Calculo 2 - Anton resolucoes

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Multiple Integrals
Exercise Set 14.1
1.
∫ 1
0
∫ 2
0
(x+ 3) dy dx =
∫ 1
0
(2x+ 6) dx = 7.
2.
∫ 3
1
∫ 1
−1
(2x− 4y) dy dx =
∫ 3
1
4x dx = 16.
3.
∫ 4
2
∫ 1
0
x2y dx dy =
∫ 4
2
1
3
y dy = 2.
4.
∫ 0
−2
∫ 2
−1
(x2 + y2) dx dy =
∫ 0
−2
(3 + 3y2) dy = 14.
5.
∫ ln 3
0
∫ ln 2
0
ex+y dy dx =
∫ ln 3
0
ex dx = 2.
6.
∫ 2
0
∫ 1
0
y sinx dy dx =
∫ 2
0
1
2
sinx dx =
1− cos 2
2
.
7.
∫ 0
−1
∫ 5
2
dx dy =
∫ 0
−1
3 dy = 3.
8.
∫ 6
4
∫ 7
−3
dy dx =
∫ 6
4
10 dx = 20.
9.
∫ 1
0
∫ 1
0
x
(xy + 1)2
dy dx =
∫ 1
0
(
1− 1
x+ 1
)
dx = 1− ln 2.
10.
∫ π
π/2
∫ 2
1
x cosxy dy dx =
∫ π
π/2
(sin 2x− sinx) dx = −2.
11.
∫ ln 2
0
∫ 1
0
xy ey
2x dy dx =
∫ ln 2
0
1
2
(ex − 1) dx = 1− ln 2
2
.
12.
∫ 4
3
∫ 2
1
1
(x+ y)2
dy dx =
∫ 4
3
(
1
x+ 1
− 1
x+ 2
)
dx = ln(25/24).
13.
∫ 1
−1
∫ 2
−2
4xy3 dy dx =
∫ 1
−1
0 dx = 0.
14.
∫ 1
0
∫ 1
0
xy√
x2 + y2 + 1
dy dx =
∫ 1
0
[x(x2 + 2)1/2 − x(x2 + 1)1/2] dx = 1
3
(3
√
3− 4
√
2 + 1).
675
676 Chapter 14
15.
∫ 1
0
∫ 3
2
x
√
1− x2 dy dx =
∫ 1
0
x(1− x2)1/2 dx = 1
3
.
16.
∫ π/2
0
∫ π/3
0
(x sin y − y sinx)dy dx =
∫ π/2
0
(
x
2
− π
2
18
sinx
)
dx =
π2
144
.
17. (a) x∗k = k/2 − 1/4, k = 1, 2, 3, 4; y∗l = l/2 − 1/4, l = 1, 2, 3, 4,
∫∫
R
f(x, y) dx dy ≈
4∑
k=1
4∑
l=1
f(x∗k, y
∗
l )∆Akl =
4∑
k=1
4∑
l=1
[(
k
2
− 1
4
)2
+
(
l
2
− 1
4
)](
1
2
)2
=
37
4
.
(b)
∫ 2
0
∫ 2
0
(x2 + y) dx dy =
28
3
; the error is
∣∣∣∣
37
4
− 28
3
∣∣∣∣ =
1
12
.
18. (a) x∗k = k/2 − 1/4, k = 1, 2, 3, 4; y∗l = l/2 − 1/4, l = 1, 2, 3, 4,
∫∫
R
f(x, y) dx dy ≈
4∑
k=1
4∑
l=1
f(x∗k, y
∗
l )∆Akl =
4∑
k=1
4∑
l=1
[(
k
2
− 1
4
)
− 2
(
l
2
− 1
4
)](
1
2
)2
= −4.
(b)
∫ 2
0
∫ 2
0
(x− 2y) dx dy = −4; the error is zero.
19. The solid is a rectangular box with sides of length 1, 5, and 4, so its volume is 1 · 5 · 4 = 20;
∫ 5
0
∫ 2
1
4 dx dy =
∫ 5
0
4x
]2
x=1
dy =
∫ 5
0
4 dy = 20.
(1, 0, 4)
(2, 5, 0)
x
y
z
20. Two copies of the solid will fit together to form a rectangular box whose base is a square of side 1 and whose
height is 2, so the solid’s volume is (12 · 2)/2 = 1;
∫ 1
0
∫ 1
0
(2− x− y) dx dy =
∫ 1
0
[
2x− 1
2
x2 − xy
]1
x=0
dy =
∫ 1
0
(
3
2
− y
)
dy =
[
3
2
y − 1
2
y2
]1
0
= 1.
(1, 1, 0)
(0, 0, 2)
x
y
z
Exercise Set 14.1 677
21.
(0, 0, 5)
(3, 0, 4)
(3, 4, 0)
(0, 4, 3)
z
x
y
22.
(2, 2, 0)
(2, 2, 8)
x
y
z
23. False. ∆Ak represents the area of such a region.
24. True.
∫∫
R
f(x, y) dA =
∫ 4
1
∫ 3
0
f(x, y) dy dx =
∫ 4
1
2x dx = x2
]4
1
= 15.
25. False.
∫∫
R
f(x, y) dA =
∫ 5
1
∫ 4
2
f(x, y) dy dx.
26. True, by equation (12).
27.
∫∫
R
f(x, y) dA =
∫ b
a
[∫ d
c
g(x)h(y) dy
]
dx =
∫ b
a
g(x)
[∫ d
c
h(y) dy
]
dx =
[∫ b
a
g(x) dx
][∫ d
c
h(y) dy
]
.
28. The integral of tanx (an odd function) over the interval [−1, 1] is zero, so the iterated integral is also zero.
29. V =
∫ 5
3
∫ 2
1
(2x+ y) dy dx =
∫ 5
3
(
2x+
3
2
)
dx = 19.
30. V =
∫ 3
1
∫ 2
0
(3x3 + 3x2y) dy dx =
∫ 3
1
(6x3 + 6x2) dx = 172.
31. V =
∫ 2
0
∫ 3
0
x2 dy dx =
∫ 2
0
3x2 dx = 8.
32. V =
∫ 3
0
∫ 4
0
5
(
1− x
3
)
dy dx =
∫ 3
0
5
(
4− 4x
3
)
dx = 30.
33.
∫ 1/2
0
∫ π
0
x cos(xy) cos2 πx dy dx =
∫ 1/2
0
cos2 πx sin(xy)
]π
0
dx =
∫ 1/2
0
cos2 πx sinπx dx = − 1
3π
cos3 πx
]1/2
0
=
1
3π
.
34. (a)
y
x
z
5
3
(0, 2, 2)
(5, 3, 0)
678 Chapter 14
(b) V =
∫ 5
0
∫ 2
0
y dy dx+
∫ 5
0
∫ 3
2
(−2y + 6) dy dx = 10 + 5 = 15.
35. fave =
1
48
∫ 6
0
∫ 8
0
xy2 dx dy =
1
48
∫ 6
0
(
1
2
x2y2
]x=8
x=0
)
dy =
1
48
∫ 6
0
32y2 dy = 48.
36. fave =
1
18
∫ 6
0
∫ 3
0
x2 + 7y dx dy =
1
18
∫ 6
0
(
1
3
x3 + 7yx
]x=3
x=0
)
dy =
1
18
∫ 6
0
9 + 21y dy = 24.
37. fave =
2
π
∫ π/2
0
∫ 1
0
y sinxy dx dy =
2
π
∫ π/2
0
(
− cosxy
]x=1
x=0
)
dy =
2
π
∫ π/2
0
(1− cos y) dy = 1− 2
π
.
38. fave =
1
3
∫ 3
0
∫ 1
0
x(x2 + y)1/2 dx dy =
∫ 3
0
1
9
[(1 + y)3/2 − y3/2] dy = 2
45
(31− 9
√
3).
39. Tave =
1
2
∫ 1
0
∫ 2
0
(
10− 8x2 − 2y2
)
dy dx =
1
2
∫ 1
0
(
44
3
− 16x2
)
dx =
(
14
3
)◦
C.
40. fave =
1
A(R)
∫ b
a
∫ d
c
k dy dx =
1
A(R)
(b− a)(d− c)k = k.
41. 1.381737122
42. 2.230985141
43. The first integral equals 1/2, the second equals −1/2. This does not contradict Theorem 14.1.3 because the
integrand is not continuous at (x, y) = (0, 0); if f(x, y) =
y − x
(x+ y)3
, then lim
x→0
f(x, 0) = lim
x→0
−1
x2
→ −∞.
44. V =
∫ 1
0
∫ π
0
xy3 sin(xy) dx dy =
∫ 1
0
[
y sin(xy)− xy2 cos(xy)
]π
x=0
dy =
∫ 1
0
[y sin(πy)− πy2 cos(πy)] dy =
=
[
3
π2
sin(πy)− 3
π
y cos(πy)− y2 sin(πy)
]1
0
=
3
π
.
45. If R is a rectangular region defined by a ≤ x ≤ b, c ≤ y ≤ d, then the volume given in equation (5) can be written
as an iterated integral: V =
∫∫
R
f(x, y) dA =
∫ b
a
(∫ d
c
f(x, y) dy
)
dx. The inner integral,
∫ d
c
f(x, y) dy, is the
area A(x) of the cross-section with x-coordinate x of the solid enclosed between R and the surface z = f(x, y). So
V =
∫ b
a
A(x) dx, as found in Section 6.2.
Exercise Set 14.2
1.
∫ 1
0
∫ x
x2
xy2 dy dx =
∫ 1
0
1
3
(x4 − x7) dx = 1
40
.
2.
∫ 3/2
1
∫ 3−y
y
y dx dy =
∫ 3/2
1
(3y − 2y2)dy = 7
24
.
3.
∫ 3
0
∫ √9−y2
0
y dx dy =
∫ 3
0
y
√
9− y2 dy = 9.
Exercise Set 14.2 679
4.
∫ 1
1/4
∫ x
x2
√
x/y dy dx =
∫ 1
1/4
∫ x
x2
x1/2y−1/2 dy dx =
∫ 1
1/4
2(x− x3/2) dx = 13
80
.
5.
∫ √2π
√
π
∫ x3
0
sin(y/x) dy dx =
∫ √2π
√
π
[−x cos(x2) + x] dx = π
2
.
6.
∫ 1
−1
∫ x2
−x2
(x2 − y) dy dx =
∫ 1
−1
2x4 dx =
4
5
.
7.
∫ 1
0
∫ x
0
y
√
x2 − y2 dy dx =
∫ 1
0
1
3
x3 dx =
1
12
.
8.
∫ 2
1
∫ y2
0
ex/y
2
dx dy =
∫ 2
1
(e− 1)y2 dy = 7(e− 1)
3
.
9. (a)
∫ 2
0
∫ x2
0
f(x, y) dy dx. (b)
∫ 4
0
∫ 2
√
y
f(x, y) dx dy.
10. (a)
∫ 1
0
∫ √x
x2
f(x, y) dy dx. (b)
∫ 1
0
∫ √y
y2
f(x, y) dx dy.
11. (a)
∫ 2
1
∫ 3
−2x+5
f(x, y) dy dx+
∫ 4
2
∫ 3
1
f(x, y) dy dx+
∫ 5
4
∫ 3
2x−7
f(x, y) dy dx.
(b)
∫ 3
1
∫ (y+7)/2
(5−y)/2
f(x, y) dx dy.
12. (a)
∫ 1
−1
∫ √1−x2
−
√
1−x2
f(x, y) dy dx. (b)
∫ 1
−1
∫ √1−y2
−
√
1−y2
f(x, y) dx dy.
13. (a)
∫ 2
0
∫ x2
0
xy dy dx =
∫ 2
0
1
2
x5 dx =
16
3
.
(b)
∫ 3
1
∫ (y+7)/2
(5−y)/2
xy dx dy =
∫ 3
1
(3y2 + 3y) dy = 38.
14. (a)
∫ 1
0
∫ √x
x2
(x+ y) dy dx =
∫ 1
0
(
x3/2 +
x
2
− x3 − x
4
2
)
dx =
3
10
.
(b)
∫ 1
−1
∫ √1−x2
−
√
1−x2
x dy dx+
∫ 1
−1
∫ √1−x2
−
√
1−x2
y dy dx =
∫ 1
−1
2x
√
1− x2 dx+ 0 = 0.
15. (a)
∫ 8
4
∫ x
16/x
x2dy dx =
∫ 8
4
(x3 − 16x) dx = 576.
(b)
∫ 4
2
∫ 8
16/y
x2 dx dy +
∫ 8
4
∫ 8
y
x2 dx dy =
∫ 8
4
[
512
3
− 4096
3y3
]
dy +
∫ 8
4
512− y3
3
dy =
640
3
+
1088
3
= 576.
16. (a)
∫ 1
0
∫ 2
1
xy2 dy dx+
∫ 2
1
∫ 2
x
xy2 dy dx =
∫ 1
0
7x/3 dx+
∫ 2
1
8x− x4
3
dx =
7
6
+
29
15
=
31
10
.
680 Chapter 14
(b)
∫ 2
1
∫ y
0
xy2 dx dy =
∫ 2
1
1
2
y4 dy =
31
10
.
17. (a)
∫ 1
−1
∫ √1−x2
−
√
1−x2
(3x− 2y) dy dx =
∫ 1
−1
6x
√
1− x2 dx = 0.
(b)
∫ 1
−1
∫ √1−y2
−
√
1−y2
(3x− 2y) dx dy =
∫ 1
−1
−4y
√
1− y2 dy = 0.
18. (a)
∫ 5
0
∫ √25−x2
5−x
y dy dx =
∫ 5
0
(5x− x2) dx = 125
6
.
(b)
∫ 5
0
∫ √25−y2
5−y
y dx dy =
∫ 5
0
y
(√
25− y2 − 5 + y
)
dy =
125
6
.
19.
∫ 4
0
∫ √y
0
x(1 + y2)−1/2 dx dy =
∫ 4
0
1
2
y(1 + y2)−1/2 dy =
√
17− 1
2
.
20.
∫ π
0
∫ x
0
x cos y dy dx =
∫ π
0
x sinx dx = π.
21.
∫ 2
0
∫ 6−y
y2
xy dx dy =
∫ 2
0
1
2
(36y − 12y2 + y3 − y5) dy = 50
3
.
22.
∫ π/4
0
∫ 1/√2
sin y
x dx dy =
∫ π/4
0
1
4
cos 2y dy =
1
8
.
23.
∫ 1
0
∫ x
x3
(x− 1) dy dx =
∫ 1
0
(−x4 + x3 + x2 − x) dx = − 7
60
.
24.
∫ 1/√2
0
∫ 2x
x
x2 dy dx+
∫ 1
1/
√
2
∫ 1/x
x
x2 dy dx =
∫ 1/√2
0
x3 dx+
∫ 1
1/
√
2
(x− x3)dx = 1
8
.
25.
∫ 2
0
∫ y2
0
sin(y3) dx dy =
∫ 2
0
y2 sin(y3) dy =
1− cos 8
3
.
26.
∫ 1
0
∫ e
ex
x dy dx =
∫ 1
0
(ex− xex) dx = e
2
− 1.
27. (a)
–1–2 0.5 1.5
x
y
1
2
3
4(b) (−1.8414, 0.1586), (1.1462, 3.1462).
Exercise Set 14.2 681
(c)
∫∫
R
x dA ≈
∫ 1.1462
−1.8414
∫ x+2
ex
x dy dx =
∫ 1.1462
−1.8414
x(x+ 2− ex) dx ≈ −0.4044.
(d)
∫∫
R
x dA ≈
∫ 3.1462
0.1586
∫ ln y
y−2
x dx dy =
∫ 3.1462
0.1586
[
ln2 y
2
− (y − 2)
2
2
]
dy ≈ −0.4044.
28. (a) 1 2 3
5
15
25
x
y
R
(b) (1, 3), (3, 27).
(c)
∫ 3
1
∫ 4x3−x4
3−4x+4x2
x dy dx =
∫ 3
1
x[(4x3 − x4)− (3− 4x+ 4x2)] dx = 224
15
.
29. A =
∫ π/4
0
∫ cos x
sin x
dy dx =
∫ π/4
0
(cosx− sinx) dx =
√
2− 1.
30. A =
∫ 1
−4
∫ −y2
3y−4
dx dy =
∫ 1
−4
(−y2 − 3y + 4) dy = 125
6
.
31. A =
∫ 3
−3
∫ 9−y2
1−y2/9
dx dy =
∫ 3
−3
8
(
1− y
2
9
)
dy = 32.
32. A =
∫ 1
0
∫ cosh x
sinh x
dy dx =
∫ 1
0
(coshx− sinhx) dx = 1− e−1.
33. False. The expression on the right side doesn’t make sense. To evaluate an integral of the form
∫ 2x
x2
g(y) dy, x
must have a fixed value. But then we can’t use x as a variable in defining g(y) =
∫ 1
0
f(x, y) dx.
34. True. This is Theorem 14.2.2(a).
35. False. For example, if f(x, y) = x then
∫∫
R
f(x, y) dA =
∫ 1
−1
∫ 1
x2
x dy dx =
∫ 1
−1
xy
]1
y=x2
dx =
∫ 1
−1
x(1 − x2) dx =
[
1
2
x2 − 1
4
x4
]1
−1
= 0, but 2
∫ 1
0
∫ 1
x2
x dy dx =
∫ 1
0
xy
]1
y=x2
dx =
∫ 1
0
x(1− x2) dx =
[
1
2
x2 − 1
4
x4
]1
0
=
1
4
.
36. False. For example, if R is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, then the area of R is 1, but
∫∫
R
xy dA =
∫ 1
0
∫ 1
0
xy dy dx =
∫ 1
0
1
2
xy2
]1
y=0
dx =
∫ 1
0
1
2
x dx =
1
4
x2
]1
0
=
1
4
.
37.
∫ 4
0
∫ 6−3x/2
0
(
3− 3x
4
− y
2
)
dy dx =
∫ 4
0
[(
3− 3x
4
)(
6− 3x
2
)
− 1
4
(
6− 3x
2
)2]
dx = 12.
682 Chapter 14
38.
∫ 2
0
∫ √4−x2
0
√
4− x2 dy dx =
∫ 2
0
(4− x2) dx = 16
3
.
39. V =
∫ 3
−3
∫ √9−x2
−
√
9−x2
(3− x) dy dx =
∫ 3
−3
(
6
√
9− x2 − 2x
√
9− x2
)
dx = 27π.
40. V =
∫ 1
0
∫ x
x2
(x2 + 3y2) dy dx =
∫ 1
0
(2x3 − x4 − x6) dx = 11
70
.
41. V =
∫ 3
0
∫ 2
0
(9x2 + y2) dy dx =
∫ 3
0
(
18x2 +
8
3
)
dx = 170.
42. V =
∫ 1
−1
∫ 1
y2
(1− x) dx dy =
∫ 1
−1
(
1
2
− y2 + y
4
2
)
dy =
8
15
.
43. V =
∫ 3/2
−3/2
∫ √9−4x2
−
√
9−4x2
(y + 3) dy dx =
∫ 3/2
−3/2
6
√
9− 4x2 dx = 27π
2
.
44. V =
∫ 3
0
∫ 3
y2/3
(9− x2) dx dy =
∫ 3
0
(
18− 3y2 + y
6
81
)
dy =
216
7
.
45. V = 4
∫ 1
0
∫ √1−x2
0
(1− x2 − y2) dy dx = 8
3
∫ 1
0
(1− x2)3/2 dx = π
2
.
46. V =
∫ 2
0
∫ √4−x2
0
(x2 + y2) dy dx =
∫ 2
0
[
x2
√
4− x2 + 1
3
(4− x2)3/2
]
dx = 2π.
47.
∫ √2
0
∫ 2
y2
f(x, y) dx dy.
48.
∫ 8
0
∫ x/2
0
f(x, y) dy dx.
49.
∫ e2
1
∫ 2
ln x
f(x, y) dy dx.
50.
∫ 1
0
∫ e
ey
f(x, y) dx dy.
51.
∫ π/2
0
∫ sin x
0
f(x, y) dy dx.
52.
∫ 1
0
∫ √x
x2
f(x, y) dy dx.
53.
∫ 4
0
∫ y/4
0
e−y
2
dx dy =
∫ 4
0
1
4
ye−y
2
dy =
1− e−16
8
.
54.
∫ 1
0
∫ 2x
0
cos(x2) dy dx =
∫ 1
0
2x cos(x2) dx = sin 1.
Exercise Set 14.2 683
55.
∫ 2
0
∫ x2
0
ex
3
dy dx =
∫ 2
0
x2ex
3
dx =
e8 − 1
3
.
56.
∫ ln 3
0
∫ 3
ey
x dx dy =
1
2
∫ ln 3
0
(9− e2y) dy = 9 ln 3− 4
2
.
57. (a)
∫ 4
0
∫ 2
√
x
sin
(
πy3
)
dy dx; the inner integral is non-elementary.
∫ 2
0
∫ y2
0
sin
(
πy3
)
dx dy =
∫ 2
0
y2 sin
(
πy3
)
dy = − 1
3π
cos
(
πy3
) ]2
0
= 0.
(b)
∫ 1
0
∫ π/2
sin−1 y
sec2(cosx) dx dy; the inner integral is non-elementary.
∫ π/2
0
∫ sin x
0
sec2(cosx) dy dx =
∫ π/2
0
sec2(cosx) sinx dx = tan 1.
58. V = 4
∫ 2
0
∫ √4−x2
0
(x2 + y2) dy dx = 4
∫ 2
0
(
x2
√
4− x2 + 1
3
(4− x2)3/2
)
dx =
=
∫ π/2
0
(
64
3
+
64
3
sin2 θ − 128
3
sin4 θ
)
dθ =
64
3
π
2
+
64
3
π
4
− 128
3
π
2
1 · 3
2 · 4 = 8π (by substituting x = 2 sin θ).
59. The region is symmetric with respect to the y-axis, and the integrand is an odd function of x, hence the answer
is zero.
60. This is the volume in the first octant under the surface z =
√
1− x2 − y2, so 1/8 of the volume of the sphere of
radius 1, thus
π
6
.
61. Area of triangle is 1/2, so fave = 2
∫ 1
0
∫ 1
x
1
1 + x2
dy dx = 2
∫ 1
0
[
1
1 + x2
− x
1 + x2
]
dx =
π
2
− ln 2.
62. Area =
∫ 2
0
(3x−x2−x) dx = 4
3
, so fave =
3
4
∫ 2
0
∫ 3x−x2
x
(x2−xy) dy dx = 3
4
∫ 2
0
(
−2x3 + 2x4 − x
5
2
)
dx = −3
4
8
15
=
−2
5
.
63. Tave =
1
A(R)
∫∫
R
(5xy + x2) dA. The diamond has corners (±2, 0), (0,±4) and thus has area A(R) = 41
2
2(4) =
16m2. Since 5xy is an odd function of x (as well as y),
∫∫
R
5xy dA = 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
∫ 4−2x
0
x2 dy dx =
1
4
∫ 2
0
(4− 2x)x2 dx = 1
4
[
4
3
x3 − 1
2
x4
]2
0
=
(
2
3
)◦
C.
64. The area of the lens is πR2 = 4π and the average thickness Tave is Tave =
4
4π
∫ 2
0
∫ √4−x2
0
(
1− x
2 + y2
4
)
dy dx =
1
π
∫ 2
0
1
6
(4− x2)3/2 dx = 8
3π
∫ π/2
0
sin4 θ dθ =
8
3π
1 · 3
2 · 4
π
2
=
1
2
in (by substituting x = 2 cos θ).
65. y = sinx and y = x/2 intersect at x = 0 and x = a ≈ 1.895494, so V =
∫ a
0
∫ sin x
x/2
√
1 + x+ y dy dx ≈ 0.676089.
684 Chapter 14
67. See Example 7. Given an iterated integral
∫ d
c
∫ h2(y)
h1(y)
f(x, y) dx dy, draw the type II region R defined by c ≤ y ≤ d,
h1(y) ≤ x ≤ h2(y). If R is also a type I region, try to determine the numbers a and b and functions g1(x)
and g2(x) such that R is also described by a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x). Then
∫ d
c
∫ h2(y)
h1(y)
f(x, y) dx dy =
∫ b
a
∫ g2(x)
g1(x)
f(x, y) dy dx. This isn’t always possible: R may not be a type I region. Even if it is, it may not be
possible to find formulas for g1(x) and g2(x).
Exercise Set 14.3
1.
∫ π/2
0
∫ sin θ
0
r cos θ dr dθ =
∫ π/2
0
1
2
sin2 θ cos θ dθ =
1
6
.
2.
∫ π
0
∫ 1+cos θ
0
r dr dθ =
∫ π
0
1
2
(1 + cos θ)2 dθ =
3π
4
.
3.
∫ π/2
0
∫ a sin θ
0
r2 dr dθ =
∫ π/2
0
a3
3
sin3 θ dθ =
2
9
a3.
4.
∫ π/6
0
∫ cos 3θ
0
r dr dθ =
∫ π/6
0
1
2
cos2 3θ dθ =
π
24
.
5.
∫ π
0
∫ 1−sin θ
0
r2 cos θ dr dθ =
∫ π
0
1
3
(1− sin θ)3 cos θ dθ = 0.
6.
∫ π/2
0
∫ cos θ
0
r3 dr dθ =
∫ π/2
0
1
4
cos4 θ dθ =
3π
64
.
7. A =
∫ 2π
0
∫ 1−cos θ
0
r dr dθ =
∫ 2π
0
1
2
(1− cos θ)2 dθ = 3π
2
.
8. A = 4
∫ π/2
0
∫ sin 2θ
0
r dr dθ = 2
∫ π/2
0
sin2 2θ dθ =
π
2
.
9. A =
∫ π/2
π/4
∫ 1
sin 2θ
r dr dθ =
∫ π/2
π/4
1
2
(1− sin2 2θ) dθ = π
16
.
10. A = 2
∫ π/3
0
∫ 2
sec θ
r dr dθ =
∫ π/3
0
(4− sec2 θ) dθ = 4π
3
−
√
3.
11. A =
∫ 5π/6
π/6
∫ 4 sin θ
2
f(r, θ) r dr dθ.
Exercise Set 14.3 685
x
y
2
4
r = 2
r = 4 sin!
12. A =
∫ 3π/2
π/2
∫ 1
1+cos θ
f(r, θ)r dr dθ.
x
y
2
4
r = 1
r = 1 + cos!
13. V = 8
∫ π/2
0
∫ 3
1
r
√
9− r2 dr dθ.
14. V = 2
∫ π/2
0
∫ 2 sin θ
0
r2dr dθ.
15. V = 2
∫ π/2
0
∫ cos θ
0
(1− r2)r dr dθ.
16. V = 4
∫ π/2
0
∫ 3
1
dr dθ.
17. V = 8
∫ π/2
0
∫ 3
1
r
√
9− r2 dr dθ = 128
3
√
2
∫ π/2
0
dθ =
64
3
√
2π.
18. V = 2
∫ π/2
0
∫ 2 sin θ
0
r2dr dθ =
16
3
∫ π/2
0
sin3 θ dθ =
32
9
.
19. V = 2
∫ π/2
0
∫ cos θ
0
(1− r2)r dr dθ = 1
2
∫ π/2
0
(2 cos2 θ − cos4 θ) dθ = 5π
32
.
20. V = 4
∫ π/2
0
∫ 3
1
dr dθ = 8
∫ π/2
0
dθ = 4π.
21. V =
∫ π/2
0
∫ 3 sin θ
0
r2 sin θ dr dθ = 9
∫ π/2
0
sin4 θ dθ =
27π
16
.
22. V = 4
∫ π/2
0
∫ 2
2 cos θ
√
4− r2 r dr dθ + 4
∫ π
π/2
∫ 2
0
√
4− r2 r dr dθ =
∫ π/2
0
32
3
(1− cos2 θ)3/2θ dθ +
∫ π
π/2
32
3
dθ =
64
9
+
16π
3
.
686 Chapter 14
23.
∫ 2π
0
∫ 3
0
sin(r2)r dr dθ =
1
2
(1− cos 9)
∫ 2π
0
dθ = π(1− cos 9).
24.
∫ π/2
0
∫ 3
0
r
√
9− r2 dr dθ = 9
∫ π/2
0
dθ =
9π
2
.
25.
∫ π/4
0
∫ 2
0
1
1 + r2
r dr dθ =
1
2
ln 5
∫ π/4
0
dθ =
π
8
ln 5.
26.
∫ π/2
π/4
∫ 2 cos θ
0
2r2 sin θ dr dθ =
16
3
∫ π/2
π/4
cos3 θ sin θ dθ =
1
3
.
27.
∫ π/2
0
∫ 1
0
r3 dr dθ =
1
4
∫ π/2
0
dθ =
π
8
.
28.
∫ 2π
0
∫ 2
0
e−r
2
r dr dθ =
1
2
(1− e−4)
∫ 2π
0
dθ = (1− e−4)π.
29.
∫ π/2
0
∫ 2 cos θ
0
r2dr dθ =
8
3
∫ π/2
0
cos3 θ dθ =
16
9
.
30.
∫ π/2
0
∫ 1
0
cos(r2)r dr dθ =
1
2
sin 1
∫ π/2
0
dθ =
π
4
sin 1.
31.
∫ π/2
0
∫ a
0
r
(1 + r2)3/2
drdθ =
π
2
(
1− 1√
1 + a2
)
.
32.
∫ π/4
0
∫ sec θ tan θ
0
r2 dr dθ =
1
3
∫ π/4
0
sec3 θ tan3 θ dθ =
2(
√
2 + 1)
45
.
33.
∫ π/4
0
∫ 2
0
r√
1 + r2
dr dθ =
π
4
(
√
5− 1).
34.
∫ 3π/2
π/2
∫ 4
0
3r2 cos θ dr dθ =
∫ 3π/2
π/2
64 cos θ dθ = −128.
35. True. It can be defined by the inequalities 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2.
36. False. The volume is
∫∫
R
f(r, θ) dA. The extra factor r isn’t introduced until we write this as an iterated integral
as in Theorem 14.3.3.
37. False. The integrand in the iterated integral should be multiplied by r:
∫∫
R
f(r, θ) dA =
∫ π/2
0
∫ 2
1
f(r, θ) r dr dθ.
38. False. The region is described by 0 ≤ θ ≤ π, 0 ≤ r ≤ sin θ, so A =
∫ π
0
∫ sin θ
0
r dr dθ.
39. V =
∫ 2π
0
∫ a
0
hr dr dθ =
∫ 2π
0
h
a2
2
dθ = πa2h.
40. V =
∫ 2π
0
∫ R
0
D(r)r dr dθ =
∫ 2π
0
∫ R
0
ke−rr dr dθ = −2πk(1 + r)e−r
]R
0
= 2πk[1− (R+ 1)e−R].
Exercise Set 14.4 687
41.
∫ tan−1(2)
tan−1(1/3)
∫ 2
0
r3 cos2 θ dr dθ = 4
∫ tan−1(2)
tan−1(1/3)
cos2 θ dθ = 2
∫ tan−1(2)
tan−1(1/3)
(1+cos(2θ)) dθ =
[
2θ+2 cos θ sin θ
]tan−1(2)
tan−1(1/3)
=
2 tan−1(2)+2 · 1√
5
· 2√
5
−2 tan−1(1/3)−2 · 3√
10
· 1√
10
= 2
(
tan−1(2)− tan−1(1/3)
)
+
1
5
= 2 tan−1(1)+
1
5
=
π
2
+
1
5
.
42. A =
∫ φ
0
∫ 2a sin θ
0
r dr dθ = 2a2
∫ φ
0
sin2 θ dθ = a2φ− 1
2
a2 sin 2φ.
43. (a) V = 8
∫ π/2
0
∫ a
0
c
a
(a2 − r2)1/2 r dr dθ = − 4c
3a
π(a2 − r2)3/2
]a
0
=
4
3
πa2c.
(b) V ≈ 4
3
π(6378.1370)26356.5231 km3 ≈ 1.0831682 · 1012 km3 = 1.0831682 · 1021 m3.
44. V = 2
∫ π/2
0
∫ a sin θ
0
c
a
(a2 − r2)1/2r dr dθ = 2
3
a2c
∫ π/2
0
(1− cos3 θ) dθ = (3π − 4)a
2c
9
.
45. A = 4
∫ π/4
0
∫ a√2 cos 2θ
0
r dr dθ = 4a2
∫ π/4
0
cos 2θ dθ = 2a2.
46. A =
∫ π/4
π/6
∫ 4 sin θ
√
8 cos 2θ
r dr dθ+
∫ π/2
π/4
∫ 4 sin θ
0
r dr dθ =
∫ π/4
π/6
(8 sin2 θ− 4 cos 2θ) dθ+
∫ π/2
π/4
8 sin2 θ dθ =
4π
3
+ 2
√
3− 2.
Exercise Set 14.4
1. z =
√
9− y2, zx = 0, zy = −y/
√
9− y2, z2x + z2y + 1 = 9/(9− y2), S =
∫ 2
0
∫ 3
−3
3√
9− y2
dy dx =
∫ 2
0
3π dx = 6π.
2. z = 8− 2x− 2y, z2x + z2y + 1 = 4 + 4 + 1 = 9, S =
∫ 4
0
∫ 4−x
0
3 dy dx =
∫ 4
0
3(4− x)dx = 24.
3. z2 = 4x2 + 4y2, 2zzx = 8x so zx = 4x/z; similarly zy = 4y/z so z
2
x + z
2
y + 1 = (16x
2 + 16y2)/z2 + 1 = 5,
S =
∫ 1
0
∫ x
x2
√
5 dy dx =
√
5
∫ 1
0
(x− x2) dx =
√
5
6
.
4. zx = 2, zy = 2y, z
2
x + z
2
y + 1 = 5 + 4y
2, S =
∫ 1
0
∫ y
0
√
5 + 4y2 dx dy =
∫ 1
0
y
√
5 + 4y2 dy =
27− 5
√
5
12
.
5. z2 = x2 +y2, zx = x/z, zy = y/z, z
2
x+z
2
y +1 = (x
2 +y2)/z2 +1 = 2, S =
∫∫
R
√
2 dA = 2
∫ π/2
0
∫ 2 cos θ
0
√
2 r dr dθ =
4
√
2
∫ π/2
0
cos2 θ dθ =
√
2π.
6. zx = −2x, zy = −2y, z2x + z2y + 1 = 4x2 + 4y2 + 1, S =
∫∫
R
√
4x2 + 4y2 + 1 dA =
∫ 2π
0
∫ 1
0
r
√
4r2 + 1 dr dθ =
1
12
(5
√
5− 1)
∫ 2π
0
dθ =
π
6
(5
√
5− 1).
7. zx = y, zy = x, z
2
x + z
2
y + 1 = x
2 + y2 + 1, S =
∫∫
R
√
x2 + y2 + 1 dA =
∫ π/6
0
∫ 3
0
r
√
r2 + 1 dr dθ =
1
3
(10
√
10 −
1)
∫ π/6
0
dθ =
π
18
(10
√
10− 1).
688 Chapter 14
8. zx = x, zy = y, z
2
x+z
2
y+1 = x
2+y2+1, S =
∫∫
R
√
x2 + y2 + 1 dA =
∫ 2π
0
∫ √8
0
r
√
r2 + 1 dr dθ =
26
3
∫ 2π
0
dθ =
52π
3
.
9. On the sphere, zx = −x/z and zy = −y/z so z2x+z2y +1 = (x2 +y2 +z2)/z2 = 16/(16−x2−y2). The planes z = 1
and z = 2 intersect the sphere along the circles x2 + y2 = 15 and x2 + y2 = 12, so S =
∫∫
R
4√
16− x2 − y2
dA =
∫ 2π
0
∫ √15
√
12
4r√
16− r2
dr dθ = 4
∫ 2π
0
dθ = 8π.
10. On the sphere, zx = −x/z and zy = −y/z so z2x + z2y + 1 = (x2 + y2 + z2)/z2 = 8/(8− x2 − y2); the cone cuts the
sphere in the circle x2 + y2 = 4; S =
∫ 2π
0
∫ 2
0
2
√
2r√
8− r2
dr dθ = (8− 4
√
2)
∫ 2π
0
dθ = 8(2−
√
2)π.
11. (a)
z
x y (b)
x
y
z
(c)
x
y
z
12. (a)
y
x
z
(b)
z
y
x
(c)
z
y
x
13. (a) x = u, y = v, z =
5
2
+
3
2
u− 2v. (b) x = u, y = v, z = u2.
14. (a) x = u, y = v, z =
v
1 + u2
. (b) x = u, y = v, z =
1
3
v2 − 5
3
.
15. (a) x =
√
5 cosu, y =
√
5 sinu, z = v; 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1.
(b) x = 2 cosu, y = v, z = 2 sinu; 0 ≤ u ≤ 2π, 1 ≤ v ≤ 3.
16. (a) x = u, y = 1− u, z = v;−1 ≤ v ≤ 1 (b) x = u, y = 5 + 2v, z = v; 0 ≤ u ≤ 3.
17. x = u, y = sinu cos v, z = sinu sin v.
18. x = u, y = eu cos v, z = eu sin v.
19. x = r cos θ, y = r sin θ, z =
1
1 + r2
.
Exercise Set 14.4 689
20. x = r cos θ, y = r sin θ, z = e−r
2
.
21. x = r cos θ, y = r sin θ, z = 2r2 cos θ sin θ.
22. x = r cos θ, y = r sin θ, z = r2(cos2 θ − sin2 θ).
23. x = r cos θ, y = r sin θ, z =
√
9− r2; r ≤
√
5.
24. x = r cos θ, y = r sin θ, z = r; r ≤ 3.
25. x =
1
2
ρ cos θ, y =
1
2
ρ sin θ, z =
√
3
2
ρ.
26. x = 3 cos θ, y = 3 sin θ, z = 3 cotφ.
27. z = x− 2y; a plane.
28. y = x2 + z2, 0 ≤ y ≤ 4; part of a circular paraboloid.
29. (x/3)2 + (y/2)2 = 1; 2 ≤ z ≤ 4; part of an elliptic cylinder.
30. z = x2 + y2; 0 ≤ z ≤ 4; part of a circular paraboloid.
31. (x/3)2 + (y/4)2 = z2; 0 ≤ z ≤ 1; part of an elliptic cone.
32. x2 + (y/2)2 + (z/3)2 = 1; an ellipsoid.
33. (a) I: x = r cos θ, y = r sin θ, z = r, 0 ≤ r ≤ 2; II: x = u, y = v, z =
√
u2 + v2; 0 ≤ u2 + v2 ≤ 4.
34. (a) I: x = r cos θ, y = r sin θ, z = r2, 0 ≤ r ≤
√
2; II: x = u, y = v, z = u2 + v2; u2 + v2 ≤ 2.
35. (a) 0 ≤ u ≤ 3, 0 ≤ v ≤ π. (b) 0 ≤ u ≤ 4, −π/2 ≤ v ≤ π/2.
36. (a) 0 ≤ u ≤ 6, −π ≤ v ≤ 0. (b) 0 ≤ u ≤ 5, π/2 ≤ v ≤ 3π/2.
37. (a) 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π. (b) 0 ≤ φ ≤ π, 0 ≤ θ ≤ π.
38. (a) π/2 ≤ φ ≤ π, 0 ≤ θ ≤ 2π. (b) 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/2.
39. u = 1, v = 2, ru × rv = −2i− 4j + k; 2x+ 4y − z = 5.
40. u = 1, v = 2, ru × rv = −4i− 2j + 8k; 2x+ y − 4z = −6.
41. u = 0, v = 1, ru × rv = 6k; z = 0.
42. ru × rv = 2i− j− 3k; 2x− y − 3z = −4.
43. ru × rv =
1√
2
i− 1√
2
j +
1
2
k; x− y + 1√
2
z =
π
√
2
8
.
44. ru × rv = 2i− ln 2k; 2x− (ln 2)z = 0.
45. ru = cos v i + sin v j + 2u k, rv = −u sin v i + u cos v j, ‖ru × rv‖ = u
√
4u2 + 1; S =
∫ 2π
0
∫ 2
1
u
√
4u2 + 1 du dv =
π
6
(17
√
17− 5
√
5).
690 Chapter 14
46. ru = cos v i + sin v j + k, rv = −u sin v i + u cos v j, ‖ru × rv‖ =
√
2u; S =
∫ π/2
0
∫ 2v
0
√
2u du dv =
√
2
12
π3.
47. False. For example, if f(x, y) = 1 then the surface has the same area as R,
∫∫
R
dA, not
∫∫
R
√
2 dA.
48. True. q× r =
〈
−∂z
∂x
,−∂z
∂y
, 1
〉
, so
∫∫
R
‖q× r‖ dA =
∫∫
R
√(
∂z
∂x
)2
+
(
∂z
∂y
)2
+ 1 dA = S, by equation (2).
49. True, as explained before Definition 14.4.1.
50. True. ‖〈1, 0, a〉 × 〈0, 1, b〉‖ = ‖〈−a,−b, 1〉‖ =
√
a2 + b2 + 1 =
√(
∂z
∂x
)2
+
(
∂z
∂y
)2
+ 1, so the area of the surface
is
∫∫
R
√(
∂z
∂x
)2
+
(
∂z
∂y
)2
+ 1 dA =
∫∫
R
‖〈1, 0, a〉 × 〈0, 1, b〉‖ dA = ‖〈1, 0, a〉 × 〈0, 1, b〉‖ ·
∫∫
R
dA =
= ‖〈1, 0, a〉 × 〈0, 1, b〉‖·(area of R).
51. r(u, v) = a cosu sin vi+a sinu sin vj+a cos vk, ‖ru×rv‖ = a2 sin v, S =
∫ π
0
∫ 2π
0
a2 sin v du dv = 2πa2
∫ π
0
sin v dv =
4πa2.
52. r = r cosui + r sinuj + vk, ‖ru × rv‖ = r; S =
∫ h
0
∫ 2π
0
r du dv = 2πrh.
53. zx =
h
a
x√
x2 + y2
, zy =
h
a
y√
x2 + y2
, z2x + z
2
y + 1 =
h2x2 + h2y2
a2(x2 + y2)
+ 1 =
a2 + h2
a2
, S =
∫ 2π
0
∫ a
0
√
a2 + h2
a
r dr dθ =
1
2
a
√
a2 + h2
∫ 2π
0
dθ = πa
√
a2 + h2.
54. (a) Revolving a point (a0, 0, b0) of the xz-plane around the z-axis generates a circle, an equation of which is
r = a0 cosui + a0 sinuj + b0k, 0 ≤ u ≤ 2π. A point on the circle (x− a)2 + z2 = b2 which generates the torus can
be written r = (a+ b cos v)i+ b sin vk, 0 ≤ v ≤ 2π. Set a0 = a+ b cos v and b0 = a+ b sin v and use the first result:
any point on the torus can thus be written in the form r = (a+ b cos v) cosui+ (a+ b cos v) sinuj+ b sin vk, which
yields the result.
55. ru = −(a+b cos v) sinu i+(a+b cos v) cosu j, rv = −b sin v cosu i−b sin v sinu j+ b cos v k, ‖ru×rv‖ = b(a+b cos v);
S =
∫ 2π
0
∫ 2π
0
b(a+ b cos v) du dv = 4π2ab.
56. ‖ru × rv‖ =
√
u2 + 1;S =
∫ 4π
0
∫ 5
0
√
u2 + 1 du dv = 4π
∫ 5
0
√
u2 + 1 du ≈ 174.7199011.
57. z = −1 when v ≈ 0.27955, z = 1 when v ≈ 2.86204, ‖ru × rv‖ = | cos v|; S ≈
∫ 2π
0
∫ 2.86204
0.27955
| cos v| dv du ≈ 9.099.
58. (a) x = v cosu, y = v sinu, z= f(v), for example. (b) x = v cosu, y = v sinu, z = 1/v2.
Exercise Set 14.5 691
(c)
z
y
x
59.
(x
a
)2
+
(y
b
)2
+
(z
c
)2
= 1, ellipsoid.
60.
(x
a
)2
+
(y
b
)2
−
(z
c
)2
= 1, hyperboloid of one sheet.
61. −
(x
a
)2
−
(y
b
)2
+
(z
c
)2
= 1, hyperboloid of two sheets.
Exercise Set 14.5
1.
∫ 1
−1
∫ 2
0
∫ 1
0
(x2 + y2 + z2) dx dy dz =
∫ 1
−1
∫ 2
0
(1/3 + y2 + z2) dy dz =
∫ 1
−1
(10/3 + 2z2) dz = 8.
2.
∫ 1/2
1/3
∫ π
0
∫ 1
0
zx sinxy dz dy dx =
∫ 1/2
1/3
∫ π
0
1
2
x sinxy dy dx =
∫ 1/2
1/3
1
2
(1− cosπx) dx = 1
12
+
√
3− 2
4π
.
3.
∫ 2
0
∫ y2
−1
∫ z
−1
yz dx dz dy =
∫ 2
0
∫ y2
−1
(yz2 + yz) dz dy =
∫ 2
0
(
1
3
y7 +
1
2
y5 − 1
6
y
)
dy =
47
3
.
4.
∫ π/4
0
∫ 1
0
∫ x2
0
x cos y dz dx dy =
∫ π/4
0
∫ 1
0
x3 cos y dx dy =
∫ π/4
0
1
4
cos y dy =
√
2
8
.
5.
∫ 3
0
∫ √9−z2
0
∫ x
0
xy dy dx dz =
∫ 3
0
∫ √9−z2
0
1
2
x3dx dz =
∫ 3
0
1
8
(81− 18z2 + z4) dz = 81
5
.
6.
∫ 3
1
∫ x2
x
∫ ln z
0
xey dy dz dx =
∫ 3
1
∫ x2
x
(xz − x) dz dx =
∫ 3
1
(
1
2
x5 − 3
2
x3 + x2
)
dx =
118
3
.
7.
∫ 2
0
∫ √4−x2
0
∫ 3−x2−y2
−5+x2+y2
x dz dy dx =
∫ 2
0
∫ √4−x2
0
[2x(4− x2)− 2xy2] dy dx =
∫ 2
0
4
3
x(4− x2)3/2 dx = 128
15
.
8.
∫ 2
1
∫ 2
z
∫ √3y
0
y
x2 + y2
dx dy dz =
∫ 2
1
∫ 2
z
π
3
dy dz =
∫ 2
1
π
3
(2− z) dz = π
6
.
9.
∫ π
0
∫ 1
0
∫ π/6
0
xy sin yz dz dy dx =
∫ π
0
∫ 1
0
x[1− cos(πy/6)] dy dx =
∫ π
0
(1− 3/π)x dx = π(π − 3)
2
.
10.
∫ 1
−1
∫ 1−x2
0
∫ y
0
y dz dy dx =
∫ 1
−1
∫ 1−x2
0
y2 dy dx =
∫ 1
−1
1
3
(1− x2)3 dx = 32
105
.
692 Chapter 14
11.
∫ √2
0
∫ x
0
∫ 2−x2
0
xyz dz dy dx =
∫ √2
0
∫ x
0
1
2
xy(2− x2)2dy dx =
∫ √2
0
1
4
x3(2− x2)2 dx = 1
6
.
12.
∫ π/2
π/6
∫ π/2
y
∫ xy
0
cos(z/y) dz dx dy =
∫ π/2
π/6
∫ π/2
y
y sinx dx dy =
∫ π/2
π/6
y cos y dy =
5π − 6
√
3
12
.
13.
∫ 3
0
∫ 2
1
∫ 1
−2
√
x+ z2
y
dz dy dx ≈ 9.425.
14. 8
∫ 1
0
∫ √1−x2
0
∫ √1−x2−y2
0
e−x
2−y2−z2 dz dy dx ≈ 2.381.
15. V =
∫ 4
0
∫ (4−x)/2
0
∫ (12−3x−6y)/4
0
dz dy dx =
∫ 4
0
∫ (4−x)/2
0
1
4
(12− 3x− 6y) dy dx =
∫ 4
0
3
16
(4− x)2 dx = 4.
16. V =
∫ 1
0
∫ 1−x
0
∫ √y
0
dz dy dx =
∫ 1
0
∫ 1−x
0
√
y dy dx =
∫ 1
0
2
3
(1− x)3/2 dx = 4
15
.
17. V = 2
∫ 2
0
∫ 4
x2
∫ 4−y
0
dz dy dx = 2
∫ 2
0
∫ 4
x2
(4− y) dy dx = 2
∫ 2
0
(
8− 4x2 + 1
2
x4
)
dx =
256
15
.
18. V =
∫ 1
0
∫ y
0
∫ √1−y2
0
dz dx dy =
∫ 1
0
∫ y
0
√
1− y2 dx dy =
∫ 1
0
y
√
1− y2 dy = 1
3
.
19. The projection of the curve of intersection onto the xy-plane is x2 + y2 = 1,
(a)
∫ 1
−1
∫ √1−x2
−
√
1−x2
∫ 4−3y2
4x2+y2
f(x, y, z) dz dy dx. (b)
∫ 1
−1
∫ √1−y2
−
√
1−y2
∫ 4−3y2
4x2+y2
f(x, y, z) dz dx dy.
20. The projection of the curve of intersection onto the xy-plane is 2x2 + y2 = 4,
(a)
∫ √2
−
√
2
∫ √4−2x2
−
√
4−2x2
∫ 8−x2−y2
3x2+y2
f(x, y, z) dz dy dx. (b)
∫ 2
−2
∫ √(4−y2)/2
−
√
(4−y2)/2
∫ 8−x2−y2
3x2+y2
f(x, y, z) dz dx dy.
21. Let f(x, y, z) = 1 in Exercise 19(a). V =
∫ 1
−1
∫ √1−x2
−
√
1−x2
∫ 4−3y2
4x2+y2
dz dy dx = 4
∫ 1
0
∫ √1−x2
0
∫ 4−3y2
4x2+y2
dz dy dx.
22. Let f(x, y, z) = 1 in Exercise 20(a). V =
∫ √2
−
√
2
∫ √4−2x2
−
√
4−2x2
∫ 8−x2−y2
3x2+y2
dz dy dx = 4
∫ √2
0
∫ √4−2x2
0
∫ 8−x2−y2
3x2+y2
dz dy dx.
23. V = 2
∫ 3
−3
∫ √9−x2/3
0
∫ x+3
0
dz dy dx.
24. V = 8
∫ 1
0
∫ √1−x2
0
∫ √1−x2
0
dz dy dx.
Exercise Set 14.5 693
25. (a)
(0, 0, 1)
(0, 1, 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
26. (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
27. True, by changing the order of integration in Theorem 14.5.1.
28. False. For example, consider the simple xy-solid G defined by −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, 0 ≤ z ≤ x2 + y2.
Cross-sections of G parallel to the xy-plane with z > 0 are neither type I nor type II regions, so the triple integral
over G can’t be expressed as an integral whose outermost integration is performed with respect to z. (As shown
in Theorem 14.5.2, the triple integral can be expressed as an iterated integral whose innermost integration is
performed with respect to z.)
29. False. The middle integral (with respect to y) should be
∫ √1−x2
0
.
30. False. For example, let G be described by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, and let f(x, y, z) = 2x. Then
∫∫∫
G
2x dV =
∫ 1
0
∫ 1
0
∫ 1
0
2x dz dy dx =
∫ 1
0
∫ 1
0
2x dy dx =
∫ 1
0
2x dx = x2
]1
0
= 1 = volume of G.
31.
∫ b
a
∫ d
c
∫ `
k
f(x)g(y)h(z)dz dy dx =
∫ b
a
∫ d
c
f(x)g(y)
[∫ `
k
h(z) dz
]
dy dx =
[∫ b
a
f(x)
[∫ d
c
g(y) dy
]
dx
][∫ `
k
h(z) dz
]
=
[∫ b
a
f(x) dx
][∫ d
c
g(y)dy
][∫ `
k
h(z) dz
]
.
32. (a)
[∫ 1
−1
x dx
] [∫ 1
0
y2 dy
] [∫ π/2
0
sin z dz
]
= (0)(1/3)(1) = 0.
(b)
[∫ 1
0
e2x dx
] [∫ ln 3
0
ey dy
][∫ ln 2
0
e−zdz
]
= [(e2 − 1)/2](2)(1/2) = (e2 − 1)/2.
33. V =
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
dz dy dx = 1/6, fave = 6
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
(x+ y + z) dz dy dx =
3
4
.
34. The integrand is an odd function of each of x, y, and z, so the average is zero.
694 Chapter 14
35. The volume V =
3π√
2
, and thus
rave =
√
2
3π
∫∫∫
G
√
x2 + y2 + z2 dV =
√
2
3π
∫ 1/√2
−1/
√
2
∫ √1−2x2
−
√
1−2x2
∫ 6−7x2−y2
5x2+5y2
√
x2 + y2 + z2 dz dy dx ≈ 3.291.
36. V = 1, dave =
1
V
∫ 1
0
∫ 1
0
∫ 1
0
√
(x− z)2 + (y − z)2 + z2 dx dy dz ≈ 0.771.
37. (a)
∫ a
0
∫ b(1−x/a)
0
∫ c(1−x/a−y/b)
0
dz dy dx,
∫ b
0
∫ a(1−y/b)
0
∫ c(1−x/a−y/b)
0
dz dx dy,
∫ c
0
∫ a(1−z/c)
0
∫ b(1−x/a−z/c)
0
dy dx dz,
∫ a
0
∫ c(1−x/a)
0
∫ b(1−x/a−z/c)
0
dy dz dx,
∫ c
0
∫ b(1−z/c)
0
∫ a(1−y/b−z/c)
0
dx dy dz,
∫ b
0
∫ c(1−y/b)
0
∫ a(1−y/b−z/c)
0
dx dz dy.
(b) Use the first integral in part (a) to get
∫ a
0
∫ b(1−x/a)
0
c
(
1− x
a
− y
b
)
dy dx =
∫ a
0
1
2
bc
(
1− x
a
)2
dx =
1
6
abc.
38. V = 8
∫ a
0
∫ b√1−x2/a2
0
∫ c√1−x2/a2−y2/b2
0
dz dy dx = 8
∫ a
0
∫ b√1−x2/a2
0
c
√
1− x
2
a2
− y
2
b2
dy dx =
=
8c
b
∫ a
0
∫ b√1−x2/a2
0
√
b2
(
1− x
2
a2
)
− y2 dy dx =
=
8c
b
∫ a
0
[
y
2
√
b2
(
1− x
2
a2
)
− y2 + b
2
2
(
1− x
2
a2
)
sin−1
y√
b2(1− x2/a2)
]b√1−x2/a2
y=0
dx =
=
8c
b
∫ a
0
b2
2
(
1− x
2
a2
)
π
2
dx = 2πbc
∫ a
0
(
1− x
2
a2
)
dx = 2πbc
[
x− x
3
3a2
]a
0
=
4πabc
3
, by Endpaper Integral Table
Formula 74.
39. (a)
∫ 2
0
∫ √4−x2
0
∫ 5
0
f(x, y, z) dz dy dx (b)
∫ 9
0
∫ 3−√x
0
∫ 3−√x
y
f(x, y, z) dz dy dx
(c)
∫ 2
0
∫ 4−x2
0
∫ 8−y
y
f(x, y, z) dz dy dx
40. (a)
∫ 3
0
∫ √9−x2
0
∫ √9−x2−y2
0
f(x, y, z) dz dy dx (b)
∫ 4
0
∫ x/2
0
∫ 2
0
f(x, y, z) dz dy dx
(c)
∫ 2
0
∫ 4−x2
0
∫ 4−y
x2
f(x, y, z) dz dy dx
41. See discussion after Theorem 14.5.2.
Exercise Set 14.6
1.
∫ 2π
0
∫ 1
0
∫ √1−r2
0
zr dz dr dθ =
∫ 2π
0
∫ 1
0
1
2
(1− r2)r dr dθ =
∫ 2π
0
1
8
dθ =
π
4
.
2.
∫ π/2
0
∫ cos θ
0
∫ r2
0
r sin θ dz dr dθ =
∫ π/2
0
∫ cos θ
0
r3 sin θ dr dθ =
∫ π/2
0
1
4
cos4 θ sin θ dθ =
1
20
.
Exercise Set 14.6 695
3.
∫ π/2
0
∫ π/2
0
∫ 1
0
ρ3 sinφ cosφdρ dφ dθ =
∫ π/2
0
∫ π/2
0
1
4
sinφ cosφdφ dθ =
∫ π/2
0
1
8
dθ =
π
16
.
4.
∫ 2π
0
∫ π/4
0
∫ a secφ
0
ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/4
0
1
3
a3 sec3 φ sinφdφ dθ =
∫ 2π
0
1
6
a3 dθ =
πa3
3
.
5. f(r, θ, z) = z
z
y
x 6. f(r, θ, z) = sin θ
0.0
0.5
1.0
x
0.0
0.2
0.4
y
0.0
0.5
1.0
z
7. f(ρ, θ, φ) = ρ cosφ
z
y
x 8. f(ρ, θ, φ) = 1
(0, 0, )a
z
yx
9. V =
∫ 2π
0
∫ 3
0
∫ 9
r2
r dz dr dθ =
∫ 2π
0
∫ 3
0
r(9− r2) dr dθ =
∫ 2π
0
81
4
dθ =
81π
2
.
10. V =
∫ 2π
0
∫ 1/√2
0
∫ √1−r2
r
r dz dr dθ =
∫ 2π
0
∫ 1/√2
0
r
√
1− r2 − r2 dr dθ =
∫ 2π
0
1
6
(2−
√
2)dθ =
π
3
(2−
√
2).
11. r2 + z2 = 20 intersects z = r2 in a circle of radius 2; the volume consists of two portions, one inside the
cylinder r = 2 and one outside that cylinder: V =
∫ 2π
0
∫ 2
0
∫ r2
−
√
20−r2
r dzdr dθ +
∫ 2π
0
∫ √20
2
∫ √20−r2
−
√
20−r2
r dz dr dθ =
∫ 2π
0
∫ 2
0
r
(
r2 +
√
20− r2
)
dr dθ +
∫ 2π
0
∫ √20
2
2r
√
20− r2 dr dθ = 4
3
(10
√
5− 13)
∫ 2π
0
dθ +
128
3
∫ 2π
0
dθ =
152
3
π +
80
3
π
√
5.
12. z = hr/a intersects z = h in a circle of radius a, V =
∫ 2π
0
∫ a
0
∫ h
hr/a
r dz dr dθ =
∫ 2π
0
∫ a
0
h
a
(ar − r2) dr dθ =
∫ 2π
0
1
6
a2h dθ =
πa2h
3
.
13. V =
∫ 2π
0
∫ π/3
0
∫ 4
0
ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/3
0
64
3
sinφdφ dθ =
32
3
∫ 2π
0
dθ =
64π
3
.
14. V =
∫ 2π
0
∫ π/4
0
∫ 2
1
ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/4
0
7
3
sinφdφ dθ =
7
6
(2−
√
2)
∫ 2π
0
dθ =
7π
3
(2−
√
2).
15. In spherical coordinates the sphere and the plane z = a are ρ = 2a and ρ = a secφ, respectively. They intersect at
φ = π/3, V =
∫ 2π
0
∫ π/3
0
∫ a secφ
0
ρ2 sinφdρ dφ dθ +
∫ 2π
0
∫ π/2
π/3
∫ 2a
0
ρ2 sinφdρ dφ dθ =
696 Chapter 14
=
∫ 2π
0
∫ π/3
0
1
3
a3 sec3 φ sinφdφ dθ +
∫ 2π
0
∫ π/2
π/3
8
3
a3 sinφdφ dθ =
1
2
a3
∫ 2π
0
dθ +
4
3
a3
∫ 2π
0
dθ =
11πa3
3
.
16. V =
∫ 2π
0
∫ π/2
π/4
∫ 3
0
ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/2
π/4
9 sinφdφ dθ =
9
√
2
2
∫ 2π
0
dθ = 9
√
2π.
17.
∫ π/2
0
∫ a
0
∫ a2−r2
0
r3 cos2 θ dz dr dθ =
∫ π/2
0
∫ a
0
(a2r3 − r5) cos2 θ dr dθ = 1
12
a6
∫ π/2
0
cos2 θ dθ =
πa6
48
.
18.
∫ π
0
∫ π/2
0
∫ 1
0
e−ρ
3
ρ2 sinφdρ dφ dθ =
1
3
(1− e−1)
∫ π
0
∫ π/2
0
sinφdφ dθ =
π
3
(1− e−1).
19.
∫ π/2
0
∫ π/4
0
∫ √8
0
ρ4 cos2 φ sinφdρ dφ dθ =
32π
15
(2
√
2− 1).
20.
∫ 2π
0
∫ π
0
∫ 3
0
ρ3 sinφdρ dφ dθ = 81π.
21. False. The factor r2 should be just r.
22. True. If G is the spherical wedge then the volume of G is
∫∫∫
G
1 dV =
∫ θ2
θ1
∫ φ2
φ1
∫ ρ2
ρ1
ρ2 sinφdρ dφ dθ, by equation
(9).
23. True. The region is described by 0 ≤ φ ≤ π/4, 0 ≤ θ ≤ 2π, 1 ≤ ρ ≤ 3, so the volume is
∫∫∫
G
1 dV =
∫ π/4
0
∫ 2π
0
∫ 3
1
ρ2 sinφdρ dθ dφ.
24. False. The “sin θ” and “cos θ” in the iterated integral are reversed.
25. (a)
∫ 2
−2
∫ 4
1
∫ π/3
π/6
r tan3 θ√
1 + z2
dθ dr dz =
(∫ 2
−2
1√
1 + z2
dz
)(∫ 4
1
r dr
)(∫ π/3
π/6
tan3 θ dθ
)
=
= 2 ln(2 +
√
5) · 15
2
·
(
4
3
− 1
2
ln 3
)
=
5
2
(8− 3 ln 3) ln(2 +
√
5) ≈ 16.97774195.
(b) G is the cylindrical wedge π/6 ≤ θ ≤ π/3, 1 ≤ r ≤ 4, −2 ≤ z ≤ 2. Since dx dy dz = dV = r dθ dr dz, the
integrand in rectangular coordinates is
1
r
· r tan
3 θ√
1 + z2
=
(y/x)3√
1 + z2
, so f(x, y, z) =
y3
x3
√
1 + z2
.
x
yz
1
4
4
2
!2
Exercise Set 14.7 697
26.
∫ π/2
0
∫ π/4
0
1
18
cos37 θ cosφdφ dθ =
√
2
36
∫ π/2
0
cos37 θ dθ =
4,294,967,296
755,505,013,725
√
2 ≈ 0.008040.
27. (a) V = 2
∫ 2π
0
∫ a
0
∫ √a2−r2
0
r dz dr dθ =
4πa3
3
. (b) V =
∫ 2π
0
∫ π
0
∫ a
0
ρ2 sinφdρ dφ dθ =
4πa3
3
.
28. (a)
∫ 2
0
∫ √4−x2
0
∫ √4−x2−y2
0
xyz dz dy dx =
∫ 2
0
∫ √4−x2
0
1
2
xy(4− x2 − y2) dy dx = 1
8
∫ 2
0
x(4− x2)2 dx = 4
3
.
(b)
∫ π/2
0
∫ 2
0
∫ √4−r2
0
r3z sin θ cos θ dz dr dθ =
∫ π/2
0
∫ 2
0
1
2
(4r3 − r5) sin θ cos θ dr dθ = 8
3
∫ π/2
0
sin θ cos θ dθ =
4
3
.
(c)
∫ π/2
0
∫ π/2
0
∫ 2
0
ρ5 sin3 φ cosφ sin θ cos θ dρ dφ dθ =
∫ π/2
0
∫ π/2
0
32
3
sin3 φ cosφ sin θ cos θ dφ dθ =
=
8
3
∫ π/2
0
sin θ cos θ dθ =
4
3
.
29. V =
∫ π/2
0
∫ π/3
π/6
∫ 2
0
ρ2 sinφdρ dφ dθ =
∫ π/2
0
∫ π/3
π/6
8
3
sinφdφ dθ =
4
3
(
√
3− 1)
∫ π/2
0
dθ =
2π
3
(
√
3− 1).
30. (a) The sphere and cone intersect in a circle of radius ρ0 sinφ0, V =
∫ θ2
θ1
∫ ρ0 sinφ0
0
∫ √ρ20−r2
r cotφ0
r dz dr dθ =
∫ θ2
θ1
∫ ρ0 sinφ0
0
(
r
√
ρ20 − r2 − r2 cotφ0
)
dr dθ =
∫ θ2
θ1
1
3
ρ30(1− cos3 φ0 − sin3 φ0 cotφ0) dθ =
=
1
3
ρ30(1− cos3 φ0 − sin2 φ0 cosφ0)(θ2 − θ1) =
1
3
ρ30(1− cosφ0)(θ2 − θ1).
(b) From part (a), the volume of the solid bounded by θ = θ1, θ = θ2, φ = φ1, φ = φ2, and ρ = ρ0 is
1
3
ρ30(1− cosφ2)(θ2 − θ1)−
1
3
ρ30(1− cosφ1)(θ2 − θ1) =
1
3
ρ30(cosφ1 − cosφ2)(θ2 − θ1), so the volume of the spherical
wedge between ρ = ρ1 and ρ = ρ2 is ∆V =
1
3
ρ32(cosφ1 − cosφ2)(θ2 − θ1) −
1
3
ρ31(cosφ1 − cosφ2)(θ2 − θ1) =
1
3
(ρ32 − ρ31)(cosφ1 − cosφ2)(θ2 − θ1).
(c)
d
dφ
cosφ = − sinφ so from the Mean-Value Theorem cosφ2 − cosφ1 = −(φ2 − φ1) sinφ∗ where φ∗ is between
φ1 and φ2. Similarly
d
dρ
ρ3 = 3ρ2 so ρ32−ρ31 = 3ρ∗2(ρ2−ρ1) where ρ∗ is between ρ1 and ρ2. Thus cosφ1− cosφ2 =
sinφ∗∆φ and ρ32 − ρ31 = 3ρ∗2∆ρ so ∆V = ρ∗2 sinφ∗∆ρ∆φ∆θ.
31. The fact that none of the limits involves θ means that the solid is obtained by rotating a region in the xz-plane
about the z-axis, between two angles θ1 and θ2. If the integral is expressed in cylindrical coordinates, then the
plane region must be either a type I region or a type II region (with the role of y replaced by z); see Definition
14.2.1. If the integral is expressed in spherical coordinates, then the plane region may be a simple polar region
(with the roles of θ and r replaced by φ and ρ); see Definition 14.3.1. Or it may be described by inequalities of
the form ρ1 ≤ ρ ≤ ρ2, φ1(ρ) ≤ φ ≤ φ2(ρ) for some numbers ρ1 ≤ ρ2 and functions φ1(ρ) ≤ φ2(ρ).
Exercise Set 14.7
1.
∂(x, y)
∂(u, v)
=
∣∣∣∣
1 4
3 −5
∣∣∣∣ = −17.
698 Chapter 14
2.
∂(x, y)
∂(u, v)
=
∣∣∣∣
1 4v
4u −1
∣∣∣∣ = −1− 16uv.
3.
∂(x, y)
∂(u, v)
=
∣∣∣∣
cosu − sin v
sinu cos v
∣∣∣∣ = cosu cos v + sinu sin v = cos(u− v).
4.
∂(x, y)
∂(u, v)
=
∣∣∣∣∣∣∣∣∣
2(v2 − u2)
(u2 + v2)2
− 4uv
(u2 + v2)2
4uv
(u2 + v2)2
2(v2 − u2)
(u2 + v2)2
∣∣∣∣∣∣∣∣∣
= 4/(u2 + v2)2.
5. x =
2
9
u+
5
9
v, y = −1
9
u+
2
9
v;
∂(x, y)
∂(u, v)
=
∣∣∣∣
2/9 5/9
−1/9 2/9
∣∣∣∣ =
1
9
.
6. x = lnu, y = uv;
∂(x, y)
∂(u, v)
=
∣∣∣∣
1/u 0
v u
∣∣∣∣ = 1.
7. x =
√
u+ v√
2
, y =
√
v − u√
2
;
∂(x, y)
∂(u, v)
=
∣∣∣∣∣∣∣∣
1
2
√
2
√
u+ v
1
2
√
2
√
u+ v
− 1
2
√
2
√
v − u
1
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
− v
1/2
2u3/2
1
2u1/2v1/2
∣∣∣∣∣∣∣∣∣
=
1
2v
.
9.
∂(x, y, z)
∂(u, v, w)
=
∣∣∣∣∣∣
3 1 0
1 0 −2
0 1 1
∣∣∣∣∣∣
= 5.
10.
∂(x, y, z)
∂(u, v, w)
=
∣∣∣∣∣∣
1− v −u 0
v − vw u− uw −uv
vw 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 0
0 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 −1/2
1/2 −1/2 0
∣∣∣∣∣∣
= −1
4
.
13. False. It is the area of the parallelogram.
14. False. If the mapping is not one-to-one, then the integral may be larger than the area. For example, let x = u,
y = (v− 3)2. Then R is the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 4, with area 8, but ∂(x, y)
∂(u, v)
=
∣∣∣∣
1 0
0 2(v − 3)
∣∣∣∣ = 2(v− 3),
so
∫ 5
1
∫ 2
0
∣∣∣∣
∂(x, y)
∂(u, v)
∣∣∣∣ du dv =
∫ 5
1
∫ 2
0
2|v − 3| du dv =
∫ 5
1
4|v − 3| dv =
∫ 3
1
4(3 − v) dv +
∫ 5
3
4(v − 3) dv = (12v −
2v2)
]3
1
+ (2v2 − 12v)
]5
3
= 8 + 8 = 16.
15. False. The Jacobian is
∂(x, y)
∂(u, v)
=
∣∣∣∣
cos θ −r sin θ
sin θ r cos θ
∣∣∣∣ = r(cos2 θ + sin2 θ) = r.
Exercise Set 14.7 699
16. True. See the solution of Exercise 14.7.48(b).
17.
x
y
(0, 2)
(–1, 0) (1, 0)
(0, 0)
18. 1 2 3
1
2
3
4
x
y
(0, 0) (4, 0)
(3, 4)
19.
–3 3
–3
3
x
y
(2, 0)
(0, 3)
20. 1 2
1
2
x
y
21. x =
1
5
u+
2
5
v, y = −2
5
u+
1
5
v,
∂(x, y)
∂(u, v)
=
1
5
;
1
5
∫∫
S
u
v
dAuv =
1
5
∫ 3
1
∫ 4
1
u
v
du dv =
3
2
ln 3.
22. x =
1
2
u+
1
2
v, y =
1
2
u− 1
2
v,
∂(x, y)
∂(u, v)
= −1
2
;
1
2
∫∫
S
veuv dAuv =
1
2
∫ 4
1
∫ 1
0
veuv du dv =
1
2
(e4 − e− 3).
23. 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
sinu cos v dAuv = 2
∫ 1
0
∫ u
0
sinu cos v dv du = 1− 1
2
sin 2.
24. 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
∫ 4
1
∫ 3
1
v2du dv = 21.
25. x = 3u, y = 4v,
∂(x, y)
∂(u, v)
= 12; S is the regionin the uv-plane enclosed by the circle u2 + v2 = 1. Use polar
coordinates to obtain
∫∫
S
12
√
u2 + v2(12) dAuv = 144
∫ 2π
0
∫ 1
0
r2 dr dθ = 96π.
26. 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−(4u
2+4v2)(2) dAuv = 2
∫ 2π
0
∫ 1
0
re−4r
2
dr dθ =
π
2
(1− e−4).
27. 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)
=
700 Chapter 14
∣∣∣∣
1/2 0
0 1/3
∣∣∣∣ = 1/6, use polar coordinates to get
1
6
∫∫
S
sin(u2 + v2) dAuv =
1
6
∫ π/2
0
∫ 1
0
r sin r2 dr dθ =
=
π
24
(− cos r2)
]1
0
=
π
24
(1− cos 1).
28. u = x/a, v = y/b, x = au, y = bv;
∂(x, y)
∂(u, v)
= ab; A = ab
∫ 2π
0
∫ 1
0
r dr dθ = πab.
29. 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
6
dVuvw =
1
54
∫ 2π
0
∫ π
0
∫ 6
0
(ρ sinφ cos θ)2ρ2 sinφdρ dφ dθ =
1
54
∫ 2π
0
∫ π
0
∫ 6
0
ρ4 sin3 φ cos2 θ dρ dφ dθ =
192π
5
.
30. Let G1 be the region u
2 + v2 + w2 ≤ 1, with x = au, y = bv, z = cw, ∂(x, y, z)
∂(u, v, w)
= abc; then use spherical
coordinates in uvw-space:
∫∫∫
G
(y2 +z2) dVxyz = abc
∫∫∫
G1
(b2v2 + c2w2) dVuvw =
∫ 2π
0
∫ π
0
∫ 1
0
abc(b2 sin2 φ sin2 θ+
c2 cos2 φ)ρ4 sinφdρ dφ dθ =
∫ 2π
0
abc
15
(4b2 sin2 θ + 2c2) dθ =
4
15
πabc(b2 + c2).
31. u = θ =



cot−1(x/y) if y 6= 0
0 if y = 0, x > 0
π if y = 0, x < 0
, v = r =
√
x2 + y2. Other answers are possible.
32. u = r =
√
x2 + y2, v =
1
2
+
θ
π
=



1
2
+
1
π
tan−1(y/x) if x 6= 0
1 if x = 0, y > 0
0 if x = 0, y < 0
. Other answers are possible.
33. u =
3
7
x− 2
7
y, v = −1
7
x+
3
7
y. Other answers are possible.
34. u = −x+ 4
3
y, v = y. Other answers are possible.
35. 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
v
dAuv =
1
8
∫ 5
2
∫ 2
0
u
v
du dv =
1
4
ln
5
2
.
36. Let u = y+x, v = y−x, then x = 1
2
(u−v), y = 1
2
(u+v) so
∂(x, y)
∂(u, v)
=
1
2
; −1
2
∫∫
S
uv dAuv = −
1
2
∫ 2
0
∫ 1
0
uv du dv =
−1
2
.
37. 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 = π/4; thus
1
2
∫∫
S
sinu
cos v
dAuv =
1
2
∫ π/4
0
∫ v
0
sinu
cos v
du dv =
1
2
[
ln(
√
2 + 1)− π
4
]
.
Exercise Set 14.7 701
38. Let u = y − x, v = y + x, then x = 1
2
(v − u), y = 1
2
(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; thus 1
2
∫∫
S
eu/vdAuv =
1
2
∫ 4
1
∫ v
−v
eu/vdu dv =
15
4
(e− e−1).
39. Let u =
y
x
, v =
x
y2
, then x =
1
u2v
, y =
1
uv
so
∂(x, y)
∂(u, v)
=
1
u4v3
;
∫∫
S
1
u4v3
dAuv =
∫ 4
1
∫ 2
1
1
u4v3
du dv =
35
256
.
40. 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 = 6
∫ 2π
0
∫ 1
0
(9− 3r cos θ − 2r sin θ)r dr dθ = 54π.
41. x = u, y =
w
u
, z = v +
w
u
,
∂(x, y, z)
∂(u, v, w)
= − 1
u
;
∫∫∫
S
v2w
u
dVuvw =
∫ 4
2
∫ 1
0
∫ 3
1
v2w
u
du dv dw = 2 ln 3.
42. 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 =
∫ 2
1
∫ 3
1
∫ 4
1
1
2
√
uvw
dw dv du = 4(
√
2− 1)(
√
3− 1).
43. (b)
∂(x, y)
∂(u, v)
· ∂(u, v)
∂(x, y)
=
∣∣∣∣
xu xv
yu yv
∣∣∣∣ ·
∣∣∣∣
ux uy
vx vy
∣∣∣∣ =
∣∣∣∣
xuux + xvvx xuuy + xvvy
yuux + yvvx yuuy + yvvy
∣∣∣∣ =
∣∣∣∣
xx xy
yx yy
∣∣∣∣ =
∣∣∣∣
1 0
0 1
∣∣∣∣ = 1.
44.
∂(u, v)
∂(x, y)
= 3xy4 = 3v so
∂(x, y)
∂(u, v)
=
1
3v
;
1
3
∫∫
S
sinu
v
dAuv =
1
3
∫ 2
1
∫ 2π
π
sinu
v
du dv = −2
3
ln 2.
45.
∂(u, v)
∂(x, y)
= 8xy so
∂(x, y)
∂(u, v)
=
1
8xy
; xy
∣∣∣∣
∂(x, y)
∂(u, v)
∣∣∣∣ = xy ·
1
8xy
=
1
8
so
1
8
∫∫
S
dAuv =
1
8
∫ 16
9
∫ 4
1
du dv =
21
8
.
46.
∂(u, v)
∂(x, y)
= −2(x2 +y2), so ∂(x, y)
∂(u, v)
= − 1
2(x2 + y2)
; (x4−y4)exy
∣∣∣∣
∂(x, y)
∂(u, v)
∣∣∣∣ =
x4 − y4
2(x2 + y2)
exy =
1
2
(x2−y2)exy = 1
2
veu,
so
1
2
∫∫
S
veu dAuv =
1
2
∫ 4
3
∫ 3
1
veu du dv =
7
4
(e3 − e).
47. Set u = x+y+ 2z, v = x−2y+ z, w = 4x+y+ z, then ∂(u, v, w)
∂(x, y, z)
=
∣∣∣∣∣∣
1 1 2
1 −2 1
4 1 1
∣∣∣∣∣∣
= 18, and V =
∫∫∫
R
dx dy dz =
∫ 6
−6
∫ 2
−2
∫ 3
−3
∂(x, y, z)
∂(u, v, w)
du dv dw = 6 · 4 · 12 · 1
18
= 16.
48. (a)
∂(x, y, z)
∂(r, θ, z)
=
∣∣∣∣∣∣
cos θ −r sin θ 0
sin θ r cos θ 0
0 0 1
∣∣∣∣∣∣
= r.
(b)
∂(x, y, z)
∂(ρ, φ, θ)
=
∣∣∣∣∣∣
sinφ cos θ ρ cosφ cos θ −ρ sinφ sin θ
sinφ sin θ ρ cosφ sin θ ρ sinφ cos θ
cosφ −ρ sinφ 0
∣∣∣∣∣∣
= ρ2 sinφ.
702 Chapter 14
49. The main motivation is to change the region of integration to one that has a simple description in either rectangular,
polar, cylindrical, or spherical coordinates.
50. First consider the case in which R is defined by a ≤ u(x, y) ≤ b, c ≤ v(x, y) ≤ d, for some functions u and v. If we
can solve for x and y in terms of u and v, then we can write
∫∫
R
f(x, y) dAxy =
∫∫
S
f(x(u, v), y(u, v))
∣∣∣∣
∂(x, y)
∂(u, v)
∣∣∣∣ dAuv,
where S is the rectangle a ≤ u ≤ b, c ≤ v ≤ d. For the more general case in which the boundary curves of R are
level curves of more than 2 functions, we can pick 2 of these functions, say u(x, y) and v(x, y), try to solve for x
and y in terms of u and v, and rewrite all of the inequalities in terms of u and v. This gives a region S in the
uv-plane, with one boundary curve which is a horizontal line segment and one which is a vertical line segment.
If we are very lucky, the other boundary curves may also be fairly simple and we may be able to compute the
resulting integral over S. See Examples 2 and 3.
Exercise Set 14.8
1. M =
∫ 1
0
∫ √x
0
(x + y) dy dx =
13
20
, Mx =
∫ 1
0
∫ √x
0
(x + y)y dy dx =
3
10
, My =
∫ 1
0
∫ √x
0
(x + y)x dy dx =
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
)
.
2. M =
∫ π
0
∫ sin x
0
y dy dx =
π
4
, x =
π
2
from the symmetry of the density and the region, Mx =
∫ π
0
∫ sin x
0
y2 dy dx =
4
9
, y =
Mx
M
=
16
9π
; mass
π
4
, center of gravity
(
π
2
,
16
9π
)
.
3. M =
∫ π/2
0
∫ a
0
r3 sin θ cos θ dr dθ =
a4
8
, x = y from the symmetry of the density and the region,
My =
∫ π/2
0
∫ a
0
r4 sin θ cos2 θ dr dθ =
a5
15
, x =
8a
15
; mass
a4
8
, center of gravity
(
8a
15
,
8a
15
)
.
4. M =
∫ π
0
∫ 1
0
r3 dr dθ =
π
4
, x = 0 from the symmetry of density and region, Mx =
∫ π
0
∫ 1
0
r4 sin θ dr dθ =
2
5
,
y =
8
5π
; mass
π
4
, center of gravity
(
0,
8
5π
)
.
5. M =
∫∫
R
δ(x, y) dA =
∫ 1
0
∫ 1
0
|x + y − 1| dx dy =
∫ 1
0
[∫ 1−x
0
(1− x− y) dy +
∫ 1
1−x
(x+ y − 1) dy
]
dx =
1
3
. x =
3
∫ 1
0
∫ 1
0
xδ(x, y) dy dx = 3
∫ 1
0
[∫ 1−x
0
x(1− x− y) dy +
∫ 1
1−x
x(x+ y − 1) dy
]
dx =
1
2
. By symmetry, y =
1
2
as
well; center of gravity
(
1
2
,
1
2
)
.
6. x =
1
M
∫∫
G
xδ(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.
7. V = 1, x =
∫ 1
0
∫ 1
0
∫ 1
0
x dz dy dx =
1
2
, similarly y = z =
1
2
; centroid
(
1
2
,
1
2
,
1
2
)
.
8. V = πr2h = 2π, x = y = 0 by symmetry,
∫∫∫
G
z dz dy dx =
∫ 2
0
∫ 2π
0
∫ 1
0
rz dr dθ dz = 2π, centroid = (0, 0, 1).
Exercise Set 14.8 703
9. True. This is the definition of “centroid”; see Section 6.7.
10. False. For example, suppose the lamina is the annulus 1 ≤ r ≤ 2 with constant density 1. The centroid is the
origin, which is not part of the annulus, so the density is 0 there. But the mass is not 0.
11. False. The coordinates are the first moments about the y- and x-axes, divided by the mass.
12. False. Density in 3-space has units of mass per unit volume.13. Let x = r cos θ, y = r sin θ, and dA = r dr dθ in formulas (11) and (12).
14. x = 0 from the symmetry of the region, A =
∫ 2π
0
∫ a(1+sin θ)
0
r dr dθ =
3πa2
2
, y =
1
A
∫ 2π
0
∫ a(1+sin θ)
0
r2 sin θ dr dθ =
2
3πa2
· 5πa
3
4
=
5a
6
; centroid
(
0,
5a
6
)
.
15. x = y from the symmetry of the region, A =
∫ π/2
0
∫ sin 2θ
0
r dr dθ =
π
8
, x =
1
A
∫ π/2
0
∫ sin 2θ
0
r2 cos θ dr dθ =
8
π
· 16
105
=
128
105π
; centroid
(
128
105π
,
128
105π
)
.
16. x = 0 from the symmetry of the region, A =
1
2
π(b2−a2), y = 1
A
∫ π
0
∫ b
a
r2 sin θ dr dθ =
1
A
2
3
(b3−a3) = 4(b
3 − a3)
3π(b2 − a2) ;
centroid
(
0,
4(b3 − a3)
3π(b2 − a2)
)
.
17. y = 0 from the symmetry of the region, A =
1
2
πa2, x =
1
A
∫ π/2
−π/2
∫ a
0
r2 cos θ dr dθ =
1
A
2
3
a3 =
4a
3π
; centroid
(
4a
3π
, 0
)
.
18. 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.
19. x = y = z from the symmetry of the region, V = 1/6, x =
1
V
∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
x dz dy dx = 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 =
∫ 1
−1
∫ 1−y2
0
∫ 1−z
0
dx dz dy =
4
5
, x =
1
V
∫ 1
−1
∫ 1−y2
0
∫ 1−z
0
x dx dz dy =
5
14
, y = 0 by symmetry, z =
1
V
∫ 1
−1
∫ 1−y2
0
∫ 1−z
0
z dx dz dy =
2
7
; the centroid is
(
5
14
, 0,
2
7
)
.
21. x = 1/2 and y = 0 from the symmetry of the region, V =
∫ 1
0
∫ 1
−1
∫ 1
y2
dz dy dx =
4
3
, z =
1
V
∫∫∫
G
z dV =
3
4
· 4
5
=
3
5
;
centroid
(
1
2
, 0,
3
5
)
.
704 Chapter 14
22. x = y from the symmetry of the region, V =
∫ 2
0
∫ 2
0
∫ xy
0
dz dy dx = 4, x =
1
V
∫∫∫
G
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 = πa3/6, x =
1
V
∫ a
0
∫ √a2−x2
0
∫ √a2−x2−y2
0
x dz dy dx =
1
V
∫ a
0
∫ √a2−x2
0
x
√
a2 − x2 − y2 dy dx = 1
V
∫ π/2
0
∫ a
0
r2
√
a2 − r2 cos θ dr dθ = 6
πa3
· πa
4
16
=
3a
8
; this gives us the
centroid
(
3a
8
,
3a
8
,
3a
8
)
.
24. x = y = 0 from the symmetry of the region, V = 2πa3/3, z =
1
V
∫ a
−a
∫ √a2−x2
−
√
a2−x2
∫ √a2−x2−y2
0
z dz dy dx =
1
V
∫ a
−a
∫ √a2−x2
−
√
a2−x2
1
2
(a2 − x2 − y2) dy dx = 1
V
∫ 2π
0
∫ a
0
1
2
(a2 − r2)r dr dθ = 3
2πa3
· πa
4
4
=
3a
8
; centroid
(
0, 0,
3a
8
)
.
25. M =
∫ a
0
∫ a
0
∫ a
0
(a− x) dz dy dx = a
4
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
· a
5
6
=
a
3
; mass
a4
2
, center of gravity
(a
3
,
a
2
,
a
2
)
.
26. M =
∫ a
−a
∫ √a2−x2
−
√
a2−x2
∫ h
0
(h − z) dz dy dx = π
2
a2h2, x = y = 0 from the symmetry of density and region, z =
1
M
∫∫∫
G
z(h− z) dV = 2
πa2h2
· πa
2h3
6
=
h
3
; mass
πa2h2
2
, center of gravity
(
0, 0,
h
3
)
.
27. M =
∫ 1
−1
∫ 1
0
∫ 1−y2
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
yz2 dV = 6 · 1
12
=
1
2
; mass
1
6
, center of gravity
(
0,
16
35
,
1
2
)
.
28. M =
∫ 3
0
∫ 9−x2
0
∫ 1
0
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 =
∫ 1
0
∫ 1
0
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 =
∫ 1
0
∫ 1
0
kx dy dx =
k
2
, x =
1
M
∫∫
R
kx2 dA =
2
k
· k
3
=
2
3
,
center of gravity
(
2
3
,
1
2
)
.
Exercise Set 14.8 705
30. (a) x = y = z from the symmetry of density and region, M =
∫ 1
0
∫ 1
0
∫ 1
0
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 =
∫ 1
0
∫ 1
0
∫ 1
0
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 =
∫ π
0
∫ sin x
0
∫ 1/(1+x2+y2)
0
dz dy dx ≈ 0.666633, x = 1
V
∫∫∫
G
x dV ≈ 1.177406, y = 1
V
∫∫∫
G
y dV ≈
0.353554, z =
1
V
∫∫∫
G
z dV ≈ 0.231557.
32. (b) Use polar coordinates for x and y to get V =
∫∫∫
G
dV =
∫ 2π
0
∫ a
0
∫ 1/(1+r2)
0
r dz dr dθ = π ln(1 + a2),
z =
1
V
∫∫∫
G
z dV =
a2
2(1 + a2) ln(1 + a2)
. Thus lim
a→0+
z =
1
2
; lim
a→+∞
z = 0. Also, lim
a→0+
z =
1
2
; lim
a→+∞
z = 0.
(c) Solve z = 1/4 for a to obtain a ≈ 1.980291. 2 4 6
a
z
(1.980, 1/4)
1/4
1/2
33. M =
∫ 2π
0
∫ 3
0
∫ 3
r
(3− z)r dz dr dθ =
∫ 2π
0
∫ 3
0
1
2
r(3− r)2 dr dθ = 27
8
∫ 2π
0
dθ =
27π
4
.
34. M =
∫ 2π
0
∫ a
0
∫ h
0
kzr dz dr dθ =
∫ 2π
0
∫ a
0
1
2
kh2r dr dθ =
1
4
ka2h2
∫ 2π
0
dθ =
πka2h2
2
.
35. M =
∫ 2π
0
∫ π
0
∫ a
0
kρ3 sinφdρ dφ dθ =
∫ 2π
0
∫ π
0
1
4
ka4 sinφdφ dθ =
1
2
ka4
∫ 2π
0
dθ = πka4.
36. M =
∫ 2π
0
∫ π
0
∫ 2
1
ρ sinφdρ dφ dθ =
∫ 2π
0
∫ π
0
3
2
sinφdφ dθ = 3
∫ 2π
0
dθ = 6π.
37. x̄ = ȳ = 0 from the symmetry of the region, V =
∫ 2π
0
∫ 1
0
∫ √2−r2
r2
r dz dr dθ =
∫ 2π
0
∫ 1
0
(r
√
2− r2 − r3) dr dθ =
π
6
(8
√
2− 7), z̄ = 1
V
∫ 2π
0
∫ 1
0
∫ √2−r2
r2
zr dz dr dθ =
6
(8
√
2− 7)π
· 7π
12
=
7
16
√
2− 14
; centroid
(
0, 0,
7
16
√
2− 14
)
.
38. x̄ = ȳ = 0 from the symmetry of the region, V = 8π/3, z̄ =
1
V
∫ 2π
0
∫ 2
0
∫ 2
r
zr dz dr dθ =
3
8π
· 4π = 3
2
; centroid
(
0, 0,
3
2
)
.
706 Chapter 14
39. ȳ = 0 from the symmetry of the region, V = 2
∫ π/2
0
∫ 2 cos θ
0
∫ r2
0
r dz dr dθ = 3π/2,
x̄ =
2
V
∫ π/2
0
∫ 2 cos θ
0
∫ r2
0
r2 cos θ dz dr dθ
4
3π
(π) = 4/3, z̄ =
1
V
∫ π/2
0
∫ 2 cos θ
0
∫ r2
0
zr dz dr dθ =
4
3π
(5π/6) = 10/9;
centroid (4/3, 0, 10/9).
40. M =
∫ π/2
0
∫ 2 cos θ
0
∫ 4−r2
0
zr dz dr dθ =
∫ π/2
0
∫ 2 cos θ
0
1
2
r(4−r2)2 dr dθ = 16
3
∫ π/2
0
(1−sin6 θ) dθ = (16/3)(11π/32) =
11π/6.
41. x̄ = ȳ = z̄ from the symmetry of the region, V = πa3/6, z̄ =
1
V
∫ π/2
0
∫ π/2
0
∫ a
0
ρ3 cosφ sinφdρ dφ dθ =
6
πa3
· πa
4
16
=
3a
8
; centroid
(
3a
8
,
3a
8
,
3a
8
)
.
42. x̄ = ȳ = 0 from the symmetry of the region, V =
∫ 2π
0
∫ π/3
0
∫ 4
0
ρ2 sinφdρ dφ dθ =
64π
3
,
z̄ =
1
V
∫ 2π
0
∫ π/3
0
∫ 4
0
ρ3 cosφ sinφdρ dφ dθ =
3
64π
· 48π = 9
4
; centroid
(
0, 0,
9
4
)
.
43. M =
∫ 2π
0
∫ π/4
0
∫ 1
0
ρ3 sinφdρ dφ dθ =
∫ 2π
0
∫ π/4
0
1
4
sinφdφ dθ =
1
8
(2−
√
2)
∫ 2π
0
dθ =
π
4
(2−
√
2).
44. x̄ = ȳ = 0 from the symmetry of density and region, M =
∫ 2π
0
∫ 1
0
∫ 1−r2
0
(r2 + z2)r dz dr dθ =
π
4
, z̄ =
1
M
∫ 2π
0
∫ 1
0
∫ 1−r2
0
z(r2 + z2)r dz dr dθ =
4
π
· 11π
120
=
11
30
; center of gravity
(
0, 0,
11
30
)
.
45. x̄ = ȳ = 0 from the symmetry of density and region, M =
∫ 2π
0
∫ 1
0
∫ r
0
zr dz dr dθ =
π
4
,
z̄ =
1
M
∫ 2π
0
∫ 1
0
∫ r
0
z2r dz dr dθ =
4
π
· 2π
15
=
8
15
; center of gravity
(
0, 0,
8
15
)
.
46. x̄ = ȳ = 0 from the symmetry of density and region, M =
∫ 2π
0
∫ π/2
0
∫ a
0
kρ3 sinφdρ dφ dθ =
πka4
2
, z̄ =
1
M
∫ 2π
0
∫ π/2
0
∫ a
0
kρ4 sinφ cosφdρ dφ dθ =
2
πka4
· πka
5
5
=
2a
5
; center of gravity
(
0, 0,
2a
5
)
.
47. x̄ = z̄ = 0 from the symmetry of the region, V = 54π/3−16π/3 = 38π/3, ȳ = 1
V
∫ π
0
∫ π
0
∫ 3
2
ρ3 sin2 φ sin θ dρ dφ dθ =
1
V
∫ π
0
∫ π
0
65
4
sin2 φ sin θ dφ dθ =
1
V
∫ π
0
65π
8
sin θ dθ =
3
38π
· 65π
4
=
195
152
; centroid
(
0,
195
152
, 0
)
.
48. M =
∫ 2π
0
∫ π
0
∫ R
0
δ0e
−(ρ/R)3ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π
0
1
3
(1− e−1)R3δ0 sinφdφ dθ =
4π
3
(1− e−1)δ0R3.
49. Ix =
∫ a0
∫ b
0
y2δ dy dx =
δab3
3
, Iy =
∫ a
0
∫ b
0
x2δ dy dx =
δa3b
3
, Iz = Ix + Iy =
δab(a2 + b2)
3
.
50. Ix =
∫ 2π
0
∫ a
0
r3 sin2 θ δ dr dθ =
δπa4
4
; Iy =
∫ 2π
0
∫ a
0
r3 cos2 θ δ dr dθ =
δπa4
4
= Ix; Iz = Ix + Iy =
δπa4
2
.
Exercise Set 14.8 707
51. Iz =
∫ 2π
0
∫ a
0
∫ h
0
r2δ r dz dr dθ = δ
∫ 2π
0
∫ a
0
∫ h
0
r3dz dr dθ =
1
2
δπa4h.
52. Iy =
∫ 2π
0
∫ a
0
∫ h
0
(r2 cos2 θ + z2)δr dz dr dθ = δ
∫ 2π
0
∫ a
0
(hr3 cos2 θ +
1
3
h3r) dr dθ =
= δ
∫ 2π
0
(
1
4
a4h cos2 θ +
1
6
a2h3
)
dθ = δ
(π
4
a4h+
π
3
a2h3
)
.
53. Iz =
∫ 2π
0
∫ a2
a1
∫ h
0
r2δ r dz dr dθ = δ
∫ 2π
0
∫ a2
a1
∫ h
0
r3 dz dr dθ =
1
2
δπh(a42 − a41).
54. Iz =
∫ 2π
0
∫ π
0
∫ a
0
(ρ2 sin2 φ)δ ρ2 sinφdρ dφ dθ = δ
∫ 2π
0
∫ π
0
∫ a
0
ρ4 sin3 φdρ dφ dθ =
8
15
δπa5.
55. (a) The solid generated by Rk as it revolves about L is a cylinder of height ∆yk and radius x
∗
k +
1
2
∆xk from
which a cylinder of height ∆yk and radius x
∗
k −
1
2
∆xk has been removed, so its volume is π(x
∗
k +
1
2
∆xk)
2 ∆yk −
π(x∗k −
1
2
∆xk)
2 ∆yk = 2πx
∗
k ∆xk ∆yk = 2πx
∗
k ∆Ak.
(b) From part (a), V =
∫∫
R
2πx dA = 2π
∫∫
R
x dA. From equation (13), this equals 2π · x · [area of R].
56. (a) V =
[
1
2
πa2
] [
2π
(
a+
4a
3π
)]
=
1
3
π(3π + 4)a3.
(b) The distance between the centroid and the line is
√
2
2
(
a+
4a
3π
)
, so V =
[
1
2
πa2
] [
2π
√
2
2
(
a+
4a
3π
)]
=
1
6
√
2π(3π + 4)a3.
57. x = k so V = πab · 2πk = 2π2abk.
58. y = 4 from the symmetry of the region; A =
∫ 2
−2
∫ 8−x2
x2
dy dx =
64
3
. So V =
64
3
· 2π · 4 = 512π
3
.
59. The region generates a cone of volume
1
3
πab2 when it is revolved about the x-axis, the area of the region is
1
2
ab
so
1
3
πab2 =
1
2
ab · 2πy, y = b
3
. A cone of volume
1
3
πa2b is generated when the region is revolved about the y-axis
so
1
3
πa2b =
1
2
ab · 2πx, x = a
3
. The centroid is
(
a
3
,
b
3
)
.
60. The centroid of the circle which generates the tube travels a distance s =
∫ 4π
0
√
sin2 t+ cos2 t+
1
16
dt =
√
17π,
so V = π
(
1
2
)2√
17π =
√
17π2
4
.
61. It is the point P in the plane of the lamina such that the lamina will balance on any knife-edge passing through
P . (If P is in the lamina, then the lamina will also balance on a point of support at P .)
708 Chapter 14
Chapter 14 Review Exercises
3. (a)
∫∫
R
dA (b)
∫∫∫
G
dV (c)
∫∫
R
√(
∂z
∂x
)2
+
(
∂z
∂y
)2
+ 1 dA
4. (a) x = a sinφ cos θ, y = a sinφ sin θ, z = a cosφ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.
(b) x = a cos θ, y = a sin θ, z = z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h.
5.
∫ 1
0
∫ 1+√1−y2
1−
√
1−y2
f(x, y) dx dy
6.
∫ 2
0
∫ 2x
x
f(x, y) dy dx+
∫ 3
2
∫ 6−x
x
f(x, y) dy dx
7. (a) The transformation sends (1, 0) to (a, c) and (0, 1) to (b, d). There are two possibilities: either (a, c) = (2, 1)
and (b, d) = (1, 2) or (a, c) = (1, 2) and (b, d) = (2, 1). So either a = 2, b = 1, c = 1, d = 2 or a = 1, b = 2, c = 2,
d = 1.
(b) For either transformation in part (a),
∣∣∣∣
∂(x, y)
∂(u, v)
∣∣∣∣ = 3, so the area is
∫∫
R
dA =
∫ 1
0
∫ 1
0
∣∣∣∣
∂(x, y)
∂(u, v)
∣∣∣∣ du dv =
∫ 1
0
∫ 1
0
3 du dv = 3. The diagonals of R cut it into 4 congruent right triangles. One of these has vertices (0, 0),
(
3
2
,
3
2
)
, and (2, 1), so its bases have lengths
3
2
√
2 and
1
2
√
2 and its area is
1
2
· 3
2
√
2 · 1
2
√
2 =
3
4
; hence R has area
4 · 3
4
= 3.
8. If 0 < x, y < π then 0 < sin
√
xy ≤ 1, with equality only on the hyperbola xy = π2/4, so 0 =
∫ π
0
∫ π
0
0 dy dx <
∫ π
0
∫ π
0
sin
√
xy dy dx <
∫ π
0
∫ π
0
1 dy dx = π2.
9.
∫ 1
1/2
2x cos(πx2) dx =
1
π
sin(πx2)
]1
1/2
= − 1√
2π
.
10.
∫ 2
0
x2
2
ey
3
]2y
x=−y
dy =
3
2
∫ 2
0
y2ey
3
dy =
1
2
ey
3
]2
0
=
1
2
(
e8 − 1
)
.
11.
∫ 1
0
∫ 2
2y
exey dx dy
12.
∫ π
0
∫ x
0
sinx
x
dy dx
Chapter 14 Review Exercises 709
13.
6
1
x
y
y = sin x
y = tan (x/2)
14. 0
p /2
p /6
r = a
r = a(1 + cos u)
15. 2
∫ 8
0
∫ y1/3
0
x2 sin y2 dx dy =
2
3
∫ 8
0
y sin y2 dy = −1
3
cos y2
]8
0
=
1
3
(1− cos 64) ≈ 0.20271.
16.
∫ π/2
0
∫ 2
0
(4− r2)r dr dθ = 2π.
17. sin 2θ = 2 sin θ cos θ =
2xy
x2 + y2
, and r = 2a sin θ is the circle x2 + (y− a)2 = a2, so
∫ a
0
∫ a+√a2−x2
a−
√
a2−x2
2xy
x2 + y2
dy dx =
∫ a
0
x
[
ln
(
a+
√
a2 − x2
)
− ln
(
a−
√
a2 − x2
)]
dx = a2.
18.
∫ π/2
π/4
∫ 2
0
4r2(cos θ sin θ) r dr dθ = −4 cos 2θ
]π/2
π/4
= 4.
19.
∫ 2
0
∫ 2−y/2
(y/2)1/3
dx dy =
∫ 2
0
(
2− y
2
−
(y
2
)1/3)
dy =
(
2y − y
2
4
− 3
2
(y
2
)4/3)]2
0
=
3
2
.
20. A = 6
∫ π/6
0
∫ cos 3θ
0
r dr dθ = 3
∫ π/6
0
cos2 3θ =
π
4
.
21.
∫ 2π
0
∫ 2
0
∫ 16
r4
r2 cos2 θ r dz dr dθ =
∫ 2π
0
cos2 θ dθ
∫ 2
0
r3(16− r4) dr = 32π.
22.
∫ π/2
0
∫ π/2
0
∫ 1
0
1
1 + ρ2
ρ2 sinφdρ dφ dθ =
(
1− π
4
) π
2
∫ π/2
0
sinφdφ =
(
1− π
4
) π
2
(− cosφ)
]π/2
0
=
(
1− π
4
) π
2
.
23. (a)
∫ 2π
0
∫ π/3
0
∫ a
0
(ρ2 sin2 φ)ρ2 sinφdρ dφ dθ =
∫ 2π
0
∫ π/3
0
∫ a
0
ρ4 sin3 φdρ dφ dθ.
(b)
∫ 2π
0
∫ √3a/2
0
∫ √a2−r2
r/
√
3
r2 dz rdr dθ =
∫ 2π
0
∫ √3a/2
0
∫ √a2−r2
r/
√
3
r3 dz dr dθ.
(c)
∫ √3a/2
−
√
3a/2
∫ √(3a2/4)−x2
−
√
(3a2/4)−x2
∫ √a2−x2−y2
√
x2+y2/
√
3
(x2 + y2) dz dy dx.
24. (a)
∫ 4
0
∫ √4x−x2
−
√
4x−x2
∫ 4x
x2+y2
dz dy dx (b)
∫ π/2
−π/2
∫ 4 cos θ
0
∫ 4r cos θ
r2
r dz dr dθ
25. V =
∫ 2π
0
∫ a/√3
0
∫ a
√
3r
r dz dr dθ = 2π
∫ a/√3
0
r(a−
√
3r) dr =
πa3
9
.
710 Chapter 14
26. The intersection of the two surfaces projects onto the yz-plane as 2y2 + z2 = 1, so
V = 4
∫ 1/√2
0
∫ √1−2y2
0
∫ 1−y2
y2+z2
dx dz dy = 4
∫ 1/√2
0
∫ √1−2y2
0
(1−2y2−z2) dz dy = 4
∫ 1/√2
0
2
3
(1−2y2)3/2 dy =
√
2π
4
.
27. The triangular region R is described by 0 ≤ x ≤ 1, −x ≤ y ≤ x. Hence S =
∫∫
R
√
z2x + z
2
y + 1 dA =
∫ 1
0
∫ x
−x
√
(4x)2 + 32 + 1 dy dx =
∫ 1
0
∫ x
−x
√
16x2 + 10 dy dx =
∫ 1
0
2x
√
16x2 + 10 dx =
1
24
(16x2 + 10)3/2
]1
0
=
1
12
(13
√
26− 5
√
10) ≈ 4.20632.
28. ‖ru × rv‖ =
√
2u2 + 2v2 + 4, S =
∫∫
u2+v2≤4
√
2u2 + 2v2 + 4 dA =
∫ 2π
0
∫ 2
0
√
2r2 + 4 r dr dθ =
8π
3
(3
√
3− 1).
29. (ru × rv)
∣∣∣
u=1
v=2
= 〈−2,−4, 1〉, tangent plane 2x+ 4y − z = 5.
30. u = −3, v = 0, (ru × rv)
∣∣∣
u=−3
v=0
= 〈−18, 0,−3〉, tangent plane 6x+ z = −9.
32. x =
1
10
u +
3
10
v and y = − 3
10
u +
1
10
v, hence |J(u, v)| =
∣∣∣∣∣
(
1
10
)2
+
(
3
10
)2∣∣∣∣∣ =
1
10
, and
∫∫
R
x− 3y
(3x+ y)2
dA =
1
10
∫ 3
1
∫ 4
0
u
v2
du dv =
1
10
∫ 3
1
1
v2
dv
∫ 4
0
u du =
1
10
2
3
8 =
8
15
.
33. (a) Add u and w to get x = ln(u + w) − ln 2; subtract w from u to get y = 1
2
u − 1
2
w, substitute these values
into v = y + 2z to get z = −1
4
u+
1
2
v +
1
4
w. Hence xu =
1
u+ w
, xv = 0, xw =
1
u+ w
; yu =
1
2
, yv = 0, yz = −
1
2
;
zu = −
1
4
, zv =
1
2
, zw =
1
4
, and thus
∂(x, y, z)
∂(u, v, w)
=
1
2(u+ w)
.
(b) V =
∫∫∫
G
dV =
∫ 3
1
∫ 2
1
∫ 4
0
1
2(u+ w)
dw dv du =
1
2
(7 ln 7− 5 ln 5− 3 ln 3) = 1
2
ln
823543
84375
≈ 1.139172308.
34. V =
4
3
πa3, d̄ =
3
4πa3
∫∫∫
ρ≤a
ρ dV =
3
4πa3
∫ π
0
∫ 2π
0
∫ a
0
ρ3 sinφdρ dθ dφ =
3
4πa3
· 2 · 2π · a
4
4
=
3
4
a.
35. A =
∫ 4
−4
∫ 2+y2/8
y2/4
dx dy =
∫ 4
−4
(
2− y
2
8
)
dy =
32
3
; ȳ = 0 by symmetry;
∫ 4
−4
∫ 2+y2/8
y2/4
x dx dy =
∫ 4
−4
(
2 +
1
4
y2 − 3
128
y4
)
dy =
256
15
, x̄ =
3
32
256
15
=
8
5
; centroid
(
8
5
, 0
)
.
36. A = πab/2, x̄ = 0 by symmetry,
∫ a
−a
∫ b√1−x2/a2
0
y dy dx =
1
2
∫ a
−a
b2
(
1− x
2
a2
)
dx =
2ab2
3
, centroid
(
0,
4b
3π
)
.
37. V =
1
3
πa2h, x̄ = ȳ = 0 by symmetry,
∫ 2π
0
∫ a
0
∫ h−rh/a
0
rz dz dr dθ = π
∫ a
0
rh2
(
1− r
a
)2
dr =
πa2h2
12
, centroid
(
0, 0,
h
4
)
.
Chapter 14 Making Connections 711
38. V =
∫ 2
−2
∫ 4
x2
∫ 4−y
0
dz dy dx =
∫ 2
−2
∫ 4
x2
(4 − y) dy dx =
∫ 2
−2
(
8− 4x2 + 1
2
x4
)
dx =
256
15
,
∫ 2
−2
∫ 4
x2
∫ 4−y
0
y dz dy dx =
∫ 2
−2
∫ 4
x2
(4y− y2) dy dx =
∫ 2
−2
(
1
3
x6 −2x4 + 32
3
)
dx =
1024
35
,
∫ 2
−2
∫ 4
x2
∫ 4−y
0
z dz dy dx =
∫ 2
−2
∫ 4
x2
1
2
(4− y)2 dy dx =
∫ 2
−2
(
−x
6
6
+ 2x4 − 8x2 + 32
3
)
dx =
2048
105
, x̄ = 0 by symmetry, centroid
(
0,
12
7
,
8
7
)
.
Chapter 14 Making Connections
1. (a) I2 =
[∫ +∞
0
e−x
2
dx
] [∫ +∞
0
e−y
2
dy
]
=
∫ +∞
0
[∫ +∞
0
e−x
2
dx
]
e−y
2
dy =
∫ +∞
0
∫ +∞
0
e−x
2
e−y
2
dx dy =
=
∫ +∞
0
∫ +∞
0
e−(x
2+y2) dx dy.
(b) I2 =
∫ π/2
0
∫ +∞
0
e−r
2
r dr dθ =
1
2
∫ π/2
0
dθ =
π
4
.
(c) Since I > 0, I =
√
π
4
=
√
π
2
.
2. The two quarter-circles with center at the origin and of radius A and
√
2A lie inside and outside of the square
with corners (0, 0), (A, 0), (A,A), (0, A), so the following inequalities hold:
∫ π/2
0
∫ A
0
1
(1 + r2)2
r dr dθ ≤
∫ A
0
∫ A
0
1
(1 + x2 + y2)2
dx dy ≤
∫ π/2
0
∫ √2A
0
1
(1 + r2)2
r dr dθ.
The integral on the left can be evaluated as
πA2
4(1 +A2)
and the integral on the right equals
2πA2
4(1 + 2A2)
. Since both
of these quantities tend to
π
4
as A→ +∞, it follows by sandwiching that
∫ +∞
0
∫ +∞
0
1
(1 + x2 + y2)2
dx dy =
π
4
.
3. (a) 1.173108605 (b)
∫ π
0
∫ 1
0
re−r
4
dr dθ = π
∫ 1
0
re−r
4
dr ≈ 1.173108605.
4. (a) At any point outside the closed sphere {x2 + y2 + z2 ≤ 1} the integrand is negative, so to maximize
the integral it suffices to include all points inside the sphere; hence the maximum value is taken on the region
G = {x2 + y2 + z2 ≤ 1}.
(b) 1.675516
(c)
∫ 2π
0
∫ π
0
∫ 1
0
(1− ρ2)ρ2 sinφdρ dφ dθ = 8π
15
.
5. (a) Let S1 be the set of points (x, y, z) which satisfy the equation x
2/3 + y2/3 + z2/3 = a2/3, and let S2 be the
set of points (x, y, z) where x = a(sinφ cos θ)3, y = a(sinφ sin θ)3, z = a cos3 φ, 0 ≤ φ ≤ π, 0 ≤ θ < 2π. If (x, y, z)
is a point of S2 then x
2/3 + y2/3 + z2/3 = a2/3[(sinφ cos θ)3 + (sinφ sin θ)3 + cos3 φ] = 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 = x
1/3, y1 = y
1/3, z1 = z
1/3, a1 = a
1/3.
Then x21 + y
2
1 + z
2
1 = a
2
1, so in spherical coordinates x1 = a1 sinφ cos θ, y1 = a1 sinφ sin θ, z1 = a1 cosφ, with
θ = tan−1
(
y1
x1
)
= tan−1
(y
x
)1/3
, φ = cos−1
z1
a1
= cos−1
(z
a
)1/3
. Then x = x31 = a
3
1(sinφ cos θ)
3 = a(sinφ cos θ)3,
similarly y = a(sinφ sin θ)3, z = a cosφ so (x, y, z) belongs to S2. Thus S1 = S2.
(b) Let a = 1 and r = (cos θ sinφ)3 i + (sin θ sinφ)3 j + cos3 φk, then S = 8
∫ π/2
0
∫ π/2
0
‖rθ × rφ‖ dφ dθ =
712 Chapter 14
72
∫ π/2
0
∫ π/2
0
sin θ cos θ sin4 φ cosφ
√
cos2 φ+ sin2 φ sin2 θ cos2 θ dθ dφ ≈ 4.4506.
6.
∂(x, y, z)
∂(ρ, φ, θ)
=
∣∣∣∣∣∣
sin3 φ cos3 θ 3ρ sin2 φ cosφ cos3 θ −3ρ sin3 φ cos2 θ sin θ
sin3 φ sin3 θ 3ρ sin2 φ cosφ sin3 θ 3ρ sin3 φ sin2 θ cos θ
cos3 φ −3ρ cos2 φ sinφ 0
∣∣∣∣∣∣
= 9ρ2 cos2 θ sin2 θ cos2 φ sin5 φ, so
V = 9
∫ 2π
0
∫ π
0
∫ a
0
ρ2 cos2 θ sin2 θ cos2 φ sin5 φdρ dφ dθ =
4
35
πa3.
Topics in Vector Calculus
Exercise Set 15.1
1. (a) III, because the vector field is independent of y and the direction is that of the negative x-axis for negative
x, and positive for positive.
(b) IV, because the y-component is constant, and the x-component varies periodically with x.
2. (a) I, since the vector field is constant.
(b) II, since the vector field points away from the origin.
3. (a) True. (b) True. (c) True.
4. (a) False, the lengths are equal to 1. (b) False, the y-component is zero. (c) False, the x-component is zero.
5.
x
y
6.
x
y
7.
x
y
8.
x
y
9.
x
y
10.
x
y
11. False, the k-component is nonzero.
12. False, the power of ‖r‖ should be 3.
13. True (this example is the curl of F).
14. False, φ is the divergence of F.
713
714 Chapter 15
15. (a) ∇φ = φxi + φyj =
y
1 + x2y2
i +
x
1 + x2y2
j = F, so F is conservative for all x, y.
(b) ∇φ = φxi + φyj + φzk = 2xi− 6yj + 8zk = F so F is conservative for all x, y and z.
16. (a) ∇φ = φxi + φyj = (6xy − y3)i + (4y + 3x2 − 3xy2)j = F, so F is conservative for all x, y.
(b) ∇φ = φxi + φyj + φzk = (sin z + y cosx)i + (sinx+ z cos y)j + (x cos z + sin y)k = F, so F is conservative for
all x, y, and z.
17. div F = 2x+ y, curl F = zi.
18. div F = z3 + 8y3x2 + 10zy, curl F = 5z2i + 3xz2j + 4xy4k.
19. div F = 0, curl F = (40x2z4 − 12xy3)i + (14y3z + 3y4)j− (16xz5 + 21y2z2)k.
20. div F = yexy + sin y + 2 sin z cos z, curl F = −xexyk.
21. div F =
2√
x2 + y2 + z2
, curl F = 0.
22. div F =
1
x
+ xzexyz +
x
x2 + z2
, curl F = −xyexyzi + z
x2 + z2
j + yzexyzk.
23. ∇ · (F×G) = ∇ · (−(z + 4y2)i + (4xy + 2xz)j + (2xy − x)k) = 4x.
24. ∇ · (F×G) = ∇ · ((x2yz2 − x2y2)i− xy2z2j + xy2zk) = −xy2.
25. ∇ · (∇× F) = ∇ · (− sin(x− y)k) = 0.
26. ∇ · (∇× F) = ∇ · (−zeyzi + xexzj + 3eyk) = 0.
27. ∇× (∇× F) = ∇× (xzi− yzj + yk) = (1 + y)i + xj.
28. ∇× (∇× F) = ∇× ((x+ 3y)i− yj− 2xyk) = −2xi + 2yj− 3k.
31. Let F = f i + gj + hk ; div (kF) = k
∂f
∂x
+ k
∂g
∂y
+ k
∂h
∂z
= k div F.
32. Let F = f i + gj + hk ; curl (kF) = k
(
∂h
∂y
− ∂g
∂z
)
i + k
(
∂f
∂z
− ∂h
∂x
)
j + k
(
∂g
∂x
− ∂f
∂y
)
k = k curl F.
33. Let F = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i +Q(x, y, z)j +R(x, y, z)k, then
div (F+G) =
(
∂f
∂x
+
∂P
∂x
)
+
(
∂g
∂y
+
∂Q
∂y
)
+
(
∂h
∂z
+
∂R
∂z
)
=
(
∂f
∂x
+
∂g
∂y
+
∂h
∂z
)
+
(
∂P
∂x
+
∂Q
∂y
+
∂R
∂z
)
= div F+
div G.
34. Let F = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i +Q(x, y, z)j +R(x, y, z)k, then
curl (F+G) =
[
∂
∂y
(h+R)− ∂
∂z
(g +Q)
]
i+
[
∂
∂z
(f + P )− ∂
∂x
(h+R)
]
j+
[
∂
∂x
(g +Q)− ∂
∂y
(f + P )
]
k; expand
and rearrange terms to get curl F + curl G.
35. Let F = f i + gj + hk ; div (φF) =
(
φ
∂f
∂x
+
∂φ
∂x
f
)
+
(
φ
∂g
∂y
+
∂φ
∂y
g
)
+
(
φ
∂h
∂z
+
∂φ
∂z
h
)
= φ
(
∂f
∂x
+
∂g
∂y
+
∂h
∂z
)
+
(
∂φ
∂x
f +
∂φ
∂y
g +
∂φ
∂z
h
)
= φ div F +∇φ · F.
Exercise Set 15.1 715
36. Let F = f i+ gj+hk ; curl (φF) =
[
∂
∂y
(φh)− ∂
∂z
(φg)
]
i+
[
∂
∂z
(φf)− ∂
∂x
(φh)
]
j+
[
∂
∂x
(φg)− ∂
∂y
(φf)
]
k; use the
product rule to expand each of the partial derivatives, rearrange to get φ curl F +∇φ× F.
37. Let F = f i + gj + hk ; div(curl F) =
∂
∂x
(
∂h
∂y
− ∂g
∂z
)
+
∂
∂y
(
∂f
∂z
− ∂h
∂x
)
+
∂
∂z
(
∂g
∂x
− ∂f
∂y
)
=
∂2h
∂x∂y
− ∂
2g
∂x∂z
+
∂2f
∂y∂z
− ∂
2h
∂y∂x
+
∂2g
∂z∂x
− ∂
2f
∂z∂y
= 0, assuming equality of mixed second partial derivatives, which follows from the
continuity assumptions.
38. curl (∇φ) =
(
∂2φ
∂y∂z
− ∂
2φ
∂z∂y
)
i+
(
∂2φ
∂z∂x
− ∂
2φ
∂x∂z
)
j+
(
∂2φ
∂x∂y
− ∂
2φ
∂y∂x
)
k = 0, assuming equality of mixed second
partial derivatives, which follows from the continuity assumptions.
39. ∇ · (kF) = k∇ · F, ∇ · (F + G) = ∇ · F +∇ · G, ∇ · (φF) = φ∇ · F +∇φ · F, ∇ · (∇× F) = 0.
40. ∇× (kF) = k∇× F, ∇× (F + G) = ∇× F +∇×G, ∇× (φF) = φ∇× F +∇φ× F, ∇× (∇φ) = 0.
41. (a) curl r = 0i + 0j + 0k = 0.
(b) ∇‖r‖ = ∇
√
x2 + y2 + z2 =
x√
x2 + y2 + z2
i +
y√
x2 + y2 + z2
j +
z√
x2 + y2 + z2
k =
r
‖r‖ .
42. (a) div r = 1 + 1 + 1 = 3.
(b) ∇ 1‖r‖ = ∇(x
2 + y2 + z2)−1/2 = − xi + yj + zk
(x2 + y2 + z2)3/2
= − r‖r‖3 .
43. (a) ∇f(r) = f ′(r) ∂r
∂x
i + f ′(r)
∂r
∂y
j + f ′(r)
∂r
∂z
k = f ′(r)∇r = f
′(r)
r
r.
(b) div[f(r)r] = f(r)div r +∇f(r) · r = 3f(r) + f
′(r)
r
r · r = 3f(r) + rf ′(r).
44. (a) curl[f(r)r] = f(r)curl r +∇f(r)× r = f(r)0 + f
′(r)
r
r× r = 0 + 0 = 0.
(b) ∇2f(r) = div[∇f(r)] = div
[
f ′(r)
r
r
]
=
f ′(r)
r
div r +∇f
′(r)
r
· r = 3f
′(r)
r
+
rf ′′(r)− f ′(r)
r3
r · r = 2f
′(r)
r
+
f ′′(r).
45. f(r) = 1/r3, f ′(r) = −3/r4, div(r/r3) = 3(1/r3) + r(−3/r4) = 0.
46. Multiply 3f(r) + rf ′(r) = 0 through by r2 to obtain 3r2f(r) + r3f ′(r) = 0, d[r3f(r)]/dr = 0, r3f(r) = C, f(r) =
C/r3, so F = Cr/r3 (an inverse-square field).
47. (a) At the point (x, y) the slope of the line along which the vector −yi+ xj lies is −x/y; the slope of the tangent
line to C at (x, y) is dy/dx, so dy/dx = −x/y.
(b) ydy = −xdx, y2/2 = −x2/2 +K1, x2 + y2 = K.
716 Chapter 15
48. dy/dx = x, y = x2/2 +K.
x
y
49. dy/dx = 1/x, y = lnx+K.
x
y
50. dy/dx = −y/x,(1/y)dy = (−1/x)dx, ln y = − lnx+K1, y = eK1e− ln x = K/x.
x
y
Exercise Set 15.2
1. (a)
∫
C
ds = length of line segment = 1. (b) 0, because sinxy = 0 along C.
2. (a)
∫
C
ds = length of line segment = 2. (b) 0, because x is constant and dx = 0.
3. Since F and r are parallel, F · r = ‖F‖‖r‖, and since F is constant,
∫
C
F · dr =
∫
C
d(F · r) =
∫
C
d(‖F‖‖r‖) =
√
2
∫ 4
−4
√
2dt = 16.
4.
∫
C
F · r = 0, since F is perpendicular to the curve.
5. By inspection the tangent vector in part (a) is given by T = j, so F ·T = F · j = sinx on C. But x = −π/2 on C,
thus sinx = −1, F ·T = −1 and
∫
C
F · dr =
∫
C
(−1)ds.
6. (a) Let α be the angle between F and T. Since ‖F‖ = 1, cosα = ‖F‖‖T‖ cosα = F · T, and
∫
C
F · T ds =
Exercise Set 15.2 717
∫
C
cosα(s) ds. From Figure 15.2.12(b) it is apparent that α is close to zero on most of the parabola, thus cosα ≈ 1
though cosα ≤ 1. Hence
∫
C
cosα(s) ds ≤
∫
C
ds and the first integral is close to the second.
(b) From Example 8(b)
∫
C
cosαds =
∫
C
F · dr ≈ 5.83629, and
∫
C
ds =
∫ 2
−1
√
1 + (2t)2 dt ≈ 6.125726619.
7.
∫
C
F · dr =
∫ 5
4
(8 · 0 + 8 · 1) dt = 8.
8.
∫
C
F · dr =
∫ 4
1
(2 · 1 + 5 · 0) dt = 6.
9.
∫
C
F · dr =
∫ 11
4
(0 · 0 + 2(−2) · 1) dt = −28.
10.
∫
C
F · dr =
∫ 6
−1
(−8(−t) · (−1) + 3(0) · 0) dt = −140.
11. (a) ds =
√(
dx
dt
)2
+
(
dy
dt
)2
dt, so
∫ 1
0
(2t−
√
t2)
√
4 + 4t2 dt =
∫ 1
0
2t
√
1 + t2 dt =
2
3
(1 + t2)3/2
]1
0
=
2
3
(2
√
2− 1).
(b)
∫ 1
0
(2t−
√
t2)2 dt = 1. (c)
∫ 1
0
(2t−
√
t2)2t dt =
2
3
.
12. (a)
∫ 1
0
t(3t2)(6t3)2
√
1 + 36t2 + 324t4 dt =
864
5
. (b)
∫ 1
0
t(3t2)(6t3)2 dt =
54
5
.
(c)
∫ 1
0
t(3t2)(6t3)26t dt =
648
11
. (d)
∫ 1
0
t(3t2)(6t3)218t2 dt = 162.
13. (a) C : x = t, y = t, 0 ≤ t ≤ 1;
∫ 1
0
6t dt = 3.
(b) C : x = t, y = t2, 0 ≤ t ≤ 1;
∫ 1
0
(3t+ 6t2 − 2t3)dt = 3.
(c) C : x = t, y = sin(πt/2), 0 ≤ t ≤ 1;
∫ 1
0
[3t+ 2 sin(πt/2) + πt cos(πt/2)− (π/2) sin(πt/2) cos(πt/2)]dt = 3.
(d) C : x = t3, y = t, 0 ≤ t ≤ 1;
∫ 1
0
(9t5 + 8t3 − t)dt = 3.
14. (a) C : x = t, y = t, z = t, 0 ≤ t ≤ 1;
∫ 1
0
(t+ t− t) dt = 1
2
.
(b) C : x = t, y = t2, z = t3, 0 ≤ t ≤ 1;
∫ 1
0
(t2 + t3(2t)− t(3t2)) dt = − 1
60
.
(c) C : x = cosπt, y = sinπt, z = t, 0 ≤ t ≤ 1;
∫ 1
0
(−π sin2 πt+ πt cosπt− cosπt) dt = −π
2
− 2
π
.
718 Chapter 15
15. False, a line integral is independent of the orientation of the curve.
16. False, it’s a scalar.
17. True, see (26).
18. True, since ∇f is normal to C; see (30).
19.
∫ 3
0
√
1 + t
1 + t
dt =
∫ 3
0
(1 + t)−1/2 dt = 2.
20.
√
5
∫ 1
0
1 + 2t
1 + t2
dt =
√
5(π/4 + ln 2).
21.
∫ 1
0
3(t2)(t2)(2t3/3)(1 + 2t2) dt = 2
∫ 1
0
t7(1 + 2t2) dt = 13/20.
22.
√
5
4
∫ 2π
0
e−t dt =
√
5(1− e−2π)/4.
23.
∫ π/4
0
(8 cos2 t− 16 sin2 t− 20 sin t cos t)dt = 1− π.
24.
∫ 1
−1
(
2
3
t− 2
3
t5/3 + t2/3
)
dt = 6/5.
25. C : x = (3− t)2/3, y = 3− t, 0 ≤ t ≤ 3;
∫ 3
0
1
3
(3− t)2dt = 3.
26. C : x = t2/3, y = t, −1 ≤ t ≤ 1;
∫ 1
−1
(
2
3
t2/3 − 2
3
t1/3 + t7/3
)
dt = 4/5.
27. C : x = cos t, y = sin t, 0 ≤ t ≤ π/2;
∫ π/2
0
(− sin t− cos2 t)dt = −1− π/4.
28. C : x = 3− t, y = 4− 3t, 0 ≤ t ≤ 1;
∫ 1
0
(−37 + 41t− 9t2)dt = −39/2.
29.
∫ 1
0
(−3)e3tdt = 1− e3.
30.
∫ π/2
0
(sin2 t cos t− sin2 t cos t+ t4(2t)) dt = π
6
192
.
31. (a)
∫ ln 2
0
(
e3t + e−3t
)√
e2t + e−2t dt =
63
64
√
17 +
1
4
ln(4 +
√
17)− 1
4
tanh−1
(
1
17
√
17
)
.
(b)
∫ π/2
0
[
et sin t cos t− (sin t− t) sin t+ (1 + t2)
]
dt =
1
24
π3 +
1
5
eπ/2 +
1
4
π +
6
5
.
32. (a)
∫ π/2
0
cos21 t sin9 t
√
(−3 cos2 t sin t)2 + (3 sin2 t cos t)2 dt = 3
∫ π/2
0
cos22 t sin10 t dt =
61,047
4,294,967,296
π.
Exercise Set 15.2 719
(b)
∫ e
1
(
t5 ln t+ 7t2(2t) + t4(ln t)
1
t
)
dt =
5
36
e6 +
59
16
e4 − 491
144
.
33. (a) C1 : (0, 0) to (1, 0);x = t, y = 0, 0 ≤ t ≤ 1, C2 : (1, 0) to (0, 1);x = 1 − t, y = t, 0 ≤ t ≤ 1, C3 :
(0, 1) to (0, 0);x = 0, y = 1− t, 0 ≤ t ≤ 1,
∫ 1
0
(0)dt+
∫ 1
0
(−1)dt+
∫ 1
0
(0)dt = −1.
(b) C1 : (0, 0) to (1, 0);x = t, y = 0, 0 ≤ t ≤ 1, C2 : (1, 0) to (1, 1);x = 1, y = t, 0 ≤ t ≤ 1, C3 : (1, 1) to (0, 1);x =
1−t, y = 1, 0 ≤ t ≤ 1, C4 : (0, 1) to (0, 0);x = 0, y = 1−t, 0 ≤ t ≤ 1,
∫ 1
0
(0)dt+
∫ 1
0
(−1)dt+
∫ 1
0
(−1)dt+
∫ 1
0
(0)dt =
−2.
34. (a) C1 : (0, 0) to (1, 1);x = t, y = t, 0 ≤ t ≤ 1, C2 : (1, 1) to (2, 0);x = 1 + t, y = 1 − t, 0 ≤ t ≤ 1, C3 :
(2, 0) to (0, 0);x = 2− 2t, y = 0, 0 ≤ t ≤ 1,
∫ 1
0
(0)dt+
∫ 1
0
2dt+
∫ 1
0
(0)dt = 2.
(b) C1 : (−5, 0) to (5, 0);x = t, y = 0,−5 ≤ t ≤ 5, C2 : x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ π,
∫ 5
−5
(0)dt +
∫ π
0
(−25)dt = −25π.
35. C1 : x = t, y = z = 0, 0 ≤ t ≤ 1,
∫ 1
0
0 dt = 0; C2 : x = 1, y = t, z = 0, 0 ≤ t ≤ 1,
∫ 1
0
(−t) dt = −1
2
; C3 : x =
1, y = 1, z = t, 0 ≤ t ≤ 1,
∫ 1
0
3 dt = 3;
∫
C
x2z dx− yx2 dy + 3 dz = 0− 1
2
+ 3 =
5
2
.
36. C1 : (0, 0, 0) to (1, 1, 0);x = t, y = t, z = 0, 0 ≤ t ≤ 1, C2 : (1, 1, 0) to (1, 1, 1);x = 1, y = 1, z = t, 0 ≤ t ≤ 1, C3 :
(1, 1, 1) to (0, 0, 0);x = 1− t, y = 1− t, z = 1− t, 0 ≤ t ≤ 1,
∫ 1
0
(−t3)dt+
∫ 1
0
3 dt+
∫ 1
0
−3dt = −1/4.
37.
∫ π
0
(0)dt = 0.
38.
∫ 1
0
(e2t − 4e−t)dt = e2/2 + 4e−1 − 9/2.
39.
∫ 1
0
e−tdt = 1− e−1
40.
∫ π/2
0
(7 sin2 t cos t+ 3 sin t cos t)dt = 23/6.
41. Represent the circular arc by x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π/2.
∫
C
x
√
yds = 9
√
3
∫ π/2
0
√
sin t cos t dt = 6
√
3.
42. δ(x, y) = k
√
x2 + y2 where k is the constant of proportionality,
∫
C
k
√
x2 + y2ds = k
∫ 1
0
et(
√
2et) dt =
√
2k
∫ 1
0
e2t dt = (e2 − 1)k/
√
2.
43.
∫
C
kx
1 + y2
ds = 15k
∫ π/2
0
cos t
1 + 9 sin2 t
dt = 5k tan−1 3.
44. δ(x, y, z) = kz where k is the constant of proportionality,
∫
C
k z ds =
∫ 4
1
k(4
√
t)(2 + 1/t) dt = 136k/3.
720 Chapter 15
45. C : x = t2, y = t, 0 ≤ t ≤ 1; W =
∫ 1
0
3t4dt = 3/5.
46. W =
∫ 3
1
(t2 + 1− 1/t3 + 1/t)dt = 92/9 + ln 3.
47. W =
∫ 1
0
(t3 + 5t6)dt = 27/28.
48. C1 : (0, 0, 0) to (1, 3, 1); x = t, y = 3t, z = t, 0 ≤ t ≤ 1, C2 : (1, 3, 1) to (2,−1, 4); x = 1 + t, y = 3− 4t, z = 1 + 3t,
0 ≤ t ≤ 1, W =
∫ 1
0
(4t+ 8t2)dt+
∫ 1
0
(−11− 17t− 11t2)dt = −37/2.
49. C : x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ π/2,
∫ π/2
0
(
−1
4
sin t+ cos t
)
dt = 3/4.
50. C1 : (0, 3) to (6, 3);x = 6t, y = 3, 0 ≤ t ≤ 1, C2 : (6, 3) to (6, 0);x = 6, y = 3 − 3t, 0 ≤ t ≤ 1,
∫ 1
0
6
36t2 + 9
dt +
∫ 1
0
−12
36 + 9(1− t)2 dt =
1
3
tan−1 2− 2
3
tan−1(1/2).
51. Represent the parabola by x = t, y = t2, 0 ≤ t ≤ 2.
∫
C
3x ds =
∫ 2
0
3t
√
1 + 4t2 dt = (17
√
17− 1)/4.
52. Represent the semicircle by x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π.
∫
C
x2 y ds =
∫ π
0
16 cos2 t sin t dt = 32/3.
53. (a) 2πrh = 2π(1)2 = 4π.
(b) S =
∫
C
z(t) dt, since the average height is 2.
(c) C : x = cos t, y = sin t, 0 ≤ t ≤ 2π;S =
∫ 2π
0
(2 + (1/2) sin 3t) dt = 4π.
54. C : x = a cos t, y = −a sin t, 0 ≤ t ≤ 2π,
∫
C
x dy − y dx
x2 + y2
=
∫ 2π
0
−a2 cos2 t− a2 sin2 t
a2
dt = −
∫ 2π
0
dt = −2π.
55. W =
∫
C
F · dr =
∫ 1
0
(λt2(1− t), t− λt(1− t)) · (1, λ− 2λt) dt = −λ/12, W = 1 when λ = −12.
56. The force exerted by the farmer is F =
(
150 + 20− 1
10
z
)
k =
(
170− 3
4π
t
)
k, so F · dr =
(
170− 1
10
z
)
dz, and
W =
∫ 60
0
(
170− 1
10
z
)
dz = 10,020 foot-pounds. Note that the functions x(z), y(z) are irrelevant.
57. (a) From (8), ∆sk =
∫ tk
tk−1
‖r′(t)‖ dt, thus m∆tk ≤ ∆sk ≤ M∆tk for all k. Obviously ∆sk ≤ M(max∆tk),
and since the right side of this inequality is independent of k, it follows that max∆sk ≤ M(max∆tk). Similarly
m(max∆tk) ≤ max∆sk.
(b) This follows from max∆tk ≤
1
m
max∆sk and max∆sk ≤Mmax∆tk.
Exercise Set 15.3 721
Exercise Set 15.3
1. ∂x/∂y = 0 = ∂y/∂x, conservative so ∂φ/∂x = x and ∂φ/∂y = y, φ = x2/2 + k(y), k′(y) = y, k(y) = y2/2 + K,
φ = x2/2 + y2/2 +K.
2. ∂(3y2)/∂y = 6y = ∂(6xy)/∂x, conservative so ∂φ/∂x = 3y2 and ∂φ/∂y = 6xy, φ = 3xy2 +k(y), 6xy+k′(y) = 6xy,
k′(y) = 0, k(y) = K, φ = 3xy2 +K.
3. ∂(x2y)/∂y = x2 and ∂(5xy2)/∂x = 5y2, not conservative.
4. ∂(ex cos y)/∂y = −ex sin y