Chapter 15

January 27, 2005 11:55 L24-CH15 Sheet number 1 Page number 655 black

CHAPTER 15
Multiple Integrals

EXERCISE SET 15.1
1 2 1 3 1 3
1. (x + 3)dy dx = (2x + 6)dx = 7 2. (2x − 4y)dy dx = 4x dx = 16
0 0 0 1 −1 1

4 1 4 0 2 0
1
3. x2 y dx dy = y dy = 2 4. (x2 + y 2 )dx dy = (3 + 3y 2 )dy = 14
2 0 2 3 −2 −1 −2

ln 3 ln 2 ln 3
5. ex+y dy dx = ex dx = 2
0 0 0

2 1 2
1
6. y sin x dy dx = sin x dx = (1 − cos 2)/2
0 0 0 2

0 5 0 6 7 6
7. dx dy = 3 dy = 3 8. dy dx = 10dx = 20
−1 2 −1 4 −3 4

1 1 1
x 1
9. dy dx = 1− dx = 1 − ln 2
0 0 (xy + 1)2 0 x+1

π 2 π
10. x cos xy dy dx = (sin 2x − sin x)dx = −2
π/2 1 π/2

ln 2 1 ln 2
2 1 x
11. xy ey x dy dx = (e − 1)dx = (1 − ln 2)/2
0 0 0 2

4 2 4
1 1 1
12. dy dx = − dx = ln(25/24)
3 1 (x + y)2 3 x+1 x+2

1 2 1
13. 4xy 3 dy dx = 0 dx = 0
−1 −2 −1

1 1
xy 1 √ √
14. dy dx = [x(x2 + 2)1/2 − x(x2 + 1)1/2 ]dx = (3 3 − 4 2 + 1)/3
0 0 x2 + y 2 + 1 0

1 3 1
15. x 1 − x2 dy dx = x(1 − x2 )1/2 dx = 1/3
0 2 0

π/2 π/3 π/2
x π2
16. (x sin y − y sin x)dy dx = − sin x dx = π 2 /144
0 0 0 2 18

17. (a) x∗ = k/2 − 1/4, k = 1, 2, 3, 4; yl = l/2 − 1/4, l = 1, 2, 3, 4,
k
∗

4 4 4 4
f (x, y) dxdy ≈ f (x∗ , yl )∆Akl =
k
∗
[(k/2−1/4)2 +(l/2−1/4)](1/2)2 = 37/4
R k=1 l=1 k=1 l=1
2 2
(b) (x2 + y) dxdy = 28/3; the error is |37/4 − 28/3| = 1/12
0 0

655


656 Chapter 15

18. (a) x∗ = k/2 − 1/4, k = 1, 2, 3, 4; yl = l/2 − 1/4, l = 1, 2, 3, 4,
k
∗

4 4 4 4
f (x, y) dxdy ≈ f (x∗ , yl )∆Akl =
k
∗
[(k/2 − 1/4) − 2(l/2 − 1/4)](1/2)2 = −4
R k=1 l=1 k=1 l=1
2 2
(b) (x − 2y) dxdy = −4; the error is zero
0 0

19. (a) z (b) z

(1, 0, 4) (0, 0, 5)

(0, 4, 3)

y y

(2, 5, 0) (3, 4, 0)
x x

z z
20. (a) (b) (2, 2, 8)

(0, 0, 2)

y y

(2, 2, 0)
(1, 1, 0)

x x

5 2 5
21. V = (2x + y)dy dx = (2x + 3/2)dx = 19
3 1 3

3 2 3
22. V = (3x3 + 3x2 y)dy dx = (6x3 + 6x2 )dx = 172
1 0 1

2 3 2
23. V = x2 dy dx = 3x2 dx = 8
0 0 0

3 4 3
24. V = 5(1 − x/3)dy dx = 5(4 − 4x/3)dx = 30
0 0 0

1/2 π 1/2 π
25. x cos(xy) cos2 πx dy dx = cos2 πx sin(xy) dx
0 0 0 0
1/2 1/2
1 1
= cos2 πx sin πx dx = − cos3 πx =
0 3π 0 3π


Exercise Set 15.2 657

5 2 5 3
26. (a) z (b) V = y dy dx + (−2y + 6) dy dx
0 0 0 2

(0, 2, 2)
= 10 + 5 = 15

y
3

5 (5, 3, 0)
x

π/2 1 π/2 x=1 π/2
2 2 2 2
27. fave = y sin xy dx dy = − cos xy dy = (1 − cos y) dy = 1 −
π 0 0 π 0 x=0 π 0 π

1 3 1 3
1 √
28. average = x(x2 + y)1/2 dx dy = [(1 + y)3/2 − y 3/2 ]dy = 2(31 − 9 3)/45
3 0 0 0 9
1 2 1 ◦
1 1 44 14
29. Tave = 10 − 8x2 − 2y 2 dy dx = − 16x2 dx = C
2 0 0 2 0 3 3

b d
1 1
30. fave = k dy dx = (b − a)(d − c)k = k
A(R) a c A(R)

31. 1.381737122 32. 2.230985141

b d b d
33. f (x, y)dA = g(x)h(y)dy dx = g(x) h(y)dy dx
a c a c
R
b d
= g(x)dx h(y)dy
a c

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 x 1
1 4
1. xy 2 dy dx = (x − x7 )dx = 1/40
0 x2 0 3

3/2 3−y 3/2
2. y dx dy = (3y − 2y 2 )dy = 7/24
1 y 1

√
3 9−y 2 3
3. y dx dy = y 9 − y 2 dy = 9
0 0 0

1 x 1 x 1
4. x/y dy dx = x1/2 y −1/2 dy dx = 2(x − x3/2 )dx = 13/80
1/4 x2 1/4 x2 1/4


658 Chapter 15

√ √
2π x3 2π
5. √
sin(y/x)dy dx = √
[−x cos(x2 ) + x]dx = π/2
π 0 π

1 x2 1 π x2 π
1
6. (x2 − y)dy dx = 2x4 dx = 4/5 7. cos(y/x)dy dx = sin x dx = 1
−1 −x2 −1 π/2 0 x π/2

1 x 1 1 x 1
2 2 1 3
8. ex dy dx = xex dx = (e − 1)/2 9. y x2 − y 2 dy dx = x dx = 1/12
0 0 0 0 0 0 3

2 y2 2
2
10. ex/y dx dy = (e − 1)y 2 dy = 7(e − 1)/3
1 0 1

2 x2 4 2
11. (a) f (x, y) dydx (b) √
f (x, y) dxdy
0 0 0 y
√ √
1 x 1 y
12. (a) f (x, y) dydx (b) f (x, y) dxdy
0 x2 0 y2

2 3 4 3 5 3
13. (a) f (x, y) dydx + f (x, y) dydx + f (x, y) dydx
1 −2x+5 2 1 4 2x−7
3 (y+7)/2
(b) f (x, y) dxdy
1 (5−y)/2
√ √
1 1−x2 1 1−y 2
14. (a) √ f (x, y) dydx (b) √ f (x, y) dxdy
−1 − 1−x2 −1 − 1−y 2

2 x2 2
1 5 16
15. (a) xy dy dx = x dx =
0 0 0 2 3
3 (y+7)/2 3
(b) xy dx dy = (3y 2 + 3y)dy = 38
1 −(y−5)/2 1

√
1 x 1
16. (a) (x + y)dy dx = (x3/2 + x/2 − x3 − x4 /2)dx = 3/10
0 x2 0
√ √
1 1−x2 1 1−x2 1
(b) √ x dy dx + √ y dy dx = 2x 1 − x2 dx + 0 = 0
−1 − 1−x2 −1 − 1−x2 −1

8 x 8
17. (a) x2 dy dx = (x3 − 16x)dx = 576
4 16/x 4
4 8 8 8 8
512 4096 8
512 − y 3
(b) x2 dxdy + x2 dx dy = − dy + dy
2 16/y 4 y 4 3 3y 3 4 3
640 1088
= + = 576
3 3
2 y 2
1 4
18. (a) xy 2 dx dy = y dy = 31/10
1 0 1 2
1 2 2 2 1 2
8x − x4
(b) xy 2 dydx + xy 2 dydx = 7x/3 dx + dx = 7/6 + 29/15 = 31/10
0 1 1 x 0 1 3



√
1 1−x2 1
19. (a) √ (3x − 2y)dy dx = 6x 1 − x2 dx = 0
−1 − 1−x2 −1
√
1 1−y 2 1
(b) √ (3x − 2y) dxdy = −4y 1 − y 2 dy = 0
−1 − 1−y 2 −1

√
5 25−x2 5
20. (a) y dy dx = (5x − x2 )dx = 125/6
0 5−x 0
√
5 25−y 2 5
(b) y dxdy = y 25 − y 2 − 5 + y dy = 125/6
0 5−y 0

√
4 y 4
1 √
21. x(1 + y 2 )−1/2 dx dy = y(1 + y 2 )−1/2 dy = ( 17 − 1)/2
0 0 0 2

π x π
22. x cos y dy dx = x sin x dx = π
0 0 0

2 6−y 2
1
23. xy dx dy = (36y − 12y 2 + y 3 − y 5 )dy = 50/3
0 y2 0 2
√
π/4 1/ 2 π/4
1
24. x dx dy = cos 2y dy = 1/8
0 sin y 0 4

1 x 1
25. (x − 1)dy dx = (−x4 + x3 + x2 − x)dx = −7/60
0 x3 0

√ √
1/ 2 2x 1 1/x 1/ 2 1
26. 2
x dy dx + √
2
x dy dx = x3 dx + √ (x − x3 )dx = 1/8
0 x 1/ 2 x 0 1/ 2

y
27. (a)
4

3

2

1
x
–2 –1 0.5 1.5

(b) x = (−1.8414, 0.1586), (1.1462, 3.1462)

1.1462 x+2 1.1462
(c) x dA ≈ x dydx = x(x + 2 − ex ) dx ≈ −0.4044
−1.8414 ex −1.8414
R
3.1462 ln y 3.1462
ln2 y (y − 2)2
(d) x dA ≈ x dxdy = − dy ≈ −0.4044
0.1586 y−2 0.1586 2 2
R


660 Chapter 15

28. (a) y (b) (1, 3), (3, 27)
25

15 R

5
x
1 2 3

3 4x3 −x4 3
224
(c) x dy dx = x[(4x3 − x4 ) − (3 − 4x + 4x2 )] dx =
1 3−4x+4x2 1 15

π/4 cos x π/4 √
29. A = dy dx = (cos x − sin x)dx = 2−1
0 sin x 0

1 −y 2 1
30. A = dx dy = (−y 2 − 3y + 4)dy = 125/6
−4 3y−4 −4

3 9−y 2 3
31. A = dx dy = 8(1 − y 2 /9)dy = 32
−3 1−y 2 /9 −3

1 cosh x 1
32. A = dy dx = (cosh x − sinh x)dx = 1 − e−1
0 sinh x 0

4 6−3x/2 4
33. (3 − 3x/4 − y/2) dy dx = [(3 − 3x/4)(6 − 3x/2) − (6 − 3x/2)2 /4] dx = 12
0 0 0

√
2 4−x2 2
34. 4 − x2 dy dx = (4 − x2 ) dx = 16/3
0 0 0

√
3 9−x2 3
35. V = √ (3 − x)dy dx = (6 9 − x2 − 2x 9 − x2 )dx = 27π
−3 − 9−x2 −3

1 x 1
36. V = (x2 + 3y 2 )dy dx = (2x3 − x4 − x6 )dx = 11/70
0 x2 0

3 2 3
37. V = (9x2 + y 2 )dy dx = (18x2 + 8/3)dx = 170
0 0 0

1 1 1
38. V = (1 − x)dx dy = (1/2 − y 2 + y 4 /2)dy = 8/15
−1 y2 −1

√
3/2 9−4x2 3/2
39. V = √ (y + 3)dy dx = 6 9 − 4x2 dx = 27π/2
−3/2 − 9−4x2 −3/2

3 3 3
40. V = (9 − x2 )dx dy = (18 − 3y 2 + y 6 /81)dy = 216/7
0 y 2 /3 0

√
5 25−x2 5
41. V = 8 25 − x2 dy dx = 8 (25 − x2 )dx = 2000/3
0 0 0



√
2 1−(y−1)2 2
1
42. V = 2 (x2 + y 2 )dx dy = 2 [1 − (y − 1)2 ]3/2 + y 2 [1 − (y − 1)2 ]1/2 dy,
0 0 0 3
π/2
1
let y − 1 = sin θ to get V = 2 cos3 θ + (1 + sin θ)2 cos θ cos θ dθ which eventually yields
−π/2 3
V = 3π/2
√
1 1−x2 1
8
43. V = 4 (1 − x2 − y 2 )dy dx = (1 − x2 )3/2 dx = π/2
0 0 3 0

√
2 4−x2 2
1
44. V = (x2 + y 2 )dy dx = x2 4 − x2 + (4 − x2 )3/2 dx = 2π
0 0 0 3
√
2 2 8 x/2 e2 2
45. f (x, y)dx dy 46. f (x, y)dy dx 47. f (x, y)dy dx
0 y2 0 0 1 ln x
√
1 e π/2 sin x 1 x
48. f (x, y)dx dy 49. f (x, y)dy dx 50. f (x, y)dy dx
0 ey 0 0 0 x2

4 y/4 4
1 −y2
e−y dx dy = ye dy = (1 − e−16 )/8
2
51.
0 0 0 4
1 2x 1
52. cos(x2 )dy dx = 2x cos(x2 )dx = sin 1
0 0 0

2 x2 2
3 3
53. ex dy dx = x2 ex dx = (e8 − 1)/3
0 0 0

ln 3 3 ln 3
1 1
54. x dx dy = (9 − e2y )dy = (9 ln 3 − 4)
0 ey 2 0 2

2 y2 2
55. sin(y 3 )dx dy = y 2 sin(y 3 )dy = (1 − cos 8)/3
0 0 0

1 e 1
56. x dy dx = (ex − xex )dx = e/2 − 1
0 ex 0

4 2
57. (a) √
sin πy 3 dy dx; the inner integral is non-elementary.
0 x
2 y2 2 2
1
sin πy 3 dx dy = y 2 sin πy 3 dy = − cos πy 3 =0
0 0 0 3π 0
1 π/2
(b) sec2 (cos x)dx dy ; the inner integral is non-elementary.
0 sin−1 y
π/2 sin x π/2
sec2 (cos x)dy dx = sec2 (cos x) sin x dx = tan 1
0 0 0
√
2 4−x2 2
1
58. V = 4 (x2 + y 2 ) dy dx = 4 x2 4 − x2 + (4 − x2 )3/2 dx (x = 2 sin θ)
0 0 0 3
π/2
64 64 128 64 π 64 π 128 π 1 · 3
= + sin2 θ − sin4 θ dθ = + − = 8π
0 3 3 3 3 2 3 4 3 2 2·4


662 Chapter 15

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 − y 2 , so 1/8 of the volume of
π
the sphere of radius 1, thus .
6
1 1 1
¯ 1 1 x π
61. Area of triangle is 1/2, so f = 2 dy dx = 2 − dx = − ln 2
0 x 1 + x2 0 1 + x2 1 + x2 2

2
62. Area = (3x − x2 − x) dx = 4/3, so
0
2 3x−x2 2
¯ 3
f= (x2 − xy)dy dx =
3
(−2x3 + 2x4 − x5 /2)dx = −
3 8
=−
2
4 0 x 4 0 4 15 5

1
63. Tave = (5xy + x2 ) dA. The diamond has corners (±2, 0), (0, ±4) and thus has area
A(R)
R
1
A(R) = 4 2(4) = 16m2 . Since 5xy is an odd function of x (as well as y), 5xy dA = 0. Since
2
R
x2 is an even function of both x and y,
2◦
4−2x 2 2
4 1 2 1 1 4 3 1 4
Tave = x2 dA = st x2 dydx = (4 − 2x)x2 dx = x − x = C
16 4 0 0 4 0 4 3 2 0 3
R
x,y>0

64. The area of the lens is πR2 = 4π and the average thickness Tave is
√
2 4−x2 2
4 1 1
Tave = 1 − (x2 + y 2 )/4 dydx = (4 − x2 )3/2 dx (x = 2 cos θ)
4π 0 0 π 0 6
8 π
8 1·3π 1
= sin4 θ dθ = = in
3π 0 3π 2 · 4 2 2

65. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so
a sin x
V = 1 + x + y dy dx = 0.676089
0 x/2

EXERCISE SET 15.3
π/2 sin θ π/2
1
1. r cos θdr dθ = sin2 θ cos θ dθ = 1/6
0 0 0 2

π 1+cos θ π
1
2. r dr dθ = (1 + cos θ)2 dθ = 3π/4
0 0 0 2

π/2 a sin θ π/2
a3 2
3. r2 dr dθ = sin3 θ dθ = a3
0 0 0 3 9

π/6 cos 3θ π/6
1
4. r dr dθ = cos2 3θ dθ = π/24
0 0 0 2



π 1−sin θ π
1
5. r2 cos θ dr dθ = (1 − sin θ)3 cos θ dθ = 0
0 0 0 3

π/2 cos θ π/2
1
6. r3 dr dθ = cos4 θ dθ = 3π/64
0 0 0 4

2π 1−cos θ 2π
1
7. A = r dr dθ = (1 − cos θ)2 dθ = 3π/2
0 0 0 2

π/2 sin 2θ π/2
8. A = 4 r dr dθ = 2 sin2 2θ dθ = π/2
0 0 0

π/2 1 π/2
1
9. A = r dr dθ = (1 − sin2 2θ)dθ = π/16
π/4 sin 2θ π/4 2

π/3 2 π/3 √
10. A = 2 r dr dθ = (4 − sec2 θ)dθ = 4π/3 − 3
0 sec θ 0

5π/6 4 sin θ 3π/2 1
11. A = f (r, θ) r dr dθ 12. A = f (r, θ)r dr dθ
π/6 2 π/2 1+cos θ

π/2 3 π/2 2 sin θ
13. V = 8 r 9 − r2 dr dθ 14. V = 2 r2 dr dθ
0 1 0 0

π/2 cos θ π/2 3
15. V = 2 (1 − r2 )r dr dθ 16. V = 4 dr dθ
0 0 0 1

π/2 3
128 √ π/2
64 √
17. V = 8 r 9 − r2 dr dθ = 2 dθ = 2π
0 1 3 0 3

π/2 2 sin θ π/2
16
18. V = 2 r2 dr dθ = sin3 θ dθ = 32/9
0 0 3 0

π/2 cos θ π/2
1
19. V = 2 (1 − r2 )r dr dθ = (2 cos2 θ − cos4 θ)dθ = 5π/32
0 0 2 0

π/2 3 π/2
20. V = 4 dr dθ = 8 dθ = 4π
0 1 0

π/2 3 sin θ π/2
27
21. V = r2 sin θ drdθ = 9 sin4 θ dθ = π
0 0 0 16

π/2 2 π 2
22. V = 2 4 − r2 r drdθ + 2 4 − r2 r drdθ
0 2 cos θ π/2 0
π/2 π
16 16 32 8
= (1 − cos2 θ)3/2 θ dθ + dθ = + π
0 3 π/2 3 9 3

2π 1 2π
1
e−r r dr dθ = (1 − e−1 ) dθ = (1 − e−1 )π
2
23.
0 0 2 0


664 Chapter 15

π/2 3 π/2
24. r 9 − r2 dr dθ = 9 dθ = 9π/2
0 0 0

π/4 2 π/4
1 1 π
25. r dr dθ = ln 5 dθ = ln 5
0 0 1 + r2 2 0 8

π/2 2 cos θ π/2
16
26. 2r2 sin θ dr dθ = cos3 θ sin θ dθ = 1/3
π/4 0 3 π/4

π/2 1 π/2
1
27. r3 dr dθ = dθ = π/8
0 0 4 0

2π 2 2π
1
e−r r dr dθ = (1 − e−4 ) dθ = (1 − e−4 )π
2
28.
0 0 2 0

π/2 2 cos θ π/2
8
29. r2 dr dθ = cos3 θ dθ = 16/9
0 0 3 0

π/2 1 π/2
1 π
30. cos(r2 )r dr dθ = sin 1 dθ = sin 1
0 0 2 0 4

π/2 a
r π
31. dr dθ = 1 − 1/ 1 + a2
0 0 (1 + r2 )3/2 2

π/4 sec θ tan θ
1 π/4 √
32. r2 dr dθ = sec3 θ tan3 θ dθ = 2( 2 + 1)/45
0 0 3 0

π/4 2
r π √
33. √ dr dθ = ( 5 − 1)
0 0 1+r 2 4

π/2 5 π/2
1
34. r dr dθ = (25 − 9 csc2 θ)dθ
tan−1 (3/4) 3 csc θ 2 tan−1 (3/4)

25 π 25
= − tan−1 (3/4) − 6 = tan−1 (4/3) − 6
2 2 2
2π a 2π
a2
35. V = hr dr dθ = h dθ = πa2 h
0 0 0 2

π/2 a a
c 2 4c 4 2
36. (a) V = 8 (a − r2 )1/2 r dr dθ = − π(a2 − r2 )3/2 = πa c
0 0 a 3a 0 3
4
(b) V ≈ π(6378.1370)2 6356.5231 ≈ 1,083,168,200,000 km3
3

π/2 a sin θ π/2
c 2 2
37. V = 2 (a − r2 )1/2 r dr dθ = a2 c (1 − cos3 θ)dθ = (3π − 4)a2 c/9
0 0 a 3 0

√
π/4 a 2 cos 2θ π/4
38. A = 4 r dr dθ = 4a2 cos 2θ dθ = 2a2
0 0 0



π/4 4 sin θ π/2 4 sin θ
39. A = √ r dr dθ + r dr dθ
π/6 8 cos 2θ π/4 0
π/4 π/2 √
= (8 sin2 θ − 4 cos 2θ)dθ + 8 sin2 θ dθ = 4π/3 + 2 3 − 2
π/6 π/4

φ 2a sin θ φ
1
40. A = r dr dθ = 2a2 sin2 θ dθ = a2 φ − a2 sin 2φ
0 0 0 2
+∞ +∞ +∞ +∞
e−x dx e−y dy = e−x dx e−y dy
2 2 2 2
41. (a) I 2 =
0 0 0 0
+∞ +∞ +∞ +∞
e−x e−y dx dy = e−(x
2 2 2
+y 2 )
= dx dy
0 0 0 0
π/2 +∞
1 π/2
√
e−r r dr dθ =
2
(b) I 2 = dθ = π/4 (c) I= π/2
0 0 2 0
√
42. 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 A A A π/2 2A
1 1 1
rdr dθ ≤ dx dy ≤ rdr dθ
0 0 (1 + r2 )2 0 0 (1 + x2 + y 2 )2 0 0 (1 + r2 )2
2
πA
The integral on the left can be evaluated as and the integral on the right equals
4(1 + A2 )
2
2πA π
2)
. Since both of these quantities tend to as A → +∞, it follows by sandwiching that
4(1 + 2A 4
+∞ +∞
1 π
dx dy = .
0 0 (1 + x2 + y 2 )2 4
π 1 1
re−r dr dθ = π re−r dr ≈ 1.173108605
4 4
43. (a) 1.173108605 (b)
0 0 0

2π R 2π R R
44. V = D(r)r dr dθ = ke−r r dr dθ = −2πk(1 + r)e−r = 2πk[1 − (R + 1)e−R ]
0 0 0 0 0

tan−1 (2) 2 tan−1 (2) tan−1 (2)
45. r3 cos2 θ dr dθ = 4 cos2 θ dθ = 2 (1 + cos(2θ)) dθ
tan−1 (1/3) 0 tan−1 (1/3) tan−1 (1/3)
√ √
= 2(tan−1 2 − tan−1 (1/3)) + 2/ 5 − 1/ 10

EXERCISE SET 15.4
1. (a) z (b) z (c) z

y
x x
x y y


666 Chapter 15

z z
2. (a) (b)

x

y
y

x

(c) z

y
x

5 3
3. (a) x = u, y = v, z = + u − 2v (b) x = u, y = v, z = u2
2 2
v 1 2 5
4. (a) x = u, y = v, z = (b) x = u, y = v, z = v −
1 + u2 3 3

5. (a) x = 5 cos u, y = 5 sin u, z = v; 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1
(b) x = 2 cos u, y = v, z = 2 sin u; 0 ≤ u ≤ 2π, 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

1
10. x = r cos θ, y = r sin θ, z = e−r
2
9. x = r cos θ, y = r sin θ, z =
1 + r2
11. x = r cos θ, y = r sin θ, z = 2r2 cos θ sin θ

12. x = r cos θ, y = r sin θ, z = r2 (cos2 θ − sin2 θ)
√ √
13. x = r cos θ, y = r sin θ, z = 9 − r2 ; r ≤ 5
√
1 1 3
14. x = r cos θ, y = r sin θ, z = r; r ≤ 3 15. x = ρ cos θ, y = ρ sin θ, z = ρ
2 2 2
16. x = 3 cos θ, y = 3 sin θ, z = 3 cot φ 17. z = x − 2y; a plane

18. y = x2 + z 2 , 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 + y 2 ; 0 ≤ z ≤ 4; part of a circular paraboloid

21. (x/3)2 + (y/4)2 = z 2 ; 0 ≤ z ≤ 1; part of an elliptic cone

22. x2 + (y/2)2 + (z/3)2 = 1; an ellipsoid
√
23. (a) x = r cos θ, y = r sin θ, z = r, 0 ≤ r ≤ 2; x = u, y = v, z = u2 + v 2 ; 0 ≤ u2 + v 2 ≤ 4
√
24. (a) I: x = r cos θ, y = r sin θ, z = r2 , 0 ≤ r ≤ 2; II: x = u, y = v, z = u2 + v 2 ; u2 + v 2 ≤ 2

25. (a) 0 ≤ u ≤ 3, 0 ≤ v ≤ π (b) 0 ≤ u ≤ 4, −π/2 ≤ v ≤ π/2

26. (a) 0 ≤ u ≤ 6, −π ≤ v ≤ 0 (b) 0 ≤ u ≤ 5, π/2 ≤ v ≤ 3π/2

27. (a) 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π (b) 0 ≤ φ ≤ π, 0 ≤ θ ≤ π

28. (a) π/2 ≤ φ ≤ π, 0 ≤ θ ≤ 2π (b) 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/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
√ √
√ √ 2 π 2
33. ru × rv = ( 2/2)i − ( 2/2)j + (1/2)k; x − y + z=
2 8

34. ru × rv = 2i − ln 2k; 2x − (ln 2)z = 0

35. z = 9 − y 2 , zx = 0, zy = −y/ 9 − y 2 , zx + zy + 1 = 9/(9 − y 2 ),
2 2

2 3 2
3
S= dy dx = 3π dx = 6π
0 −3 9 − y2 0

4 4−x 4
36. z = 8 − 2x − 2y, zx + zy + 1 = 4 + 4 + 1 = 9, S =
2 2
3 dy dx = 3(4 − x)dx = 24
0 0 0

37. z 2 = 4x2 + 4y 2 , 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus
1 x √ √ 1 √
zx + zy + 1 = (16x2 + 16y 2 )/z 2 + 1 = 5, S =
2 2
5 dy dx = 5 (x − x2 )dx = 5/6
0 x2 0

38. z 2 = x2 + y 2 , zx = x/z, zy = y/z, zx + zy + 1 = (z 2 + y 2 )/z 2 + 1 = 2,
2 2

√ π/2 2 cos θ √ √ π/2 √
S= 2 dA = 2 2 r dr dθ = 4 2 cos2 θ dθ = 2π
0 0 0
R

39. zx = −2x, zy = −2y, zx + zy + 1 = 4x2 + 4y 2 + 1,
2 2

2π 1
S= 4x2 + 4y 2 + 1 dA = r 4r2 + 1 dr dθ
0 0
R
1 √ 2π √
= (5 5 − 1) dθ = (5 5 − 1)π/6
12 0


668 Chapter 15

40. zx = 2, zy = 2y, zx + zy + 1 = 5 + 4y 2 ,
2 2

1 y 1 √
S= 5 + 4y 2 dx dy = y 5 + 4y 2 dy = (27 − 5 5)/12
0 0 0

41. ∂r/∂u = cos vi + sin vj + 2uk, ∂r/∂v = −u sin vi + u cos vj,
√ 2π 2 √ √
∂r/∂u × ∂r/∂v = u 4u2 + 1; S = u 4u2 + 1 du dv = (17 17 − 5 5)π/6
0 1

42. ∂r/∂u = cos vi + sin vj + k, ∂r/∂v = −u sin vi + u cos vj,
2v √
√
√ π/2
2 3
∂r/∂u × ∂r/∂v = 2u; S = 2 u du dv = π
0 0 12

43. zx = y, zy = x, zx + zy + 1 = x2 + y 2 + 1,
2 2

π/6 3
1 √ π/6 √
S= x2 + y 2 + 1 dA = r r2 + 1 dr dθ = (10 10 − 1) dθ = (10 10 − 1)π/18
0 0 3 0
R

44. zx = x, zy = y, zx + zy + 1 = x2 + y 2 + 1,
2 2
√
2π 8 2π
26
S= x2 + y2 + 1 dA = r r2 + 1 dr dθ = dθ = 52π/3
0 0 3 0
R

45. On the sphere, zx = −x/z and zy = −y/z so zx + zy + 1 = (x2 + y 2 + z 2 )/z 2 = 16/(16 − x2 − y 2 );
2 2

the planes z = 1 and z = 2 intersect the sphere along the circles x2 + y 2 = 15 and x2 + y 2 = 12;
√
2π 15 2π
4 4r
S= dA = √
√ dr dθ = 4 dθ = 8π
16 − x2 − y 2 0 12 16 − r2 0
R

46. On the sphere, zx = −x/z and zy = −y/z so zx + zy + 1 = (x2 + y 2 + z 2 )/z 2 = 8/(8 − x2 − y 2 );
2 2
2 2
the cone cuts the sphere in the circle x + y = 4;
√
2π 2
2 2r √ 2π √
S= √ dr dθ = (8 − 4 2) dθ = 8(2 − 2)π
0 0 8−r 2
0

47. r(u, v) = a cos u sin vi + a sin u sin vj + a cos vk, ru × rv = a2 sin v,
π 2π π
S= a2 sin v du dv = 2πa2 sin v dv = 4πa2
0 0 0

h 2π
48. r = r cos ui + r sin uj + vk, ru × rv = r; S = r du dv = 2πrh
0 0

h x h y 2 2 h2 x2 + h2 y 2
49. zx = , zy = , zx + zy + 1 = + 1 = (a2 + h2 )/a2 ,
a x2 + y 2 x2 + y 2a a2 (x2 + y 2 )
2π a
√ 2π
a2 + h2 1
S= r dr dθ = a a2 + h2 dθ = πa a2 + h2
0 0 a 2 0



50. (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 cos ui + a0 sin uj + b0 k, 0 ≤ u ≤ 2π. A point on the circle (x − a)2 + z 2 = 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) cos ui + (a + b cos v) sin uj + b sin vk, which yields the
result.

51. ∂r/∂u = −(a + b cos v) sin ui + (a + b cos v) cos uj,
∂r/∂v = −b sin v cos ui − b sin v sin uj + b cos vk, ∂r/∂u × ∂r/∂v = b(a + b cos v);
2π 2π
S= b(a + b cos v)du dv = 4π 2 ab
0 0

√ 4π 5 5
52. ru × rv = u2 + 1; S = u2 + 1 du dv = 4π u2 + 1 du = 174.7199011
0 0 0

53. z = −1 when v ≈ 0.27955, z = 1 when v ≈ 2.86204, ru × rv = | cos v|;
2π 2.86204
S= | cos v| dv du ≈ 9.099
0 0.27955

54. (a) Let S1 be the set of points (x, y, z) which satisfy the equation x2/3 + y 2/3 + z 2/3 = a2/3 , and
let S2 be the set of points (x, y, z) where x = a(sin φ cos θ)3 , y = a(sin φ sin θ)3 , z = a cos3 φ,
0 ≤ φ ≤ π, 0 ≤ θ < 2π.
If (x, y, z) is a point of S2 then
x2/3 + y 2/3 + z 2/3 = a2/3 [(sin φ cos θ)3 + (sin φ sin θ)3 + cos3 φ] = a2/3
so (x, y, z) belongs to S1 .
If (x, y, z) is a point of S1 then x2/3 + y 2/3 + z 2/3 = a2/3 . Let
x1 = x1/3 , y1 = y 1/3 , z1 = z 1/3 , a1 = a1/3 . Then x2 + y1 + z1 = a2 , so in spherical coordinates
1
2 2
1
x1 = a1 sin φ cos θ, y1 = a1 sin φ sin θ, z1 = a1 cos φ, with
y1 y 1/3 z1 z 1/3
θ = tan−1 = tan−1 , φ = cos−1 = cos−1 . Then
x1 x a1 a
x = x3 = a3 (sin φ cos θ)3 = a(sin φ cos θ)3 , similarly y = a(sin φ sin θ)3 , z = a cos φ so (x, y, z)
1 1
belongs to S2 . Thus S1 = S2

(b) Let a = 1 and r = (cos θ sin φ)3 i + (sin θ sin φ)3 j + cos3 φk, then
π/2 π/2
S=8 rθ × rφ dφ dθ
0 0
π/2 π/2
= 72 sin θ cos θ sin4 φ cos φ cos2 φ + sin2 φ sin2 θ cos2 θ dθ dφ ≈ 4.4506
0 0

55. (a) (x/a)2 +(y/b)2 +(z/c)2 = sin2 φ cos2 θ+sin2 φ sin2 θ+cos2 φ = sin2 φ+cos2 φ = 1, an ellipsoid
(b) r(φ, θ) = 2 sin φ cos θ, 3 sin φ sin θ, 4 cos φ ; rφ ×rθ = 2 6 sin2 φ cos θ, 4 sin2 φ sin θ, 3 cos φ sin φ ,

rφ × rθ = 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ,
2π π
S= 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ dφ dθ ≈ 111.5457699
0 0


670 Chapter 15

56. (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/v 2

(c) z

x
y

x 2 y 2 z 2
57. + + = 1, ellipsoid
a b c
x 2 y 2 z 2
58. + − = 1, hyperboloid of one sheet
a b c
x 2 y 2 z 2
59. − − + = 1, hyperboloid of two sheets
a b c

EXERCISE SET 15.5
1 2 1 1 2 1
1. (x2 + y 2 + z 2 )dx dy dz = (1/3 + y 2 + z 2 )dy dz = (10/3 + 2z 2 )dz = 8
−1 0 0 −1 0 −1

√
1/2 π 1 1/2 π
1 1/2
1 1 3−2
2. zx sin xy dz dy dx = x sin xy dy dx = (1 − cos πx)dx = +
1/3 0 0 1/3 0 2 1/3 2 12 4π

2 y2 z 2 y2 2
1 7 1 5 1 47
3. yz dx dz dy = (yz 2 + yz)dz dy = y + y − y dy =
0 −1 −1 0 −1 0 3 2 6 3

π/4 1 x2 π/4 1 π/4
1 √
4. x cos y dz dx dy = x3 cos y dx dy = cos y dy = 2/8
0 0 0 0 0 0 4
√ √
3 9−z 2 x 3 9−z 2 3
1 3 1
5. xy dy dx dz = x dx dz = (81 − 18z 2 + z 4 )dz = 81/5
0 0 0 0 0 2 0 8

3 x2 ln z 3 x2 3
1 5 3 3
6. xey dy dz dx = (xz − x)dz dx = x − x + x2 dx = 118/3
1 x 0 1 x 1 2 2
√ √
2 4−x2 3−x2 −y 2 2 4−x2
7. x dz dy dx = [2x(4 − x2 ) − 2xy 2 ]dy dx
0 0 −5+x2 +y 2 0 0
2
4
= x(4 − x2 )3/2 dx = 128/15
0 3
√
2 2 3y 2 2 2
y π π
8. dx dy dz = dy dz = (2 − z)dz = π/6
1 z 0 x2 + y 2 1 z 3 1 3



π 1 π/6 π 1 π
9. xy sin yz dz dy dx = x[1 − cos(πy/6)]dy dx = (1 − 3/π)x dx = π(π − 3)/2
0 0 0 0 0 0

1 1−x2 y 1 1−x2 1
1
10. y dz dy dx = y 2 dy dx = (1 − x2 )3 dx = 32/105
−1 0 0 −1 0 −1 3

√ √ √
2 x 2−x2 2 x 2
1 1 3
11. xyz dz dy dx = xy(2 − x2 )2 dy dx = x (2 − x2 )2 dx = 1/6
0 0 0 0 0 2 0 4

π/2 π/2 xy π/2 π/2 π/2 √
12. cos(z/y)dz dx dy = y sin x dx dy = y cos y dy = (5π − 6 3)/12
π/6 y 0 π/6 y π/6

3 2 1
√
x + z2
13. dz dy dx ≈ 9.425
0 1 −2 y

√ √
1 1−x2 1−x2 −y 2
e−x −y 2 −z 2
2
14. 8 dz dy dx ≈ 2.381
0 0 0

4 (4−x)/2 (12−3x−6y)/4 4 (4−x)/2
1
15. V = dz dy dx = (12 − 3x − 6y)dy dx
0 0 0 0 0 4
4
3
= (4 − x)2 dx = 4
0 16
√
1 1−x y 1 1−x
√ 1
2
16. V = dz dy dx = y dy dx = (1 − x)3/2 dx = 4/15
0 0 0 0 0 0 3

2 4 4−y 2 4 2
1
17. V = 2 dz dy dx = 2 (4 − y)dy dx = 2 8 − 4x2 + x4 dx = 256/15
0 x2 0 0 x2 0 2

√
1 y 1−y 2 1 y 1
18. V = dz dx dy = 1 − y 2 dx dy = y 1 − y 2 dy = 1/3
0 0 0 0 0 0

19. The projection of the curve of intersection onto the xy-plane is x2 + y 2 = 1,
√
1 1−x2 4−3y 2
(a) V = √ f (x, y, z)dz dy dx
−1 − 1−x2 4x2 +y 2
√
1 1−y 2 4−3y 2
(b) V = √ f (x, y, z)dz dx dy
−1 − 1−y 2 4x2 +y 2

20. The projection of the curve of intersection onto the xy-plane is 2x2 + y 2 = 4,
√ √
2 4−2x2 8−x2 −y 2
(a) V = √ √ f (x, y, z)dz dy dx
− 2 − 4−2x2 3x2 +y 2
√
2 (4−y 2 )/2 8−x2 −y 2
(b) V = √ f (x, y, z)dz dx dy
−2 − (4−y 2 )/2 3x2 +y 2


672 Chapter 15

21. The projection of the curve of intersection onto the xy-plane is x2 + y 2 = 1,
√
1 1−x2 4−3y 2
V =4 dz dy dx
0 0 4x2 +y 2

22. The projection of the curve of intersection onto the xy-plane is 2x2 + y 2 = 4,
√ √
2 4−2x2 8−x2 −y 2
V =4 dz dy dx
0 0 3x2 +y 2

√ √ √
3 9−x2 /3 x+3 1 1−x2 1−x2
23. V = 2 dz dy dx 24. V = 8 dz dy dx
−3 0 0 0 0 0

25. (a) z (b) z (c) z

(0, 9, 9)
(0, 0, 1)
(0, 0, 1)

y
y
(0, –1, 0) (1, 0, 0) x
(1, 2, 0)
(3, 9, 0) y
x
x

26. (a) z (b) z

(0, 0, 2)

(0, 0, 2)
(0, 2, 0)
y y

(3, 9, 0)
(2, 0, 0)
x x

(c) z

(0, 0, 4)

y

(2, 2, 0)

x

1 1−x 1−x−y 1 1−x 1−x−y
3
27. V = dz dy dx = 1/6, fave = 6 (x + y + z) dz dy dx =
0 0 0 0 0 0 4

28. The integrand is an odd function of each of x, y, and z, so the answer is zero.



3π
29. The volume V = √ , and thus
2
√
2
rave = x2 + y 2 + z 2 dV
3π
G
√ √
1/ 2
√
1−2x2 6−7x2 −y 2
2
= √ √ x2 + y 2 + z 2 dz dy dx ≈ 3.291
3π −1/ 2 − 1−2x2 5x2 +5y 2

1 1 1
1
30. V = 1, dave = (x − z)2 + (y − z)2 + z 2 dx dy dz ≈ 0.771
V 0 0 0

a b(1−x/a) c(1−x/a−y/b) b a(1−y/b) c(1−x/a−y/b)
31. (a) dz dy dx, dz dx dy,
0 0 0 0 0 0
c a(1−z/c) b(1−x/a−z/c) a c(1−x/a) b(1−x/a−z/c)
dy dx dz, dy dz dx,
0 0 0 0 0 0
c b(1−z/c) a(1−y/b−z/c) b c(1−y/b) a(1−y/b−z/c)
dx dy dz, dx dz dy
0 0 0 0 0 0

(b) Use the ﬁrst integral in Part (a) to get
a b(1−x/a) a 2
x y 1 x 1
c 1− − dy dx = bc 1 − dx = abc
0 0 a b 0 2 a 6
√ √ 2 2 2 2
a b 1−x2 /a2 c 1−x /a −y /b
32. V = 8 dz dy dx
0 0 0
√
2 4−x2 5
33. (a) f (x, y, z) dz dy dx
0 0 0
√ √
9 3− x 3− x 2 4−x2 8−y
(b) f (x, y, z) dz dy dx (c) f (x, y, z) dz dy dx
0 0 y 0 0 y
√ √
3 9−x2 9−x2 −y 2
34. (a) f (x, y, z)dz dy dx
0 0 0
4 x/2 2 2 4−x2 4−y
(b) f (x, y, z)dz dy dx (c) f (x, y, z)dz dy dx
0 0 0 0 0 x2

35. (a) At any point outside the closed sphere {x2 + y 2 + z 2 ≤ 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 + y 2 + z 2 ≤ 1}.
2π π 1
π2
(b) 4.934802202 (c) (1 − ρ2 )ρ dρ dφ dθ =
0 0 0 2

b d b d
36. f (x)g(y)h(z)dz dy dx = f (x)g(y) h(z)dz dy dx
a c k a c k

b d
= f (x) g(y)dy dx h(z)dz
a c k

b d
= f (x)dx g(y)dy h(z)dz
a c k


674 Chapter 15

1 1 π/2
37. (a) x dx y 2 dy sin z dz = (0)(1/3)(1) = 0
−1 0 0

1 ln 3 ln 2
(b) e2x dx ey dy e−z dz = [(e2 − 1)/2](2)(1/2) = (e2 − 1)/2
0 0 0

EXERCISE SET 15.6
1. (a) m1 and m3 are equidistant from x = 5, but m3 has a greater mass, so the sum is positive.
(b) Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum
is 5(0 − a) + 10(5 − a) + 20(10 − a) = 0 for equilibrium, so 250 − 35a = 0, a = 50/7. The
fulcrum should be placed 50/7 units to the right of m1 .

2. (a) The sum must be negative, since m1 , m2 and m3 are all to the left of the fulcrum, and the
magnitude of the moment of m1 about x = 4 is by itself greater than the moment of m about
x = 4 (i.e. 40 > 28), so even if we replace the masses of m2 and m3 with 0, the sum is
negative.
(b) At equilibrium, 10(0 − 4) + 3(2 − 4) + 4(3 − 4) + m(6 − 4) = 0, m = 25
1 1 1 1
1 1
3. A = 1, x = x dy dx = , y= y dy dx =
0 0 2 0 0 2

1
4. A = 2, x = x dy dx, and the region of integration is symmetric with respect to the x-axes
2
G
and the integrand is an odd function of x, so x = 0. Likewise, y = 0.
1 x 1 x
5. A = 1/2, x dA = x dy dx = 1/3, y dA = y dy dx = 1/6;
0 0 0 0
R R

centroid (2/3, 1/3)

1 x2 1 x2
6. A = dy dx = 1/3, x dA = x dy dx = 1/4,
0 0 0 0
R
1 x2
y dA = y dy dx = 1/10; centroid (3/4, 3/10)
0 0
R

1 2−x2 1 2−x2
7. A = dy dx = 7/6, x dA = x dy dx = 5/12,
0 x 0 x
R
1 2−x2
y dA = y dy dx = 19/15; centroid (5/14, 38/35)
0 x
R
√
1 1−x2
π 1 4 4
8. A = , x dA = x dy dx = ,x= , y= by symmetry
4 0 0 3 3π 3π
R

9. x = 0 from the symmetry of the region,
1 π b
2 3 4(b3 − a3 )
A= π(b2 − a2 ), y dA = r2 sin θ dr dθ = (b − a3 ); centroid x = 0, y = .
2 0 a 3 3π(b2 − a2 )
R



10. y = 0 from the symmetry of the region, A = πa2 /2,
π/2 a
4a
x dA = r2 cos θ dr dθ = 2a3 /3; centroid ,0
−π/2 0 3π
R

1 1
11. M = δ(x, y)dA = |x + y − 1| dx dy
0 0
R
1 1−x 1
1
= (1 − x − y) dy + (x + y − 1) dy dx =
0 0 1−x 3
1 1 1 1−x 1
1
x=3 xδ(x, y) dy dx = 3 x(1 − x − y) dy + x(x + y − 1) dy dx =
0 0 0 0 1−x 2
1
By symmetry, y = as well; center of gravity (1/2, 1/2)
2

1
12. x = xδ(x, y) dA, and the integrand is an odd function of x while the region is symmetric
M
G
with respect to the y-axis, thus x = 0; likewise y = 0.
√ √
1 x 1 x
13. M = (x + y)dy dx = 13/20, Mx = (x + y)y dy dx = 3/10,
0 0 0 0
√
1 x
My = (x + y)x dy dx = 19/42, x = My /M = 190/273, y = Mx /M = 6/13;
0 0
the mass is 13/20 and the center of gravity is at (190/273, 6/13).

π sin x
14. M = y dy dx = π/4, x = π/2 from the symmetry of the density and the region,
0 0
π sin x
16 π 16
Mx = y 2 dy dx = 4/9, y = Mx /M = ; mass π/4, center of gravity , .
0 0 9π 2 9π

π/2 a
15. M = r3 sin θ cos θ dr dθ = a4 /8, x = y from the symmetry of the density and the
0 0
π/2 a
region, My = r4 sin θ cos2 θ dr dθ = a5 /15, x = 8a/15; mass a4 /8, center of gravity
0 0
(8a/15, 8a/15).

π 1
16. M = r3 dr dθ = π/4, x = 0 from the symmetry of density and region,
0 0
π 1
8 8
Mx = r4 sin θ dr dθ = 2/5, y = ; mass π/4, center of gravity 0, .
0 0 5π 5π

1 1 1
1 1 1 1 1
17. V = 1, x = x dz dy dx = , similarly y = z = ; centroid , ,
0 0 0 2 2 2 2 2

2 2π 1
18. V = πr2 h = 2π, x = y = 0 by symmetry, z dz dy dx = rz dr dθ dz = 2π, centroid
0 0 0
G
= (0, 0, 1)


676 Chapter 15

19. x = y = z from the symmetry of the region, V = 1/6,
1 1−x 1−x−y
1
x= x dz dy dx = (6)(1/24) = 1/4; centroid (1/4, 1/4, 1/4)
V 0 0 0

20. The solid is described by −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2 , 0 ≤ x ≤ 1 − z;
1 1−y 2 1−z 1 1−y 2 1−z
4 1 5
V = dx dz dy = , x = x dx dz dy = , y = 0 by symmetry,
−1 0 0 5 V −1 0 0 14
1 1−y 2 1−z
1 2 5 2
z= z dx dz dy = ; the centroid is , 0, .
V −1 0 0 7 14 7

21. x = 1/2 and y = 0 from the symmetry of the region,
1 1 1
1
V = dz dy dx = 4/3, z = z dV = (3/4)(4/5) = 3/5; centroid (1/2, 0, 3/5)
0 −1 y2 V
G

22. x = y from the symmetry of the region,
2 2 xy
1
V = dz dy dx = 4, x = x dV = (1/4)(16/3) = 4/3,
0 0 0 V
G
1
z= z dV = (1/4)(32/9) = 8/9; centroid (4/3, 4/3, 8/9)
V
G

23. x = y = z from the symmetry of the region, V = πa3 /6,
√ √ 2 2 2 √
a2 −x2 a −x −y a2 −x2
1 a 1 a
x= x dz dy dx = x a2 − x2 − y 2 dy dx
V 0 0 0 V 0 0
π/2 a
1 6
= r2 a2 − r2 cos θ dr dθ = (πa4 /16) = 3a/8; centroid (3a/8, 3a/8, 3a/8)
V 0 0 πa3

24. x = y = 0 from the symmetry of the region, V = 2πa3 /3
√ √ 2 2 2 √
a2 −x2 a −x −y a2 −x2
1 a 1 a 1 2
z= z dz dy dx = (a − x2 − y 2 )dy dx
V −a −√a2 −x2 0 V −a −√a2 −x2 2
2π a
1 1 2 3
= (a − r2 )r dr dθ = (πa4 /4) = 3a/8; centroid (0, 0, 3a/8)
V 0 0 2 2πa3

a a a
25. M = (a − x)dz dy dx = a4 /2, y = z = a/2 from the symmetry of density and
0 0 0
a a a
1
region, x = x(a − x)dz dy dx = (2/a4 )(a5 /6) = a/3;
M 0 0 0
mass a4 /2, center of gravity (a/3, a/2, a/2)



√
a a2 −x2 h
1 2 2
26. M = √ (h − z)dz dy dx = πa h , x = y = 0 from the symmetry of density
−a − a2 −x2 0 2
1 2
and region, z = z(h − z)dV = (πa2 h3 /6) = h/3;
M πa2 h2
G
2 2
mass πa h /2, center of gravity (0, 0, h/3)

1 1 1−y 2
27. M = yz dz dy dx = 1/6, x = 0 by the symmetry of density and region,
−1 0 0

1 1
y= y 2 z dV = (6)(8/105) = 16/35, z = yz 2 dV = (6)(1/12) = 1/2;
M M
G G
mass 1/6, center of gravity (0, 16/35, 1/2)

3 9−x2 1
1
28. M = xz dz dy dx = 81/8, x = x2 z dV = (8/81)(81/5) = 8/5,
0 0 0 M
G
1 1
y= xyz dV = (8/81)(243/8) = 3, z = xz 2 dV = (8/81)(27/4) = 2/3;
M M
G G
mass 81/8, center of gravity (8/5, 3, 2/3)

1 1
29. (a) M = k(x2 + y 2 )dy dx = 2k/3, x = y from the symmetry of density and region,
0 0
1 3
x= kx(x2 + y 2 )dA = (5k/12) = 5/8; center of gravity (5/8, 5/8)
M 2k
R

(b) y = 1/2 from the symmetry of density and region,
1 1
1
M= kx dy dx = k/2, x = kx2 dA = (2/k)(k/3) = 2/3,
0 0 M
R
center of gravity (2/3, 1/2)

30. (a) x = y = z from the symmetry of density and region,
1 1 1
M= k(x2 + y 2 + z 2 )dz dy dx = k,
0 0 0
1
x= kx(x2 + y 2 + z 2 )dV = (1/k)(7k/12) = 7/12; center of gravity (7/12, 7/12, 7/12)
M
G

(b) x = y = z from the symmetry of density and region,
1 1 1
M= k(x + y + z)dz dy dx = 3k/2,
0 0 0
1 2
x= kx(x + y + z)dV = (5k/6) = 5/9; center of gravity (5/9, 5/9, 5/9)
M 3k
G


678 Chapter 15

π sin x 1/(1+x2 +y 2 )
31. V = dV = dz dy dx = 0.666633,
0 0 0
G

1 1 1
x= xdV = 1.177406, y = ydV = 0.353554, z = zdV = 0.231557
V V V
G G G

32. (b) Use polar coordinates for x and y to get
2π a 1/(1+r 2 )
V = dV = r dz dr dθ = π ln(1 + a2 ),
0 0 0
G

1 a2
z= zdV =
V 2(1 + a2 ) ln(1 + a2 )
G

1
Thus lim z = ; lim z = 0.
a→0+ 2 a→+∞
1
lim z = ; lim z = 0
a→0+ 2 a→+∞

(c) Solve z = 1/4 for a to obtain a ≈ 1.980291.

33. Let x = r cos θ, y = r sin θ, and dA = r dr dθ in formulas (11) and (12).

2π a(1+sin θ)
34. x = 0 from the symmetry of the region, A = r dr dθ = 3πa2 /2,
0 0
2π a(1+sin θ)
1 2
y= r2 sin θ dr dθ = (5πa3 /4) = 5a/6; centroid (0, 5a/6)
A 0 0 3πa2

π/2 sin 2θ
35. x = y from the symmetry of the region, A = r dr dθ = π/8,
0 0
π/2 sin 2θ
1 128 128 128
x= r2 cos θ dr dθ = (8/π)(16/105) = ; centroid ,
A 0 0 105π 105π 105π

36. x = 3/2 and y = 1 from the symmetry of the region,

x dA = xA = (3/2)(6) = 9, y dA = yA = (1)(6) = 6
R R

37. x = 0 from the symmetry of the region, πa2 /2 is the area of the semicircle, 2πy is the distance
traveled by the centroid to generate the sphere so 4πa3 /3 = (πa2 /2)(2πy), y = 4a/(3π)

1 2 4a 1
38. (a) V = πa 2π a + = π(3π + 4)a3
2 3π 3
√
2 4a
(b) the distance between the centroid and the line is a+ so
2 3π
√
1 2 2 4a 1√
V = πa 2π a+ = 2π(3π + 4)a3
2 2 3π 6



39. x = k so V = (πab)(2πk) = 2π 2 abk

40. y = 4 from the symmetry of the region,
2 8−x2
A= dy dx = 64/3 so V = (64/3)[2π(4)] = 512π/3
−2 x2

1
41. The region generates a cone of volume πab2 when it is revolved about the x-axis, the area of the
3
1 1 1 1
region is ab so πab2 = ab (2πy), y = b/3. A cone of volume πa2 b is generated when the
2 3 2 3
1 2 1
region is revolved about the y-axis so πa b = ab (2πx), x = a/3. The centroid is (a/3, b/3).
3 2

42. The centroid of the circle which generates the tube travels a distance
4π √ √ √
s= sin2 t + cos2 t + 1/16 dt = 17π, so V = π(1/2)2 17π = 17π 2 /4.
0

a b a b
1 1 3
43. Ix = y 2 δ dy dx = δab3 , Iy = x2 δ dy dx = δa b,
0 0 3 0 0 3
a b
1
Iz = (x2 + y 2 )δ dy dx = δab(a2 + b2 )
0 0 3

2π a 2π a
44. Ix = r3 sin2 θ δ dr dθ = δπa4 /4; Iy = r3 cos2 θ δ dr dθ = δπa4 /4 = Ix ;
0 0 0 0
4
Iz = Ix + Iy = δπa /2

EXERCISE SET 15.7
√
2π 1 1−r 2 2π 1 2π
1 1
1. zr dz dr dθ = (1 − r2 )r dr dθ = dθ = π/4
0 0 0 0 0 2 0 8

π/2 cos θ r2 π/2 cos θ π/2
1
2. r sin θ dz dr dθ = r3 sin θ dr dθ = cos4 θ sin θ dθ = 1/20
0 0 0 0 0 0 4

π/2 π/2 1 π/2 π/2 π/2
1 1
3. ρ3 sin φ cos φ dρ dφ dθ = sin φ cos φ dφ dθ = dθ = π/16
0 0 0 0 0 4 0 8

2π π/4 a sec φ 2π π/4 2π
1 3 1 3
4. ρ2 sin φ dρ dφ dθ = a sec3 φ sin φ dφ dθ = a dθ = πa3 /3
0 0 0 0 0 3 0 6

5. f (r, θ, z) = z 6. f (r, θ, z) = sin θ
z z

y
x y
x


680 Chapter 15

7. f (ρ, φ, θ) = ρ cos φ 8. f (ρ, φ, θ) = 1
z z

a

y

x

x y

2π 3 9 2π 3 2π
81
9. V = r dz dr dθ = r(9 − r2 )dr dθ = dθ = 81π/2
0 0 r2 0 0 0 4
√
2π 2 9−r 2 2π 2
10. V = 2 r dz dr dθ = 2 r 9 − r2 dr dθ
0 0 0 0 0

2 √ 2π √
= (27 − 5 5) dθ = 4(27 − 5 5)π/3
3 0

11. r2 + z 2 = 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:
√ √
2π 2 r2 2π 20 20−r 2
V= √ r dz dr dθ + √ r dz dr dθ
0 0 − 20−r 2 0 2 − 20−r 2
√
2π 2 2π 20
= r r2 + 20 − r2 dr dθ + 2r 20 − r2 dr dθ
0 0 0 2

4 √ 2π
128 2π
152 80 √
= (10 5 − 13) dθ + dθ = π+ π 5
3 0 3 0 3 3

12. z = hr/a intersects z = h in a circle of radius a,
2π a h 2π a 2π
h 1 2
V = r dz dr dθ = (ar − r2 )dr dθ = a h dθ = πa2 h/3
0 0 hr/a 0 0 a 0 6

2π π/3 4 2π π/3 2π
64 32
13. V = ρ2 sin φ dρ dφ dθ = sin φ dφ dθ = dθ = 64π/3
0 0 0 0 0 3 3 0

2π π/4 2 2π π/4
7 7 √ 2π √
14. V = ρ2 sin φ dρ dφ dθ = sin φ dφ dθ = (2 − 2) dθ = 7(2 − 2)π/3
0 0 1 0 0 3 6 0

15. In spherical coordinates the sphere and the plane z = a are ρ = 2a and ρ = a sec φ, respectively.
They intersect at φ = π/3,
2π π/3 a sec φ 2π π/2 2a
V= ρ2 sin φ dρ dφ dθ + ρ2 sin φ dρ dφ dθ
0 0 0 0 π/3 0
2π π/3 2π π/2
1 3 8 3
= a sec3 φ sin φ dφ dθ + a sin φ dφ dθ
0 0 3 0 π/3 3
2π 2π
1 3 4
= a dθ + a3 dθ = 11πa3 /3
2 0 3 0



√
2π π/2 3
2
2π π/2
9 2 2π √
16. V = ρ sin φ dρ dφ dθ = 9 sin φ dφ dθ = dθ = 9 2π
0 π/4 0 0 π/4 2 0

π/2 a a2 −r 2 π/2 a
17. r3 cos2 θ dz dr dθ = (a2 r3 − r5 ) cos2 θ dr dθ
0 0 0 0 0
π/2
1 6
= a cos2 θ dθ = πa6 /48
12 0

π π/2 1 π π/2
1
e−ρ ρ2 sin φ dρ dφ dθ = (1 − e−1 ) sin φ dφ dθ = (1 − e−1 )π/3
3
18.
0 0 0 3 0 0
√
π/2 π/4 8 √
19. ρ4 cos2 φ sin φ dρ dφ dθ = 32(2 2 − 1)π/15
0 0 0

2π π 3
20. ρ3 sin φ dρ dφ dθ = 81π
0 0 0

2 4 π/3 2 4 π/3
r tan3 θ 1
21. (a) √ dθ dr dz = √ dz r dr tan3 θ dθ
−2 1 π/6 1 + z2 −2 1 + z2 1 π/6

√ 15 4 1
= −2 ln( 5 − 2) − ln 3 ≈ 16.97774196
2 3 2
The region is a cylindrical wedge.
(b) To convert to rectangular coordinates observe that the rays θ = π/6, θ = π/3 correspond to
√ √
the lines y = x/ 3, y = 3x. Then dx dy dz = r dr dθ dz and tan θ = y/x, hence
√
4 3x 2
(y/x)3 y3
Integral = √
√ dz dy dx, so f (x, y, z) = √ .
1 x/ 3 −2 1 + z2 x3 1 + z 2
√
π/2 π/4
1 2 π/2
4,294,967,296 √
22. cos37 θ cos φ dφ dθ = cos37 θ dθ = 2 ≈ 0.008040
0 0 18 36 0 755,505,013,725
√
2π a a2 −r 2
23. (a) V = 2 r dz dr dθ = 4πa3 /3
0 0 0
2π π a
(b) V = ρ2 sin φ dρ dφ dθ = 4πa3 /3
0 0 0
√ √
2 4−x2 4−x2 −y 2
24. (a) xyz dz dy dx
0 0 0
√
2 4−x2 2
1 1
= xy(4 − x2 − y 2 )dy dx = x(4 − x2 )2 dx = 4/3
0 0 2 8 0
√
π/2 2 4−r 2
(b) r3 z sin θ cos θ dz dr dθ
0 0 0
π/2 2 π/2
1 3 8
= (4r − r5 ) sin θ cos θ dr dθ = sin θ cos θ dθ = 4/3
0 0 2 3 0
π/2 π/2 2
(c) ρ5 sin3 φ cos φ sin θ cos θ dρ dφ dθ
0 0 0
π/2 π/2 π/2
32 8
= sin3 φ cos φ sin θ cos θ dφ dθ = sin θ cos θ dθ = 4/3
0 0 3 3 0


682 Chapter 15

2π 3 3 2π 3 2π
1 27
25. M = (3 − z)r dz dr dθ = r(3 − r)2 dr dθ = dθ = 27π/4
0 0 r 0 0 2 8 0

2π a h 2π a 2π
1 2 1
26. M = k zr dz dr dθ = kh r dr dθ = ka2 h2 dθ = πka2 h2 /2
0 0 0 0 0 2 4 0

2π π a 2π π 2π
1 4 1
27. M = kρ3 sin φ dρ dφ dθ = ka sin φ dφ dθ = ka4 dθ = πka4
0 0 0 0 0 4 2 0

2π π 2 2π π 2π
3
28. M = ρ sin φ dρ dφ dθ = sin φ dφ dθ = 3 dθ = 6π
0 0 1 0 0 2 0

29. x = y = 0 from the symmetry of the region,
¯ ¯
√
2π 1 2−r 2 2π 1 √
V = r dz dr dθ = (r 2 − r2 − r3 )dr dθ = (8 2 − 7)π/6,
0 0 r2 0 0
√
1 2π 1 2−r 2
6 √
z=
¯ zr dz dr dθ = √ (7π/12) = 7/(16 2 − 14);
V 0 0 r2 (8 2 − 7)π
7
centroid 0, 0, √
16 2 − 14
30. x = y = 0 from the symmetry of the region, V = 8π/3,
¯ ¯
2π 2 2
1 3
z=
¯ zr dz dr dθ = (4π) = 3/2; centroid (0, 0, 3/2)
V 0 0 r 8π

31. x = y = z from the symmetry of the region, V = πa3 /6,
¯ ¯ ¯
π/2 π/2 a
1 6
z=
¯ ρ3 cos φ sin φ dρ dφ dθ = (πa4 /16) = 3a/8;
V 0 0 0 πa3
centroid (3a/8, 3a/8, 3a/8)
2π π/3 4
32. x = y = 0 from the symmetry of the region, V =
¯ ¯ ρ2 sin φ dρ dφ dθ = 64π/3,
0 0 0
2π π/3 4
1 3
z=
¯ ρ3 cos φ sin φ dρ dφ dθ = (48π) = 9/4; centroid (0, 0, 9/4)
V 0 0 0 64π
π/2 2 cos θ r2
33. y = 0 from the symmetry of the region, V = 2
¯ r dz dr dθ = 3π/2,
0 0 0
π/2 2 cos θ r2
2 4
x=
¯ r2 cos θ dz dr dθ = (π) = 4/3,
V 0 0 0 3π
π/2 2 cos θ r2
2 4
z=
¯ rz dz dr dθ = (5π/6) = 10/9; centroid (4/3, 0, 10/9)
V 0 0 0 3π
π/2 2 cos θ 4−r 2 π/2 2 cos θ
1
34. M = zr dz dr dθ = r(4 − r2 )2 dr dθ
0 0 0 0 0 2
π/2
16
= (1 − sin6 θ)dθ = (16/3)(11π/32) = 11π/6
3 0

π/2 π/3 2 π/2 π/3
8 4 √ π/2
35. V = ρ2 sin φ dρ dφ dθ = sin φ dφ dθ = ( 3 − 1) dθ
0 π/6 0 0 π/6 3 3 0
√
= 2( 3 − 1)π/3



2π π/4 1 2π π/4
1 1 √ 2π √
36. M = ρ3 sin φ dρ dφ dθ = sin φ dφ dθ = (2 − 2) dθ = (2 − 2)π/4
0 0 0 0 0 4 8 0

37. x = y = 0 from the symmetry of density and region,
¯ ¯
2π 1 1−r 2
M= (r2 + z 2 )r dz dr dθ = π/4,
0 0 0
2π 1 1−r 2
1
z=
¯ z(r2 +z 2 )r dz dr dθ = (4/π)(11π/120) = 11/30; center of gravity (0, 0, 11/30)
M 0 0 0

2π 1 r
38. x = y = 0 from the symmetry of density and region, M =
¯ ¯ zr dz dr dθ = π/4,
0 0 0
2π 1 r
1
z=
¯ z 2 r dz dr dθ = (4/π)(2π/15) = 8/15; center of gravity (0, 0, 8/15)
M 0 0 0

39. x = y = 0 from the symmetry of density and region,
¯ ¯
2π π/2 a
M= kρ3 sin φ dρ dφ dθ = πka4 /2,
0 0 0
2π π/2 a
1 2
z=
¯ kρ4 sin φ cos φ dρ dφ dθ = (πka5 /5) = 2a/5; center of gravity (0, 0, 2a/5)
M 0 0 0 πka4

40. x = z = 0 from the symmetry of the region, V = 54π/3 − 16π/3 = 38π/3,
¯ ¯
π π 3 π π
1 1 65
y=
¯ ρ3 sin2 φ sin θ dρ dφ dθ = sin2 φ sin θ dφ dθ
V 0 0 2 V 0 0 4
π
1 65π 3
= sin θ dθ = (65π/4) = 195/152; centroid (0, 195/152, 0)
V 0 8 38π

2π π R 2π π
1
δ0 e−(ρ/R) ρ2 sin φ dρ dφ dθ = (1 − e−1 )R3 δ0 sin φ dφ dθ
3
41. M =
0 0 0 0 0 3
4
= π(1 − e−1 )δ0 R3
3

42. (a) The sphere and cone intersect in a circle of radius ρ0 sin φ0 ,
√ 2 2
θ2 ρ0 sin φ0 ρ0 −r θ2 ρ0 sin φ0
V= r dz dr dθ = r ρ2 − r2 − r2 cot φ0 dr dθ
0
θ1 0 r cot φ0 θ1 0

θ2
1 3 1
= ρ0 (1 − cos3 φ0 − sin3 φ0 cot φ0 )dθ = ρ3 (1 − cos3 φ0 − sin2 φ0 cos φ0 )(θ2 − θ1 )
θ1 3 3 0

1 3
ρ (1 − cos φ0 )(θ2 − θ1 ).
=
3 0
(b) From Part (a), the volume of the solid bounded by θ = θ1 , θ = θ2 , φ = φ1 , φ = φ2 , and
1 1 1
ρ = ρ0 is ρ3 (1 − cos φ2 )(θ2 − θ1 ) − ρ3 (1 − cos φ1 )(θ2 − θ1 ) = ρ3 (cos φ1 − cos φ2 )(θ2 − θ1 )
3 0 3 0 3 0
so the volume of the spherical wedge between ρ = ρ1 and ρ = ρ2 is
1 1
∆V = ρ3 (cos φ1 − cos φ2 )(θ2 − θ1 ) − ρ3 (cos φ1 − cos φ2 )(θ2 − θ1 )
3 2 3 1
1 3
= (ρ − ρ3 )(cos φ1 − cos φ2 )(θ2 − θ1 )
3 2 1


684 Chapter 15

d
(c) cos φ = − sin φ so from the Mean-Value Theorem cos φ2 −cos φ1 = −(φ2 −φ1 ) sin φ∗ where
dφ
d 3
φ∗ is between φ1 and φ2 . Similarly ρ = 3ρ2 so ρ3 −ρ3 = 3ρ∗2 (ρ2 −ρ1 ) where ρ∗ is between
2 1
dρ
ρ1 and ρ2 . Thus cos φ1 −cos φ2 = sin φ ∆φ and ρ2 −ρ1 = 3ρ∗2 ∆ρ so ∆V = ρ∗2 sin φ∗ ∆ρ∆φ∆θ.
∗ 3 3

2π a h 2π a h
1
43. Iz = r2 δ r dz dr dθ = δ r3 dz dr dθ = δπa4 h
0 0 0 0 0 0 2

2π a h 2π a
1
44. Iy = (r2 cos2 θ + z 2 )δr dz dr dθ = δ (hr3 cos2 θ + h3 r)dr dθ
0 0 0 0 0 3
2π
1 4 1 π 4 π
=δ a h cos2 θ + a2 h3 dθ = δ a h + a2 h3
0 4 6 4 3

2π a2 h 2π a2 h
1
45. Iz = r2 δ r dz dr dθ = δ r3 dz dr dθ = δπh(a4 − a4 )
2 1
0 a1 0 0 a1 0 2

2π π a 2π π a
8
46. Iz = (ρ2 sin2 φ)δ ρ2 sin φ dρ dφ dθ = δ ρ4 sin3 φ dρ dφ dθ = δπa5
0 0 0 0 0 0 15

EXERCISE SET 15.8
∂(x, y) 1 4 ∂(x, y) 1 4v
1. = = −17 2. = = −1 − 16uv
∂(u, v) 3 −5 ∂(u, v) 4u −1

∂(x, y) cos u − sin v
3. = = cos u cos v + sin u sin v = cos(u − v)
∂(u, v) sin u cos v

2(v 2 − u2 ) 4uv
−
∂(x, y) (u2 + v 2 )2 (u2 + v 2 )2
4. = = 4/(u2 + v 2 )2
∂(u, v) 4uv 2(v 2 − u2 )
(u 2 + v 2 )2 (u2 + v 2 )2

2 5 1 2 ∂(x, y) 2/9 5/9 1
5. x = u + v, y = − u + v; = =
9 9 9 9 ∂(u, v) −1/9 2/9 9

∂(x, y) 1/u 0
6. x = ln u, y = uv; = =1
∂(u, v) v u

1 1
√ √ √ √
√ √ √ √ ∂(x, y) 2 2 u+v 2 2 u+v 1
7. x = u + v/ 2, y = v − u/ 2; = = √
∂(u, v) 1 1 4 v 2 − u2
− √ √ √ √
2 2 v−u 2 2 v−u

3u1/2 u3/2
−
∂(x, y) 2v 1/2 2v 3/2 1
8. x = u3/2 /v 1/2 , y = v 1/2 /u1/2 ; = =
∂(u, v) v 1/2
1 2v
−
2u3/2 2u1/2 v 1/2



3 1 0 1−v −u 0
∂(x, y, z) ∂(x, y, z)
9. = 1 0 −2 =5 10. = v − vw u − uw −uv = u2 v
∂(u, v, w) ∂(u, v, w)
0 1 1 vw uw uv

1/v −u/v 2 0
∂(x, y, z)
11. y = v, x = u/y = u/v, z = w − x = w − u/v; = 0 1 0 = 1/v
∂(u, v, w)
−1/v u/v 2 1

0 1/2 1/2
∂(x, y, z) 1
12. x = (v + w)/2, y = (u − w)/2, z = (u − v)/2, = 1/2 0 −1/2 =−
∂(u, v, w) 4
1/2 −1/2 0

y y
13. 14. (3, 4)
(0, 2) 4

3

2

1
x
(0, 0) 1 2 3 (4, 0)
(0, 0) x
(–1, 0) (1, 0)

y y
15. 3 (0, 3) 16.
2

(2, 0)
x
–3 3 1

x
–3
1 2

3 4
1 2 2 1 ∂(x, y) 1 1 u 1 u 3
17. x = u + v, y = − u + v, = ; dAuv = du dv = ln 3
5 5 5 5 ∂(u, v) 5 5 v 5 1 1 v 2
S

4 1
1 1 1 1 ∂(x, y) 1 1 1 1 4
18. x = u + v, y = u − v, =− ; veuv dAuv = veuv du dv = (e − e − 3)
2 2 2 2 ∂(u, v) 2 2 2 1 0 2
S

∂(x, y)
19. x = u + v, y = u − v, = −2; the boundary curves of the region S in the uv-plane are
∂(u, v)
1 u
1
v = 0, v = u, and u = 1 so 2 sin u cos vdAuv = 2 sin u cos v dv du = 1 − sin 2
0 0 2
S

√ ∂(x, y) 1
20. x = v/u, y = uv so, from Example 3, = − ; the boundary curves of the region S in
∂(u, v) 2u
4 3
1 1
the uv-plane are u = 1, u = 3, v = 1, and v = 4 so uv 2 dAuv = v 2 du dv = 21
2u 2 1 1
S


686 Chapter 15

∂(x, y)
21. x = 3u, y = 4v, = 12; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1.
∂(u, v)
2π 1
Use polar coordinates to obtain 12 u2 + v 2 (12) dAuv = 144 r2 dr dθ = 96π
0 0
S

∂(x, y)
22. x = 2u, y = v, = 2; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1. Use
∂(u, v)
2π 1
e−(4u re−4r dr dθ = (1 − e−4 )π/2
2
+4v 2 ) 2
polar coordinates to obtain (2) dAuv = 2
0 0
S

23. Let S be the region in the uv-plane bounded by u2 + v 2 = 1, so u = 2x, v = 3y,

∂(x, y) 1/2 0
x = u/2, y = v/3, = = 1/6, use polar coordinates to get
∂(u, v) 0 1/3
π/2 1 1
1 1 π π
sin(u2 + v 2 )du dv = r sin r2 dr dθ = (− cos r2 ) = (1 − cos 1)
6 6 0 0 24 0 24
S

2π 1
∂(x, y)
24. u = x/a, v = y/b, x = au, y = bv; = ab; A = ab r dr dθ = πab
∂(u, v) 0 0

∂(x, y, z)
25. x = u/3, y = v/2, z = w, = 1/6; S is the region in uvw-space enclosed by the sphere
∂(u, v, w)
u2 + v 2 + w2 = 36 so
2π π 6
u2 1 1
dVuvw = (ρ sin φ cos θ)2 ρ2 sin φ dρ dφ dθ
9 6 54 0 0 0
S
2π π 6
1 192
= ρ4 sin3 φ cos2 θdρ dφ dθ = π
54 0 0 0 5

∂(x, y, z)
26. Let G1 be the region u2 + v 2 + w2 ≤ 1, with x = au, y = bv, z = cw, = abc; then use
∂(u, v, w)
spherical coordinates in uvw-space:

Ix = (y 2 + z 2 )dx dy dz = abc (b2 v 2 + c2 w2 ) du dv dw
G G1
2π π 1
= abc(b2 sin2 φ sin2 θ + c2 cos2 φ)ρ4 sin φ dρ dφ dθ
0 0 0
2π
abc 2 2 4
= (4b sin θ + 2c2 )dθ = πabc(b2 + c2 )
0 15 15

27. u = θ = cot−1 (x/y), v = r = x2 + y 2

28. u = r = x2 + y 2 , v = (θ + π/2)/π = (1/π) tan−1 (y/x) + 1/2

3 2 1 3 4
29. u = x − y, v = − x + y 30. u = −x + y, v = y
7 7 7 7 3



1 1 ∂(x, y) 1
31. Let u = y − 4x, v = y + 4x, then x = (v − u), y = (v + u) so =− ;
8 2 ∂(u, v) 8
5 2
1 u 1 u 1 5
dAuv = du dv = ln
8 v 8 2 0 v 4 2
S

1 1 ∂(x, y) 1
32. Let u = y + x, v = y − x, then x = (u − v), y = (u + v) so = ;
2 2 ∂(u, v) 2
2 1
1 1 1
− uv dAuv = − uv du dv = −
2 2 0 0 2
S

1 1 ∂(x, y) 1
33. Let u = x − y, v = x + y, then x = (v + u), y = (v − u) so = ; the boundary curves of
2 2 ∂(u, v) 2
the region S in the uv-plane are u = 0, v = u, and v = π/4; thus

1 sin u 1 π/4 v
sin u 1 √
dAuv = du dv = [ln( 2 + 1) − π/4]
2 cos v 2 0 0 cos v 2
S

1 1 ∂(x, y) 1
34. Let u = y − x, v = y + x, then x = (v − u), y = (u + v) so = − ; the boundary
2 2 ∂(u, v) 2
curves of the region S in the uv-plane are v = −u, v = u, v = 1, and v = 4; thus
4 v
1 1 15
eu/v dAuv = eu/v du dv = (e − e−1 )
2 2 1 −v 4
S

∂(x, y) 1
35. Let u = y/x, v = x/y 2 , then x = 1/(u2 v), y = 1/(uv) so = 4 3;
∂(u, v) u v
4 2
1 1
dAuv = du dv = 35/256
u4 v 3 1 1 u4 v 3
S

∂(x, y)
36. Let x = 3u, y = 2v, = 6; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1
∂(u, v)
2π 1
so (9 − x − y)dA = 6(9 − 3u − 2v)dAuv = 6 (9 − 3r cos θ − 2r sin θ)r dr dθ = 54π
0 0
R S

∂(x, y, z) 1
37. x = u, y = w/u, z = v + w/u, =− ;
∂(u, v, w) u
4 1 3
v2 w v2 w
dVuvw = du dv dw = 2 ln 3
u 2 0 1 u
S

38. u = xy, v = yz, w = xz, 1 ≤ u ≤ 2, 1 ≤ v ≤ 3, 1 ≤ w ≤ 4,
∂(x, y, z) 1
x= uw/v, y = uv/w, z = vw/u, = √
∂(u, v, w) 2 uvw
2 3 4
1 √ √
V = dV = √ dw dv du = 4( 2 − 1)( 3 − 1)
1 1 1 2 uvw
G


688 Chapter 15

sin3 φ cos3 θ 3ρ sin2 φ cos φ cos3 θ −3ρ sin3 φ cos2 θ sin θ
∂(x, y, z)
39. = sin3 φ sin3 θ 3ρ sin2 φ cos φ sin3 θ 3ρ sin3 φ sin2 θ cos θ
∂(ρ, φ, θ) 3
cos φ −3ρ cos φ sin φ
2
0
= 9ρ2 cos2 θ sin2 θ cos2 φ sin5 φ,
2π π a
4
V =9 ρ2 cos2 θ sin2 θ cos2 φ sin5 φ dρ dφ dθ = πa3
0 0 0 35

40. (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 ∂x ∂v ∂x ∂x ∂u ∂x ∂v ∂x
+ = = 1, + = =0
∂u ∂x ∂v ∂x ∂x ∂u ∂y ∂v ∂y ∂y
∂y ∂u ∂y ∂v ∂y ∂y ∂u ∂y ∂v ∂y
+ = = 0, + = =1
∂u ∂x ∂v ∂x ∂x ∂u ∂y ∂v ∂y ∂y

∂(x, y) 1−v −u y
41. (a) = = u; u = x + y, v = ,
∂(u, v) v u x+y
∂(u, v) 1 1 x y 1 1
= = + = = ;
∂(x, y) −y/(x + y)2 x/(x + y)2 (x + y)2 (x + y)2 x+y u
∂(u, v) ∂(x, y)
=1
∂(x, y) ∂(u, v)

∂(x, y) v u √ √
(b) = = 2v 2 ; u = x/ y, v = y,
∂(u, v) 0 2v
√
∂(u, v) 1/ y −x/(2y 3/2 ) 1 1 ∂(u, v) ∂(x, y)
= √ = = 2; =1
∂(x, y) 0 1/(2 y) 2y 2v ∂(x, y) ∂(u, v)

∂(x, y) u v √ √
(c) = = −2uv; u = x + y, v = x − y,
∂(u, v) u −v
√ √
∂(u, v) 1/(2 x + y) 1/(2 x + y) 1 1 ∂(u, v) ∂(x, y)
= √ √ =− =− ; =1
∂(x, y) 1/(2 x − y) −1/(2 x − y) 2 x2 − y2 2uv ∂(x, y) ∂(u, v)

2 2π
∂(u, v) ∂(x, y) 1 1 sin u 1 sin u 2
42. = 3xy 4 = 3v so = ; dAuv = du dv = − ln 2
∂(x, y) ∂(u, v) 3v 3 v 3 1 π v 3
S

∂(u, v) ∂(x, y) 1 ∂(x, y) 1 1
43. = 8xy so = ; xy = xy = so
∂(x, y) ∂(u, v) 8xy ∂(u, v) 8xy 8
16 4
1 1
dAuv = du dv = 21/8
8 8 9 1
S

∂(u, v) ∂(x, y) 1
44. = −2(x2 + y 2 ) so =− 2 + y2 )
;
∂(x, y) ∂(u, v) 2(x

∂(x, y) x4 − y 4 xy 1 1
(x4 − y 4 )exy = 2 + y2 )
e = (x2 − y 2 )exy = veu so
∂(u, v) 2(x 2 2
4 3
1 1 7 3
veu dAuv = veu du dv = (e − e)
2 2 3 1 4
S


Review Exercises, Chapter 15 689

1 1 2
∂(u, v, w)
45. Set u = x + y + 2z, v = x − 2y + z, w = 4x + y + z, then = 1 −2 1 = 18, and
∂(x, y, z)
4 1 1
6 2 3
∂(x, y, z) 1
V = dx dy dz = du dv dw = 6(4)(12) = 16
−6 −2 −3 ∂(u, v, w) 18
R

46. (a) Let u = x + y, v = y, then the triangle R with vertices (0, 0), (1, 0) and (0, 1) becomes the
triangle in the uv-plane with vertices (0, 0), (1, 0), (1, 1), and
1 u 1
∂(x, y)
f (x + y)dA = f (u) dv du = uf (u) du
0 0 ∂(u, v) 0
R
1 1
(b) ueu du = (u − 1)eu =1
0 0

cos θ −r sin θ 0
∂(x, y, z) ∂(x, y, z)
47. (a) = sin θ r cos θ 0 = r, =r
∂(r, θ, z) ∂(r, θ, z)
0 0 1

sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ
∂(x, y, z) ∂(x, y, z)
(b) = sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ = ρ2 sin φ; = ρ2 sin φ
∂(ρ, φ, θ) ∂(ρ, φ, θ)
cos φ −ρ sin φ 0

REVIEW EXERCISES, CHAPTER 15
2 2
∂z ∂z
3. (a) dA (b) dV (c) 1+ + dA
∂x ∂y
R G R

4. (a) x = a sin φ cos θ, y = a sin φ sin θ, z = a cos φ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π
(b) x = a cos θ, y = a sin θ, z = z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h
√
1 1+ 1−y 2 2 2x 3 6−x
7. √ f (x, y) dx dy 8. f (x, y) dy dx + f (x, y) dy dx
0 1− 1−y 2 0 x 2 x

9. (a) (1, 2) = (b, d), (2, 1) = (a, c), so a = 2, b = 1, c = 1, d = 2
1 1 1 1
∂(x, y)
(b) dA = du dv = 3du dv = 3
0 0 ∂(u, v) 0 0
R

√
10. If 0 < x, y < π then 0 < sin xy ≤ 1, with equality only on the hyperbola xy = π 2 /4, so
π π π π
√ π π
0= 0 dy dx < sin xy dy dx < 1 dy dx = π 2
0 0 0 0 0 0

1
1 1 √
11. 2x cos(πx2 ) dx = sin(πx2 ) = −1/( 2π)
1/2 π 1/2

2 x=2y 2 2
x2 y3 3 3 1 y3 1 8
12. e dy = y 2 ey dy = e = e −1
0 2 x=−y 2 0 2 0 2


690 Chapter 15

1 2 π x
sin x
13. ex ey dx dy 14. dy dx
0 2y 0 0 x

15. y 16. p /2
1 r = a(1 + cos u)
y = sin x

y = tan (x/2) r=a

x
p /6
6 0

8 y 1/3 8 8
2 1 1
17. 2 x2 sin y 2 dx dy = y sin y 2 dy = − cos y 2 = (1 − cos 64) ≈ 0.20271
0 0 3 0 3 0 3

π/2 2
18. (4 − r2 )r dr dθ = 2π
0 0

2xy
19. sin 2θ = 2 sin θ cos θ = , and r = 2a sin θ is the circle x2 + (y − a)2 = a2 , so
x2 + y 2
√
a a+ a2 −x2 a
2xy
√ dy dx = x ln a + a2 − x2 − ln a − a2 − x2 dx = a2
0 a− a2 −x2 x2 + y 2 0

π/2 2 π/2
20. 4r2 (cos θ sin θ) r dr dθ = −4 cos 2θ =4
π/4 0 π/4

2 2−y/2 2 2
y y 1/3 y2 3 y 4/3 3
21. dx dy = 2− − dy = 2y − − =
0 (y/2)1/3 0 2 2 4 2 2 0 2

π/6 cos 3θ π/6
22. A = 6 r dr dθ = 3 cos2 3θ = π/4
0 0 0

2π 2 16 2π 2
23. r2 cos2 θ r dz dr dθ = cos2 θ dθ r3 (16 − r4 ) dr = 32π
0 0 r4 0 0

π/2 π/2 1 π/2
1 π π
24. 2
ρ2 sin φ dρ dφ dθ = 1 − sin φ dφ
0 0 0 1+ρ 4 2 0

π π π/2 π π
= 1− (− cos φ) = 1−
4 2 0 4 2
2π π/3 a 2π π/3 a
25. (a) (ρ2 sin2 φ)ρ2 sin φ dρ dφ dθ = ρ4 sin3 φ dρ dφ dθ
0 0 0 0 0 0
√ √ √ √
2π 3a/2 a2 −r 2 2π 3a/2 a2 −r 2

(b) √ r2 dz rdr dθ = √ r3 dz dr dθ
0 0 r/ 3 0 0 r/ 3
√ √ √
3a/2 (3a2 /4)−x2 a2 −x2 −y 2
(c) √ √ √ √ (x2 + y 2 ) dz dy dx
− 3a/2 − (3a2 /4)−x2 x2 +y 2 / 3


Review Exercises, Chapter 15 691

√
4 4x−x2 4x π/2 4 cos θ 4r cos θ
26. (a) √ dz dy dx (b) r dz dr dθ
0 − 4x−x2 x2 +y 2 −π/2 0 r2

√ √
2π a/ 3 a a/ 3 √ πa3
27. V = √ r dz dr dθ = 2π r(a − 3r) dr =
0 0 3r 0 9

28. The intersection of the two surfaces projects onto the yz-plane as 2y 2 + z 2 = 1, so
√ √ 2 2
1/ 2 1−2y 1−y
V =4 dx dz dy
0 0 y 2 +z 2
√ √ √ √
1/ 2 1−2y 2 1/ 2
2 2π
=4 (1 − 2y − z ) dz dy = 4
2 2
(1 − 2y 2 )3/2 dy =
0 0 0 3 4
√
29. ru × rv = 2u2 + 2v 2 + 4,
2π 2 √ 8π √
S= 2u2 + 2v 2 + 4 dA = 2 r2 + 2 r dr dθ = (3 3 − 1)
0 0 3
u2 +v 2 ≤4

√ 2 3u 2
30. ru × rv = 1 + u2 , S = 1 + u2 dv du = 3u 1 + u2 du = 53/2 − 1
0 0 0

31. (ru × rv ) u=1
= −2, −4, 1 , tangent plane 2x + 4y − z = 5
v=2

32. u = −3, v = 0, (ru × rv ) u=−3
= −18, 0, −3 , tangent plane 6x + z = −9
v=0

4 2+y 2 /8 4
y2 32
33. A = dx dy = 2− dy = ; y = 0 by symmetry;
¯
−4 y 2 /4 −4 8 3
4 2+y 2 /8 4
1 3 4 256 3 256 8 8
x dx dy = 2 + y2 − y dy = , x=
¯ = ; centroid ,0
−4 y 2 /4 −4 4 128 15 32 15 5 5

34. A = πab/2, x = 0 by symmetry,
¯
√ 2 2
a b 1−x /a
1 a 2 4b
y dy dx = b (1 − x2 /a2 )dx = 2ab2 /3, centroid 0,
−a 0 2 −a 3π

1 2
35. V = πa h, x = y = 0 by symmetry,
¯ ¯
3
2π a h−rh/a a 2
r
rz dz dr dθ = π rh2 1 − dr = πa2 h2 /12, centroid (0, 0, h/4)
0 0 0 0 a
2 4 4−y 2 4 2
1 256
36. V = dz dy dx = (4 − y)dy dx = 8 − 4x2 + x4 dx = ,
−2 x2 0 −2 x2 −2 2 15
2 4 4−y 2 4 2
1 6 32 1024
y dz dy dx = (4y − y 2 ) dy dx = x − 2x4 + dx =
−2 x2 0 −2 x2 −2 3 3 35
2 4 4−y 2 4 2
1 x6 32 2048
z dz dy dx = (4 − y)2 dy dx = − + 2x4 − 8x2 + dx =
−2 x2 0 −2 x2 2 −2 6 3 105
12 8
x = 0 by symmetry, centroid
¯ 0, ,
7 7


692 Chapter 15

π 2π a
4 3 ¯ 3 3 3 a4 3
37. V = πa , d = ρdV = ρ3 sin φ dρ dθ dφ = 2π(2) = a
3 4πa3 4πa3 0 0 0 4πa3 4 4
ρ≤a

2 2
1 3 3 1 1 3 1
38. x = u + v and y = − u + v, hence |J(u, v)| = + = , and
10 10 10 10 10 10 10

x − 3y 1 3 4
u 1 3
1 4
1 2 8
2
dA = 2
du dv = dv u du = 8=
(3x + y) 10 1 0 v 10 1 v2 0 10 3 15
R

1 1
39. (a) Add u and w to get x = ln(u + w) − ln 2; subtract w from u to get y = u − w, substitute
2 2
1 1 1 1
these values into v = y + 2z to get z = − u + v + w. Hence xu = , xv = 0, xw =
4 2 4 u+w
1 1 1 1 1 1 ∂(x, y, z) 1
; yu = , yv = 0, yz = − ; zu = − , zv = , zw = , and thus =
u+w 2 2 4 2 4 ∂(u, v, w) 2(u + w)
3 2 4
1
(b) V = = dw dv du
1 1 0 2(u + w)
G
1 823543
= (7 ln 7 − 5 ln 5 − 3 ln 3)/2 = ln ≈ 1.139172308
2 84375

Chapter 15

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Chapter 15