# 0 r 0 9 c is the boundary of r and curl f 2i 3j 4k

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Unformatted text preview: apter 16 y 24. (a) (b) (1, 3), (3, 27) 25 R 15 5 x 1 2 3 4x3 −x4 3 3 3−4x+4x2 1 π /4 1 cos x π /4 25. A = (cos x − sin x)dx = dy dx = sin x 0 3y −4 −4 (−y 2 − 3y + 4)dy = 125/6 9−y 2 3 3 27. A = dx dy = −3 1−y 2 /9 −3 1 cosh x 1 28. A = dy dx = sinh x 0 6−3x/2 4 8(1 − y 2 /9)dy = 32 (cosh x − sinh x)dx = 1 − e−1 0 4 (3 − 3x/4 − y/2) dy dx = 29. 0 2 4 − x2 dy dx = 30. 0 [(3 − 3x/4)(6 − 3x/2) − (6 − 3x/2)2 /4] dx = 12 0 √ 4−x2 2 √ 2−1 1 dx dy = −4 224 15 0 −y 2 1 26. A = 0 x[(4x3 − x4 ) − (3 − 4x + 4x2 )] dx = x dy dx = (c) (4 − x2 ) dx = 16/3 0 0 √ 9−x2 3 (3 √ − 9−x2 31. V = −3 3 − x)dy dx = −3 x 1 1 (2x3 − x4 − x6 )dx = 11/70 (x2 + 3y 2 )dy dx = 32. V = x2 0 3 0 2 3 (9x2 + y 2 )dy dx = 33. V = 0 (18x2 + 8/3)dx = 170 0 1 0 1 y2 −1 (1/2 − y 2 + y 4 /2)dy = 8/15 √ 9−4x2 3/2 (y √ − 9−4x2 35. V = −3/2 3 1 (1 − x)dx dy = 34. V = −1 3/2 + 3)dy dx = −3/2 3 y 2 /3 6 9 − 4x2 dx = 27π/2 3 (9 − x2 )dx dy = 36. V = 0 (6 9 − x2 − 2x 9 − x2 )dx = 27π (18 − 3y 2 + y 6 /81)dy = 216/7 0 Exercise Set 16.2 578 √ 25−x2 5 5 25 − x2 dy dx = 8 37. V = 8 (25 − x2 )dx = 2000/3 0 0 0 √ 2 1−(y −1)2 38. V = 2 2 0 0 1 [1 − (y − 1)2 ]3/2 + y 2 [1 − (y − 1)2 ]1/2 dy, 3 (x2 + y 2 )dx dy = 2 0 π /2 let y − 1 = sin θ to get V = 2 1 cos3 θ + (1 + sin θ)2 cos θ cos θ dθ which eventually yields 3 −π/2 V = 3π/2 √ 1−x2 1 39. V = 4 0 0 √ 4−x2 2 40. V = √ 2 0 42. f (x, y )dy dx y2 0 e 1 y /4 4 47. 0 4 e−y dx dy = 2 0 0 0 3 x2 ex dx = (e8 − 1)/3 0 0 ln 3 3 x dx dy = 50. ey 0 1 2 ln 3 (9 − e2y )dy = 0 y2 2 0 1 y 2 sin(y 3 )dy = (1 − cos 8)/3 0 e 1 (ex − xex )dx = e/2 − 1 x dy dx = 52. 0 1 (9 ln 3 − 4) 2 2 sin(y 3 )dx dy = 51. 0 x2 2x cos(x2 )dx = sin 1 2 3 ex dy dx = 49. 0 f (x, y )dy dx 0 0 x2 2 46. 0 1 cos(x2 )dy dx = 0 √ x 1 1 −y2 ye dy = (1 − e−16 )/8 4 2x 1 48. sin x f (x, y )dy dx 0 f (x, y )dy dx ln x 1 45. ey 0 2 43. 0 π /2 f (x, y )dx dy 44. e2 x/2 8 f (x, y )dx dy 0 1 4 − x2 + (4 − x2 )3/2 dx = 2π 3 x2 2 41. (1 − x2 )3/2 dx = π/2 0 2 (x2 + y 2 )dy dx = 0 0 1 8 3 (1 − x2 − y 2 )dy dx = ex 0 4 53. (a) 2 √ x 0 sin πy 3 dy dx; the inner integral is non-elementary. y2 2 2 y 2 sin π y 3 dy = − sin π y 3 dx dy = 0 0 0 1 cos π y 3 3π 2 =0 0 579 Chapter 16 π /2 1 sec2 (cos x)dx dy ; the inner integral is non-elementary. (b) sin−1 y 0 π /2 sin x π /2 sec2 (cos x)dy dx = 0 0 √ 4−x2 2 54. V = 4 2 (x2 + y 2 ) dy dx = 4 0 π /2 128 64 64 + sin2 θ − sin4 θ 3 3 3 = 0 2 1 −2 −1 −1 −2 xy 2 dy dx 1 1 1 8 x dx + 3 −1 (x = 2 sin θ) 2 xy 2 dy dx + 0 = 2 dx 64 π 64 π 128 π 1 · 3 + − = 8π 32 34 3 2 2·4 dθ = 2 xy 2 dy dx + 55. 1 4 − x2 + (4 − x2 )3/2 3 x2 0 0 −1 sec2 (cos x) sin x dx = tan 1 0 0 7 x dx + 3 2 1 8 x dx = 0 3 56. This is the volume in the ﬁrst octant under the surface z = π the sphere of radius 1, thus . 6 1 1 ¯ 57. Area of triangle is 1/2, so f = 2 x 0 1 dy dx = 2 1 + x2 1 0 1 − x2 − y 2 , so 1/8 of the volume of x π 1 − dx = − ln 2 1 + x2 1 + x2 2 2 (3x − x2 − x) dx = 4/3, so 58. Area = 0 3x−x2 2 ¯3 f= 4 0 (x2 − xy )dy dx = x 3 4 2 (−2x3 + 2x4 − x5 /2)dx = − 0 59. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so a sin x 1 + x + y dy dx = 0.676089 V= 0 x/2 EXERCISE SET 16.3 π /2 sin θ 1. π /2 r cos θdr dθ = 0 0 π 0 1+cos θ π 2. r dr dθ = 0 0 0 π /2 a sin θ 1 (1 + cos θ)2 dθ = 3π/4 2 π /2 r2 dr dθ = 3. 0 0 π /3 0 cos 3θ 4. π /3 r dr dθ = 0 0 1−sin θ π 5. 0 0 1 sin2 θ cos θ dθ = 1/6 2 0 2 a3 sin3 θ dθ = a3 3 9 1 cos2 3θ dθ = π/12 2 π r2 cos θ dr dθ = 0 1 (1 − sin θ)3 cos θ dθ = 0 3 2 38 =− 4 15 5 Exercise Set 16.3 π 580 cos θ π r3 dr dθ = 6. 0 0 0 1−cos θ 2π 2π 7. A = 0 0 π /2 sin 2θ π /2 0 0 π /2 0 π /2 1 r dr dθ = 9. A = π/4 sin 2θ π /3 π/4 (4 − sec2 θ)dθ = 4π/3 − r dr dθ = sec θ 0 π /2 4 sin θ π /2 r dr dθ = π/6 2 π 1 (−2 cos θ − cos2 θ)dθ = 2 − π/4 1+cos θ π/2 3 9 − r2 dr dθ = r 13. V = 8 0 1 π /2 2 sin θ r2 dr dθ = 14. V = 2 0 0 π /2 √ (16 sin2 θ − 4)dθ = 4π/3 + 2 3 π r dr dθ = π /2 128 √ 2 3 dr dθ = 8 1 π /2 π /2 1 2 (1 − sin4 θ)dθ = 5π/32 0 π /2 3 0 dθ = 4π 0 3 sin θ π /2 sin4 θ dθ = 27π/16 r2 sin θ dr dθ = 9 17. V = 0 0 0 π /2 2 cos θ r 18. V = 4 0 2π 1 19. 2 0 π /2 9 − r2 dr dθ = 9 0 π /2 2r2 sin θ dr dθ = 0 0 dθ = (1 − e−1 )π 0 dθ = 9π/2 1 1 r dr dθ = ln 5 2 1+r 2 2 cos θ 22. (1 − sin3 θ)dθ = 0 2 0 π /2 π /2 r π /4 2π 1 (1 − e−1 ) 2 3 21. 32 3 4 − r2 dr dθ = 0 e−r r dr dθ = 20. 0 64 √ 2π 3 sin3 θ dθ = 32/9 0 π /2 dθ = 0 (1 − r2 )r dr dθ = 16. V = 4 π /2 π /2 16 3 cos θ 15. V = 2 0 3 π/6 12. A = 2 π/2 √ 0 11. A = 2 π/4 1 (1 − sin2 2θ)dθ = π/16 2 π /3 2 10. A = 2 0 sin2 2θ dθ = π/2 r dr dθ = 2 8. A = 4 0 1 (1 − cos θ)2 dθ = 3π/2 2 r dr dθ = 0 0 1 cos4 θ dθ = 3π/32 4 16 3 π /4 dθ = 0 π ln 5 8 π /2 cos3 θ sin...
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