
ln
a +
a2 x2 ln
a
a2 x2
dx = a2
20.
/2/4
20
4r2(cos sin ) rdrd = 4cos2/2/4
= 4
21.20
20
16r4
r2 cos2 r dz dr d =20
cos2 d20
r3(16 r4) dr = 32
22.
/20
/20
10
1
1 + 22 sin ddd =
1
4
2
/20
sin d
=
1 4
2
( cos )/20
=
1 4
2
-
7/30/2019 Chapter 15 Solutions Calculus
35/37
654 Chapter 15
23. (a)
20
/30
a0
(2 sin2 )2 sin ddd =
20
/30
a0
4 sin3 ddd
(b)
20
3a/20
a2r2r/3
r2 dz rdr d =
20
3a/20
a2r2r/3
r3 dz dr d
(c) 3a/2
3a/2
(3a2/4)x2
(3a2/4)
x2
a2x2y2
x2+y2/
3
(x2 + y2) dz dy dx
24. (a)
40
4xx24xx2
4xx2+y2
dz dy dx
(b)
/2/2
4cos 0
4r cos r2
r dz drd
25.
20
2y/2(y/2)1/3
dxdy =
20
2 y
2y
2
1/3dy =
2y y
2
4 3
2
y2
4/320
=3
2
26. A = 6 /6
0
cos 3
0
r d r d = 3 /6
0
cos2 3 = /4
27. V =
20
a/30
a3r
r dz drd = 2
a/30
r(a
3r) dr =a3
9
28. The intersection of the two surfaces projects onto the yz-plane as 2y2 + z2 = 1, so
V = 4
1/20
12y20
1y2y2+z2
dx dz dy
= 4
1/20
12y20
(1 2y2 z2) dz dy = 41/20
2
3(1 2y2)3/2 dy =
2
4
29. ru rv = 2u2 + 2v2 + 4,
S =
u2+v24
2u2 + 2v2 + 4 dA =
20
20
2
r2 + 2 rdrd =8
3(3
3 1)
30. ru rv =
1 + u2, S =
20
3u0
1 + u2dvdu =
20
3u
1 + u2du = 53/2 1
31. (ru rv)u=1v=2
= 2,4, 1, tangent plane 2x + 4y z = 5
32. u =
3, v = 0, (ru
rv)u=3
v=0
=
18, 0,
3
, tangent plane 6x + z =
9
33. A =
44
2+y2/8y2/4
dxdy =
44
2 y
2
8
dy =
32
3; y = 0 by symmetry;
44
2+y2/8y2/4
xdxdy =
44
2 +
1
4y2 3
128y4
dy =256
15, x =
3
32
256
15=
8
5; centroid
8
5, 0
-
7/30/2019 Chapter 15 Solutions Calculus
36/37
Chapter 15 Supplementary Exercises 655
34. A = ab/2, x = 0 by symmetry,aa
b1x2/a20
y d y d x =1
2
aa
b2(1 x2/a2)dx = 2ab2/3, centroid
0,4b
3
35. V =1
3a2h, x = y = 0 by symmetry,
20
a0
hrh/a0 rz dz dr d =
a0 rh
2 1
r
a2
dr = a2
h2
/12, centroid (0, 0, h/4)
36. V =
22
4x2
4y0
dz dy dx =
22
4x2
(4 y)dydx =22
8 4x2 + 1
2x4
dx =256
15,
22
4x2
4y0
y dz dydx =
22
4x2
(4y y2) dydx =22
1
3x6 2x4 + 32
3
dx =
1024
3522
4x2
4y0
z dz dy dx =
22
4x2
1
2(4 y)2dydx =
22
x
6
6+ 2x4 8x2 + 32
3
dx =
2048
105
x = 0 by symmetry, centroid
0,
12
7,
8
7
37. The two quarter-circles with center at the origin and of radius A and 2A lie inside and outsideof the square with corners (0, 0), (A, 0), (A, A), (0, A), so the following inequalities hold:/20
A0
1
(1 + r2)2rdr d
A0
A0
1
(1 + x2 + y2)2dxdy
/20
2A0
1
(1 + r2)2rdrd
The integral on the left can be evaluated asA2
4(1 + A2)and the integral on the right equals
2A2
4(1 + 2A2). Since both of these quantities tend to
4as A +, it follows by sandwiching that
+0
+0
1
(1 + x2 + y2)2dxdy =
4.
38. The centroid of the circle which generates the tube travels a distance
s =
40
sin2 t + cos2 t + 1/16 dt =
17, so V = (1/2)2
17 =
172/4.
39. (a) Let S1 be the set of points (x, y, z) which satisfy the equation x2/3 + y2/3 + z2/3 = a2/3, and
let S2 be the set of points (x,y,z) where x = a(sin cos )3, y = a(sin sin )3, z = a cos3 ,
0 , 0 < 2.If (x, y, z) is a point of S2 then
x2/3 + y2/3 + z2/3 = a2/3[(sin cos )3 + (sin sin )3 + cos3 ] = a2/3
so (x, y, z) belongs to S1.If (x, y, z) is a point of S1 then x
2/3 + y2/3 + z2/3 = a2/3. Let
x1 = x1/3, y1 = y1/3, z1 = z1/3, a1 = a1/3. Then x21 + y21 + z21 = a21, so in spherical coordinatesx1 = a1 sin cos , y1 = a1 sin sin , z1 = a1 cos , with
= tan1
y1x1
= tan1
yx
1/3, = cos1
z1a1
= cos1z
a
1/3. Then
x = x31 = a31(sin cos )
3 = a(sin cos )3, similarly y = a(sin sin )3, z = a cos so (x, y, z)belongs to S2. Thus S1 = S2
-
7/30/2019 Chapter 15 Solutions Calculus
37/37
656 Chapter 15
(b) Let a = 1 and r = (cos sin )3i + (sin sin )3j + cos3 k, then
S = 8
/20
/20
r rdd
= 72
/20
/20
sin cos sin4 cos
cos2 + sin2 sin2 cos2 dd 4.4506
(c)
(x, y, z)
(,,) =
sin3 cos3 3 sin2 cos cos3 3 sin3 cos2 sin sin
3
sin3
3 sin2
cos sin3
3 sin3
sin2
cos cos3 3 cos2 sin 0
= 92 cos2 sin2 cos2 sin5 ,
V = 9
20
0
a0
2 cos2 sin2 cos2 sin5 ddd =4
35a3
40. V =4
3a3, d =
3
4a3
a
dV =3
4a3
0
20
a0
3 sin ddd =3
4a32(2)
a4
4=
3
4a
41. (a) (x/a)2+(y/b)2+(z/c)2 = sin2 cos2 +sin2 sin2 +cos2 = sin2 +cos2 = 1, an ellipsoid
(b) r(, ) = 2sin cos , 3sin sin , 4cos ; rr = 26sin2
cos , 4sin
2
sin , 3cos sin ,r r = 2
16 sin4 + 20 sin4 cos2 + 9 sin2 cos2 ,
S =
20
0
2
16sin4 + 20sin4 cos2 + 9 sin2 cos2 dd 111.5457699