1. 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

2. January 27, 2005 11:55 L24-CH15 Sheet number 2 Page number 656 black
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

3. January 27, 2005 11:55 L24-CH15 Sheet number 3 Page number 657 black
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

4. January 27, 2005 11:55 L24-CH15 Sheet number 4 Page number 658 black
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

5. January 27, 2005 11:55 L24-CH15 Sheet number 5 Page number 659 black
Exercise Set 15.2 659
√
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

6. January 27, 2005 11:55 L24-CH15 Sheet number 6 Page number 660 black
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

7. January 27, 2005 11:55 L24-CH15 Sheet number 7 Page number 661 black
Exercise Set 15.2 661
√
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

8. January 27, 2005 11:55 L24-CH15 Sheet number 8 Page number 662 black
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

9. January 27, 2005 11:55 L24-CH15 Sheet number 9 Page number 663 black
Exercise Set 15.3 663
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

10. January 27, 2005 11:55 L24-CH15 Sheet number 10 Page number 664 black
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

11. January 27, 2005 11:55 L24-CH15 Sheet number 11 Page number 665 black
Exercise Set 15.4 665
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

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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

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Exercise Set 15.4 667
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

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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