07 - CHAPTER 7 Applications of Integration Section 7.1...

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Unformatted text preview: CHAPTER 7 Applications of Integration Section 7.1 Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Area of a Region Between Two Curves . . . . . . . . . . .2 Volume: The Disk Method . . . . . . . . . . . . . . . . . 18 Volume: The Shell Method . . . . . . . . . . . . . . . . . 33 Arc Length and Surfaces of Revolution . . . . . . . . . . . 44 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Moments, Centers of Mass, and Centroids . . . . . . . . . 61 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 73 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 CHAPTER 7 Applications of Integration Section 7.1 6 Area of a Region Between Two Curves 6 2 1. A 0 0 x2 6x dx 0 x2 6x d x 2. A 2 2 2x x2 2 5 4 dx x2 2x 1 dx 3 1 3. A 0 3 x2 2x 2 0 2x 6x dx 3 x2 4x 3 dx 4. A 0 x2 x3 dx 0 0 1 5. A or 2 1 1 3 x3 x3 0 x dx x dx 6 1 x3 x dx 6. A 2 0 x 1 3 x 1 dx 6 4 7. 0 y 5 4 3 2 1 x 1 x dx 2 1 6 8. 1 1 x2 y 2 x2 1 dx 9. 0 y 6 5 42 x3 x dx 6 −2 x 2 3 2 1 x 1 2 3 4 5 −2 x 1 2 3 4 5 6 3 10. 2 y 7 6 5 4 3 2 1 −1 x3 3 x x dx 3 3 4 11. 3 2 y 3 sec x d x 12. 4 sec2 x y cos x dx (3, 6) 3 π (− 4 , 2) 2 π ( 4 , 2) (2, 2 ) 3 2 (3, 1) x 4 5 6 7 2π 3 π 3 π −1 3 2π 3 x π (− 4 , 22 ) (π , 22 ) 4 −π 4 π 4 x 2 S ection 7.1 Area of a Region Between Two Curves 3 13. (a) x x 4 y2 y y y2 6 2 4 y y 0 0 y2 2 2 −6 −4 6 4 y (0, 2) x 6 3y (− 5, − 3) −6 Intersection points: 0, 2 and 0 5, 4 3 24 0 A 5 2 x 4 3 2 y2 4 y x dx 2 dy x dx 61 6 32 3 125 6 (b) A 125 6 4 9 14. (a) y x 2 x 2 and y 6 6 2 x x 6 0⇒ x 3, 9 3x 2 0 (b) A 0 2 y dy 4 6 125 6 y y dy x ⇒x Intersection points: 2, 4 and y 10 32 3 61 6 (−3, 9) 8 6 4 (2, 4) −6 −4 −2 x −2 2 4 6 2 A 3 6 x x 2 dx 125 6 1 2x 15. f x gx A 4 x x 1 1 2 3 2 y 16. f x (3, 4) 2 2 1 y gx A x 4 3 Matches (d) (0, 1) x 1 2 3 (0, 2) Matches (a) 1 (4, 0) x 1 2 3 2 17. A 0 2 0 13 x 2 13 x 2 x2 2 4 2 2 x 2 x 1 dx 1 dx 6 5 4 3 y (2, 6) x4 8 16 8 (0, 2) 1 −2 (2, 3) x 0 (0, 1) x 1 3 4 2 0 2 4 Chapter 7 8 Applications of Integration 1 x 2 7 x 2 3 xx 8 10 dx 8 18. A 2 8 2 10 32 x 8 7x2 4 112 8 dx 8 6 4 2 y (2, 9) (8, 6) x3 8 64 10x 2 (2, 9 ) 2 (8, 0) x 2 4 6 10 80 1 7 20 18 19. The points of intersection are given by: x2 xx A 0 4 20. The points of intersection are given by: x2 4x x2 1 3x x2 x 0 3x when x 4x 3x dx 6 4x 4 4 0 0 when x gx x2 f x dx 4x d x 4 −1 −2 −3 −4 −5 1 0, 4 y 0, 3 1 dx y (0, 0) 1 2 3 (4, 0) 5 3 x A 0 3 x2 x2 0 1 x 0 x3 3 32 3 2x2 0 x3 3 9 3x2 2 27 2 3 0 5 4 3 2 (3, 4) 9 2 (0, 1) x 1 2 3 4 5 6 21. The points of intersection are given by: x2 x A 1 2 22. The points of intersection are given by: x2 4x x3 3 2x 2x 2 1 1 3x 3 1, 2 A 2 x x 2 0, 3 0 when x f x dx 3 x x2 x2 dx 2x 0 when x g x dx 4x 3x d x 2 x gx 3x 1 2 fx 0 3 1 dx 0 3 x2 x2 y 2 dx y 6 2 1 0 2x x2 2 x3 3 2 1 9 2 10 8 6 (2, 9) x3 3 32 x 2 3 0 9 2 5 4 3 (3, 5) 4 ( 1, 0) x (0, 2) 1 2 −1 1 2 3 4 5 4 3 2 x 23. The points of intersection are given by: x x A 0 y 2 1 1 x and x x 2 y 0 and 2 0 y dy 2y x x y2 0 2 1 3 2 1 0 1 (1, 1) (2, 0) x Note that if we integrate with respect to x, we need two integrals. Also, note that the region is a triangle. (0, 0) 2 3 S ection 7.1 5 5 1 Area of a Region Between Two Curves 5 24. A 1 y 1.0 0.8 1 x2 0 dx 1 x 4 5 25. The points of intersection are given by: 3x 1 3x 3 x 1 0, 3 (1, 1) x when x fx g x dx 1 x dx x A 0.6 0.4 0.2 0 3 3x (5, 0.04) x 1 2 3 4 5 1 dx y 0 3 3x 0 12 5 2 3x 9 32 x2 2 3 0 3 2 4 3 2 (3, 4) (0, 1) −2 x 1 2 3 4 26. The points of intersection are given by: 3 x x 1 1 2x 2 1 x x 0 0 1 1 3 y x3 3x2 3x 1 1 (2, 1) (1, 0) 2 x x3 x x2 xx A 2 3x2 3x 2x 1 0⇒x 1 3 0, 1, 2 1 dx 1 (0, −1) x 0 x 1 2 2 x2 2 1 2 x 1 3 x 4 0 43 0 3 4 1 2 28. The points of intersection are given by: 2y 1, 2 y 27. The points of intersection are given by: y2 y A 1 2 3 y 2 y2 3 3 y 0 when y fy g y dy y2 y 2 dy (−3, 3) 3 2y 2 1 gy y 2 y2 2 0 when y f y dy y2 y3 3 2 1 yy A 0, 3 0 dy 9 2 2 1 (4, 2) 3 2y 0 x 1 2 3 4 5 y dy y 1 3 2y 1 3y 0 (1, 3 1) 32 y 2 13 y 3 3 0 9 2 1 (0, 0) −3 −2 −1 x 1 6 Chapter 7 2 Applications of Integration 29. A 1 2 fy y2 1 g y dy 3 y (0, 2) (5, 2) 1 2 0 dy 6 1 x y3 3 2 3 4 5 6 y 1 (0, −1) 2 (2, −1) 3 30. A 0 3 0 fy y 16 3 g y dy 4 y y2 y2 3 0 dy 12 3 ( 3 ,3 7 ) 1 2 2 16 0 2y d y 1 16 y2 0 4 7 1.354 −1 x 1 2 3 31. y A 10 ⇒x x 10 2 10 y 12 1 y 32. A 0 4 4x 4 4 2 4 ln 2 x y dx (1, 4) 1 10 dy y 10 2 (0, 10) 8 6 4 (1, 10) x 0 3 10 ln y 10 ln 10 10 ln 5 33. (a) ln 2 16.0944 −4 4 ln 2 1 (0, 2) (0, 2) −2 (5, 2) x 2 4 6 8 1.227 −1 x 1 3 11 (b) The points of intersection are given by: (3, 9) x3 xx 12 3x 2 1x fx 3x 3 x2 0 when x 3 −6 (0, 0) −1 (1, 1) 0, 1, 3 f x dx 3 1 A 0 1 g x dx 1 gx x2 dx (c) Numerical approximation: 0.417 2.667 3.083 x3 0 1 3x 2 4x 2 43 x 3 3x 3x d x x2 1 x3 3x 2 3x d x 3 x3 0 x3 1 4x 2 32 x 2 3x d x 3 1 x4 4 34. (a) (−1, 2) y 32 x 2 1 0 x4 4 43 x 3 5 12 8 3 37 12 (b) The point of intersection is given by: x3 (1, 0) 2x x3 1 1 2x 0 when x g x dx 1 −2 −1 −1 −2 x 2 1 A 1 1 fx x3 1 1 (1, −2) 2x 1 dx 1 x4 4 2x d x 1 (c) Numerical approximation: 2.0 x3 1 x 1 2 S ection 7.1 35. (a) (0, 3) −6 −3 Area of a Region Between Two Curves 7 9 36. (a) (−2, 8) (4, 3) 12 −4 10 (2, 8) 4 −2 (0, 0) (b) The points of intersection are given by: x2 4x 2x x 4 (b) The points of intersection are given by: x4 2x 2 4 2 3 4 3 3 4x x2 0, 4 x2 4x 3 dx 2x 2 0 when x 2x 2 x4 x4 dx x5 5 2 0 0 when x 4x x2 8x d x 4 x2 x2 A 2 0, ± 2 A 0 4 2x 2 d x 0 2 2x 2 0 2 0 4x 2 4x 3 3 2x 3 3 4x 2 0 64 3 2 128 15 (c) Numerical approximation: 21.333 (c) Numerical approximation: 8.533 (b) The points of intersection are given by: x4 4x 2 4 1 x2 0 0 when x ± 2, ± 1 37. (a) f x x4 2 4x 2, g x x2 4 4 −4 (− 2, 0) (2, 0) 4 x4 x2 4 5x 2 x2 (−1, − 3) −5 (1, − 3) By symmetry: 1 2 (c) Numerical approximation: 5.067 2.933 8.0 A 2 0 1 x4 x4 0 4x 2 5x 2 x2 4 dx 1 4 dx 2 2 1 x2 5x 2 2 4 4 dx x4 4x 2 d x 2 2 2 x5 5 1 5 2 1 x4 x5 5 40 3 5x 3 3 8 5x 3 3 5 3 4 4x 0 2 2 32 5 4x 1 1 5 5 3 4 8 38. (a) f x x4 y 4x 2, g x x3 4x (b) The points of intersection are given by: x4 x4 xx x3 1x 0 4x 2 4x 2 x4 x3 0 4x g4 3 4x 2 2x 0 when x 1 2, 0, 1, 2 x4 0 2 −4 −3 −1 x 1 3 4 A −3 −4 x3 2 4x 4x 2 d x 4x 2 4x x3 x4 4x 4x2 dx dx f x3 (c) Numerical approximation: 8.267 0.617 0.883 9.767 1 248 30 37 60 53 60 293 30 8 Chapter 7 Applications of Integration 3 39. (a) 40. (a) 3 (−1, 1 ( ( 1, 1 ( 2 2 −3 −1 3 −1 −1 ( 3, 9 ) 5 (0, 0) 5 (b) The points of intersection are given by: 1 1 x4 x2 x2 x2 2 1 x 1 3 (b) A 0 6x x2 1 1 0 dx 3 x2 2 0 3 ln x 2 3 ln 10 0 2 x2 0 ±1 6.908 (c) Numerical approximation: 6.908 A 2 0 1 fx 1 1 g x dx x2 dx 2 x3 6 2 1 0 2 0 x2 2 arctan x 2 1 6 4 1 3 1.237 (c) Numerical approximation: 1.237 41. (a) (0, 2) −4 5 42. (a) (2, 3) 2 (0, 1) −1 −1 5 −1 5 (b) and (c) 1 x3 ≤ 1 x 2 2 on 0, 2 (b) and (c) You must use numerical integration: 4 You must use numerical integration because y 1 x3 does not have an elementary antiderivative. 2 A 0 x 4 4 x dx x 3.434 A 0 1 x 2 2 1 x3 dx 1.759 3 6 43. A 2 0 3 fx g x dx 4 3 y 44. A g 2 cos 2x 1 sin 2x 2 sin x dx 6 y 2 0 2 sin x 2 cos x ln 2 tan x d x 3 2 1 π , 3 (π , 1 ) 62 2 −π 2 3 f cos x 3 2 0 π 6 −1 x 2 21 ln cos x 0 π 2 (0, 0) π 2 x 3 4 33 4 0.614 3 4 (− π , −1) 2 π , 3 3 1.299 S ection 7.1 2 1 Area of a Region Between Two Curves x x tan dx 4 4 x 4 1 9 45. A 0 2 2 2 0 cos x cos x dx 2 cos x dx 46. A 0 2 2 42 x 2 4x 4 4 sec 1 4x 4 sec 0 2x y 3 sin x 0 4 12.566 2 2 24 2 2 4 4 1 2 2 4 2.1797 (0, 1) 2 g (2π, 1) 4 y f −1 π 2 π 2π x 3 2 1 x 1 (1, 2) 1 47. A 0 xe 1 e 2 y x2 0 dx 1 0 48. From the graph we see that f and g intersect twice at x 0 and x 1. 1 e 0.316 1 x2 1 1 2 A 0 1 gx 2x 0 f x dx 3 y (1, 3) 1 3x dx 2 1 x2 x 1 ln 3 () (0, 0) 1 1 1, e 1x 3 ln 3 1 (0, 1) 0 −1 x 1 2 x 21 0.180 49. (a) 3 (b) A 0 2 sin x 2 cos x sin 2x dx 1 cos 2x 2 2 1 2 (c) Numerical approximation: 4.0 0 0 0 2 1 2 4 50. (a) − 2 (0, 1) (π, 1) 5 4 (b) A 0 2 sin x 2 cos x cos 2x dx 1 sin 2x 2 4 0 (c) Numerical approximation: 4 4 −2 3 51. (a) 4 (b) A 1 1 1x e dx x2 3 (c) Numerical approximation: 1.323 (1, e) e (3, 0.155) 0 0 6 1x 1 e e1 3 10 Chapter 7 Applications of Integration 5 52. (a) 0 2 (b) A (5, 1.29) 6 1 4 ln x dx x 5 2 1 (c) Numerical approximation: 5.181 (1, 0) 2 ln x 2 ln 5 2 −2 53. (a) 6 (b) The integral 3 (c) A x 4 3 4.7721 A 0 −1 −1 4 x dx does not have an elementary antiderivative. 54. (a) 4 (b) The integral (1, e) 1 (c) 1.2556 A 0 −1 −1 2 x e x dx does not have an elementary antiderivative. 55. (a) 5 56. (a) 4 −3 −1 3 −4 −1 3 (b) The intersection points are difficult to determine by hand. c (b) The intersection points are difficult to determine. (c) Intersection points: 1.164035, 1.3549778 and 1.4526269, 2.1101248 1.4526269 (c) Area c 4 cos x x2 dx 6.3043 where c 1.201538. A 1.164035 3 x x 2 dx 3.0578 x 57. F x 0 1 t 2 0 1 dt t2 4 x t 0 x2 4 x 22 4 y (a) F 0 y (b) F 2 2 3 (c) F 6 y 6 5 4 3 2 62 4 6 15 6 5 4 3 2 6 5 4 3 2 −1 −1 t 1 2 3 4 5 6 −1 −1 t 1 2 3 4 5 6 −1 −1 t 1 2 3 4 5 6 S ection 7.1 x Area of a Region Between Two Curves 11 58. F x 0 12 t 2 0 2 dt 13 t 6 x 2t 0 x3 6 2x 43 6 24 56 3 (c) F 6 y y (a) F 0 y (b) F 4 36 12 48 20 20 16 16 12 12 8 8 4 4 x 1 2 3 4 5 6 1 2 3 4 5 6 x 20 16 12 8 4 x 1 2 3 4 5 6 59. F 1 cos 1 0 y 2 d 2 sin 2 2 1 sin 2 2 2 y (a) F (b) F 0 0.6366 (c) F 1 2 2 y 3 2 2 1.0868 3 2 3 2 1 2 1 2 1 2 1 2 1 −2 1 θ 1 −2 1 2 1 −2 1 θ 1 −2 1 −2 1 −2 1 2 1 θ y y 60. F y 1 4ex 2 dx 1 y 30 25 20 15 10 5 −1 8ex 2 1 8e y 2 8e 12 (a) F 0 (b) F 0 y 30 25 20 15 10 5 x 8 8e 12 3.1478 (c) F 4 y 30 25 20 15 10 5 8e2 8e 12 54.2602 1 2 3 4 −1 x 1 2 3 4 −1 1 2 3 4 x 4 61. A 2 4 2 9 x 2 7 x 2 6 12 7 dx x 6 4 5 dx 4 5 x 2 16 x 5 dx y 6 9 y = 2 x − 12 7 x 2 21x (4, 6) 5 y = − 2 x + 16 21 d x 6 4 2 72 x 4 4 7x 2 72 x 4 7 4 7 14 2 −2 −4 6 (6, 1) x 8 10 (2, −3) y=x−5 12 Chapter 7 c Applications of Integration a c b y dy c 62. A 0 c 0 b a y c a2 y 2c y a 63. 4 3 2 1 y a dy c −4 −2 −1 (0, 2) (4, 2) x 2 3 4 ay 0 (− 4, − 2) −3 −4 (0, − 2) ac 2 y ac ac 2 1 base height 2 Left boundary line: y Right boundary line: y 2 x x 2⇔x 2⇔x y y 2 2 y= cx b (b, c) A y= c b − a (x − a ) 2 2 y 4 dy 2 2 2 y 8 2 2 dy 8 16 4y x (0, 0) (a, 0) 1 3 64. A 0 1 2x 5x dx 0 3x dx 1 3 1 2x 15 dx 2 3 1 4 1 x 2 7 2 dx y 2 1 5 x 2 15 x 2 5 4 (1, 2) (0, 0) −2 −1 −1 −2 −3 2 3 4 x 5x2 2 5 2 15 2 1 0 5x2 4 45 4 45 2 (3, −2) (1, −3) 15 2 −4 65. Answers will vary. If you let x (a) Area 60 0 2 10 3 322 (b) Area 60 0 3 10 2 502 66. x 4, n 8, b 32 0 28 2 190.8 (b) Area 32 0 38 4 296.6 3 a 2 14 966 sq ft 4 14 1004 sq ft 84 2 11 6 and n 2 14 2 12 10, b 2 12 a 10 6 2 15 60. 2 20 2 23 2 25 2 26 0 2 14 4 12 2 12 4 15 2 20 4 23 2 25 4 26 0 32 2 13.5 2 14.2 2 14 2 14.2 2 15 2 13.5 0 (a) Area 381.6 sq mi 4 11 2 13.5 4 14.2 2 14 4 14.2 2 15 4 13.5 0 395.5 sq mi S ection 7.1 67. f x fx x3 3x 2 3. 1 3x 1 or y x 3 at x x4 4 3x 2 2 3x 2 2. 1 −4 −3 −2 8 6 4 2 y Area of a Region Between Two Curves 13 At 1, 1 , f 1 Tangent line: y y = 3x − 2 (1, 1) x 1 2 3 4 The tangent line intersects f x 1 f (x) = x3 −6 A 2 x3 3x 2 dx 2x 2 27 4 (− 2, − 8) −8 68. y y y 1 x3 3x 2 3 2x, 2 2 1 1 1, 1 5 4 y (2, 4) y=x+2 3 Tangent line: y 1x 1 ⇒y (− 1, 1) 2 x −1 −2 −3 2 3 4 x 2 −4 Intersection points: 2 1, 1 and 2, 4 2 y = x 3 − 2x A 1 x x4 4 1 x2 x2 2 3x2 2 x3 2 2x dx 1 x3 4 6 4 3x 1 4 2 dx 3 2 2 27 4 2x 1 69. f x fx y 1 2x 1 2 3 4 1 2 1 4 (0, 1) 1 f ( x) = 2 x +1 1 At 1, , f 1 2 Tangent line: y 1 2 1 . 2 1 x 2 1 or y 1 x2 1 dx 1 at x arctan x 1 x 2 0. x2 4 1 (1, 21 ) y=− 1x+1 2 x 1 2 1 1 3 2 2 The tangent line intersects f x 1 A 0 1 x2 2 1 1 8 22 4x2 1 1 x 2 1 ,1 2 x 0 3 4 0.0354 70. y y 1 y 2 , y 16x 4x2 y = − 2x + 2 2 1 (0, 2) y = 2 1 + 4x 2 ( 1 , 1( 2 x 1 2 1 y 2x 2x 2 1 2 −1 Tangent line: y Intersection points: 12 1 , 1 , 0, 2 2 12 2 A 0 2 1 4x 2x 2 dx arctan 2x x2 2x 0 arctan 1 1 4 1 4 3 4 0.0354 14 71. x 4 A Chapter 7 2x 2 1 Applications of Integration x 2 on x4 1, 1 2x 2 1 dx y 1≤1 1 x2 x 4 dx x5 1 5 1 72. x 3 ≥ x on 1, 0 , x 3 ≤ x on 0, 1 Both functions symmetric to origin. 0 1 1 1 x3 x2 1 1 x dx 0 x3 x dx x3 dx 0. x dx y 1 x 3 3 4 15 Thus, 2 x3 1 1 (0, 1) 1 (1, 1) You can use a single integral because x 4 2x 2 1 ≤ 1 x 2 on 1, 1 . A x 2 0 x (0, 0) ( 1, 0) (1, 0) x2 2 2 x4 4 1 0 1 2 −1 x 1 −1 (− 1, − 1) 73. Offer 2 is better because the accumulated salary (area under the curve) is larger. 3 74. Proposal 2 is better since the cumulative deficit (the area under the curve) is less. 75. 9 9 9 0 A 3 b 9 9 x2 b x3 3 2 9 3 9 x2 dx b dx x2 dx 9 0 32 b 36 10 y 18 9 6 2 b b 6 4 9 9 bx 2 x 2 6 ( 9 b, b) ( 9 b, b) 9 b b 9 9 27 2 3 32 b 9 4 9 4 b 9 3 3.330 9 76. A 2 0 9 2 x dx 9 0 9 b b 2 9x 9 9 x b bx 9 x2 2 b9 9 b 9 x2 2 9 81 0 12 y b dx x dx 9 0 b 81 2 81 2 81 2 81 2 9 2 2.636 9 (− (9 − b), b) 6 (9 −b, b) x 3 6 2 0 −6 −3 −3 −6 29 b b 9 2 S ection 7.1 1 77. Area of triangle OAB is 2 4 4 a Area of a Region Between Two Curves 15 8. 4x x2 2 a y 4 0 4 0 x dx 4a 0 a2 2 B 3 2 1 A a 2 8a 8 a 4±2 2 4 22 2 a x 1 2 3 4 O Since 0 < a < 4, select a 2 1.172. 78. Total area 2 4 2 4y y 2 dy y3 3 2 2 0 4 8 3 y 2 dy 3 y 28 0 32 3 4 1 16 3 4 42 3 4 2 a 4 a a 42 3 32 x dx 4 4 3 x 32 a 4 4 3 −1 x −1 1 2 3 4 5 a 32 a −3 4 4 4 n a 1.48 n 79. lim →0 i xi 1 x i2 x x 1 is the same as n x3 3 1 0 80. lim →0 i 4 1 x i2 2 x 4i and n 4x x x3 3 2 2 where xi 1 i and n x2 dx where xi 2 4 is the same as n 32 . 3 x 0 y x2 2 1 . 6 4 2 x2 dx y 5 0.6 0.4 0.2 0.2 0.4 f (x) x x2 3 f ( x) = 4 − x 2 (1, 0) x 2 0.6 0.8 1.0 (0, 0) (− 2, 0) −3 −1 1 1 −1 (2, 0) x 3 5 5 81. 0 7.21 0.58t 7.21 0.45t d t 0 0.13t d t 0.13 t 2 2 5 5 $1.625 billion 0 5 82. 0 7.21 0.26t 0.02t 2 7.21 0.1t 0.01t 2 d t 0 0.01t 2 0.01t 3 3 29 billion 12 0.16t d t 0.16t 2 2 5 0 $ 2.417 billion 16 Chapter 7 Applications of Integration t 83. (a) y1 Receipts (in billions) 600 500 400 300 200 100 270.3151 1.0586 R 270.3151e0.05695t (b) y2 Expenditures (in billions) 600 500 400 300 200 100 239.9704 1.0416 E t 239.9704e0.04074t t 2 4 6 8 10 12 t 2 4 6 8 10 12 Time (in years) 17 Time (in years) (c) Surplus 12 y1 y2 dt 926.4 billion dollars (Answers will vary.) 84. (a) y1 (b) Percents of total income 100 80 60 40 20 x 20 40 60 80 100 (d) No, y1 > y2 forever because 1.0586 > 1.0416. No, these models are not accurate for the future. According to news, E > R eventually. 0.0124x 2 y 0.385x 7.85 (c) Percents of total income 100 80 60 40 20 y x 20 40 60 80 100 Percents of families 100 Percents of families (d) Income inequality 0 x y1 dx 2006.7 85. 5%: P1 31%: P2 2 893,000e 0.05 t 893,000e 0.035 t 5 Difference in profits over 5 years: 0 893,000e 0.05t 893,000e 0.035t d t 893,000 e 0.05t e 0.035t 5 0.05 0.035 0 893,000 25.6805 34.0356 $ 193,156 20 28.5714 893,000 0.2163 Note: Using a graphing utility, you obtain $193,183. 86. The total area is 8 times the area of the shaded region to the right. A point x, y is on the upper boundary of the region if 2 y x2 x2 y2 y2 x2 4y y 2 4 4 4 1 y 4y 4y x2 x2 . 4 x. y2 1 y=x ( x, y ) x 1 2 We now determine where this curve intersects the line y x x2 4x 4 x Total area 8 0 1 0 x2 4 4± 2 2 16 2 2 16 x2 4 2±2 2 ⇒ x x dx 8x x3 12 x2 2 2 2 0 2 22 2 1 16 42 3 5 8 0.4379 3.503 S ection 7.1 87. The curves intersect at the point where the slope of y2 equals that of y1, 1. y2 0.08x2 k ⇒ y2 0.16x 1⇒x 1 0.16 6.25 Area of a Region Between Two Curves 17 6.25 (a) The value of k is given by y1 6.25 k y2 0.08 6.25 3.125. 2 (b) Area 2 0 6.25 y2 y1 d x 3.125 x2 2 x dx 6.25 0 k 2 0 0.08x 2 0.08x 3 3 2 3.125x 2 6.510417 5 13.02083 1 16 2 88. (a) A 2 0 1 x 2 5 9 1 3 5.5 5 x 32 x dx 5 5 5.5 1 x 0 dx 89. (a) A (b) V 6.031 2A 2 2 5.908 2 11.816 m 3 1 8 2 5.908 2 0 5 (c) 5000V 5000 11.816 59,082 pounds 25 (b) V 2A 10 5 9 2 6.031 5.5 5 6.031 m 2 12.062 m 3 60,310 pounds 91. True 3 7 6 (c) 5000 V 90. True 5000 12.062 92. False. Let f x x and g x 2x x2. f and g intersect at 1, 1 , the midpoint of 0, 2 . But b 2 93. Line: y 7 x 1 y fx a g x dx 0 x 2x x2 dx 2 3 0. A 0 sin x cos x 3 2 2.7823 7 24 3x2 14 1 3x dx 7 7 0 6 1 2 (0, 0) π 6 −1 4π 3 x ( 76π , − 1 ( 2 a 94. A a 4 0 b 1 x2 dx a2 4b a a a2 0 x2 dx a2 . 4 y y=b a2 0 x2 dx is the area of 4b a a2 4 1 of a circle 4 x 1− 2 a 2 b x Hence, A ab. a 18 Chapter 7 Applications of Integration y 95. We want to find c such that: b y = 2x − 3x 3 2x 0 3x 3 c dx b 0 c (b, c) x2 b2 But, c b2 4 34 4b 34 4x 34 4b cx 0 0 x cb 0 2b 3b3 because b, c is on the graph. 2b 8 3b3 b 12b2 9b 2 0 0 4 2 3 4 9 3b2 b c Section 7.2 1 Volume: The Disk Method 1 1. V 0 x 1 2 dx 0 x2 2x 1 dx x3 3 1 x2 x 0 3 2 2 2. V 0 4 x2 2 dx 0 x4 8x 2 16 dx x5 5 8x 3 3 2 16x 0 256 15 3 3 4 4 3. V 1 x 2 dx 1 x dx x2 2 4 1 15 2 4. V 0 9 x2 2 dx 0 9 9x x2 dx x3 3 3 18 0 1 1 5. V 0 x2 2 x3 2 dx 0 x4 x6 dx x5 5 x7 7 1 0 2 35 6. 2 8 x2 x 4 16 8 ±2 x2 4 x2 2 2 V 2 2 2 2 4 2 0 x2 4 2 2 2 dx 7. y V x2 ⇒ x 4 y 4 x4 16 2x3 3 2x2 2 12 dx 2 y 0 2 dy 0 y dy 2 2 2 x5 80 12x 0 y2 2 4 8 0 128 2 80 32 2 3 132.69 24 2 448 2 15 8. y V 0 16 4 x2 ⇒ x 16 y2 4 0 2 16 dy y2 4 9. y 16 y2 dy V x2 3 ⇒x 1 y3 2 1 y3 2 2 dy 0 0 y3 dy 0 y4 4 1 0 4 16y y3 3 128 3 S ection 7.2 4 4 Volume: The Disk Method 19 10. V 1 y2 y5 5 2y 4 4y 2 d y 1 y4 4 1 8y 3 153 5 16y 2 d y 16y 3 3 4 0 2 459 15 11. y x, y 0, x x, r x 4 (a) R x V 0 4 (b) R y V 4, r y 2 y2 y4 dy 15 y 5 2 0 x x dx 0 y dx 4 16 0 2 x2 0 8 y 3 2 1 x 16y 128 5 3 2 1 x 1 −1 2 3 4 1 −1 2 3 4 (c) R y V 4 2 y 2, r y 4 y2 2 dy 8y 2 83 y 3 0 (d) R y V 6 2 y 2, r y 6 y2 2 2 4 dy y4 dy 15 y 5 2 0 0 2 0 2 16 0 y4 dy 0 32 256 15 y 4 3 12y 2 4y 3 16y y 3 2 15 y 5 2 0 32y 192 5 2 1 x 1 −1 2 3 −1 −2 1 x 1 2 3 4 5 12. y 2x 2, y 0, x 2, r y 8 2 y2 y dy 2 4y y 4 28 (a) R y V (b) R x 16 0 2x 2, r x 2 0 4x 5 5 2 0 4 0 y 8 6 V 0 4x 4 d x y 8 6 128 5 4 2 x 2 4 −4 −2 4 2 x 2 4 −4 −2 —CONTINUED— 20 Chapter 7 Applications of Integration 12. —CONTINUED— (c) R x V 0 2 2 8, r x 2 8 64 2x 2 32 x 2 4 0 (d) R y 4 x4 dx 8x 2 y 2 8 y 2, r y 2 y 2 4 y 2 2 0 dy y dy 2 y2 4 8 y 64 32x 2 0 V 0 8 4x 4 d x 15 x 5 2 0 x4 dx 0 4 4y 16 3 4 896 15 83 x 3 4 2 32 y 3 0 8 6 6 4 2 x 2 4 −4 −2 4 2 x 4 −4 −2 13. y x2, y 4x 4x 2 x 2 intersect at 0, 0 and 2, 4 . x 2, r x 4x x2 2 (a) R x V x2 x 4 dx (b) R x V 6 2 x2, r x 6 x2 2 6 6 4x 4x x2 x2 2 dx 0 2 0 2 16x 2 0 8x 3 d x 2 8 0 x3 x4 4 y 5x 2 53 x 3 6x d x 2 16 3 x 3 y 4 2x 4 0 32 3 8 3x 2 0 64 3 5 3 4 2 1 1 −1 x 1 2 3 −2 −1 1 2 3 4 x 3 2 14. y 6 2x 6 0 x 2, y 2x 6 3 0 x 6 intersect at x 2 3, 3 and 0, 6 . (b) R x 2 (a) R x V x 2, r x 2x 4x 3 x4 y 8 6 x 6 dx 6 0 2x 3 2x x2 x2 3, r x 2 x 3 2 6 dx 3 x2 V 3 0 x x4 3 9x 2 3x 3 36x d x 3 0 x4 243 5 15 x 5 4x 3 x4 y 8 3x 2 x3 9x 2 18x d x 0 3 15 x 5 18x 2 3 108 5 4 2 x 2 4 2 x 2 −6 −4 −2 −6 −4 −2 S ection 7.2 x3 ,r x 2 4 0 2 Volume: The Disk Method 21 15. R x V 4 3 x, r x 4 x 2 1 1 2 16. R x dx y 5 4 1 0 2 0 3 V x3 2 dx x6 dx 4 x7 28 128 28 2 y x2 0 8x 4x2 15 d x 3 16 0 4x3 x4 16 x3 3 18 15x 0 3 2 1 x 1 2 3 4 16x 32 144 7 0 3 2 1 −1 −1 (2, 4) −1 x 1 2 3 4 17. R x V 4, r x 3 4 4 8 1 1 1 1 1 1 x 1 1 1 3 x 2 18. R x dx dx 3 4, r x 3 4 2 sec x 4 sec x 2 42 0 3 0 V 0 3 4 dx x 2 8 sec x 0 sec 2 x d x 3 1 x x 1 4 3 4 8 ln 1 8 ln 4 8 ln 4 32.485 8 ln sec x 8 ln 2 y tan x 3 3 3 3 tan x 0 x 0 8 ln 1 27.66 0 0 8 ln 2 y 5 2 1 −1 −1 x 1 2 3 4 2 1 π 9 π 3 x 3 2π 9 4π 9 5π 9 19. R y V 6 4 y, r y 6 y 2 dy 12y 6y 2 0 20. R y V 6, r y 4 6 y y3 3 4 0 2 6 dy 368 3 y y 6 0 2 0 4 y2 0 36 d y 4 y y 5 5 4 4 3 3 2 2 1 1 36y y3 3 208 3 36y 0 x x 1 −1 2 3 4 5 −1 1 2 3 4 5 22 Chapter 7 Applications of Integration 6 ,ry y 6 2 6 21. R y V 6 2 y 2, r y 6 y2 2 2 2 2 22. R y dy V 36 6 6 0 dy 2 2 6 y 1 2 2 0 y4 y5 5 12y 2 4y 3 32y 32 dy 2 y 2 2 y 2 ln y 1 dy y2 1 y 6 2 2 384 5 0 4 3 2 1 −1 −2 −3 x 1 2 3 5 36 36 36 y 35 6 13 3 2 ln 6 2 ln 1 3 3 2 y 6 5 4 3 2 1 2 ln 2 12 13 241.59 6 ln 3 x 1 2 3 4 5 23. R x 3 1 x x 1 x 1 1 1 , rx 2 0 dx 2 y 24. R x V 2 x4 2 x 2, r x x2 2 0 dx 3 2 1 −3 −1 x 1 2 3 y V 0 3 0 x4 0 2 1 dx 2 0 1 4x 2 4x 3 3 x4 dx x5 2 5 0 3 ln x ln 4 1 0 1 −1 2 3 x 2 128 15 −2 −3 4.355 25. R x V 1 , rx x 4 1 0 2 y 26. R x V 3 x 8 0 1 x 8 , rx 3 1 1 2 0 4 y 1 x 1 x 4 1 2 dx 1 x 1 −1 −2 2 3 4 dx 2 3 2 9 0 x 1 x dx 1 3 4 e x, r x 1 8 9 1 8 0 x 2 4 6 8 27. R x V 0 2 y 28. R x V e x 2, r x 4 0 8 y e 0 1 x2 dx 1 ex 0 4 22 dx 6 4 2 e 0 2x dx 1 ex dx 0 4 x 2 2 1 e 2x 0 2 1 2 e x 0 −2 2 4 6 x e4 e 1.358 1 168.38 S ection 7.2 29. 2x 2 x2 x x2 2x x 1 4 2 1 0 0 0 1, 2 and 2, 5 . 3 2 Volume: The Disk Method 23 x2 2x 5 2x The curves intersect at 2 V 0 2 5 2x 4 x3 x2 8 x2 x2 20 x 1 2 dx 2 3 x2 4x3 2 1 8x2 2 5 20 x 24 x 2 2x 24 d x 3 x2 2 dx 10 8 6 y 24 d x 2 (2, 5) 0 x4 152 3 4 83 x 3 125 3 1 x 2 5x 5x2 2 56 3 2 10 x 2 277 3 24 x 0 x4 83 x 3 10 x 2 2 −1 x −2 1 2 3 4 30. V 0 4 0 4 x2 4 x 16 dx 2 8 dx 4 8 4 x x2 4 5x2 2 5x 2 4 16 dx 8 1 x 2 2 dx 4 3 2 1 y (4, 2) x3 12 88 3 4 16x 0 x3 12 16x 4 −2 −1 2 4 6 8 10 x 48 31. y V 6 6 0 3x ⇒ x 1 6 3 36 0 1 6 3 2 y 6 5 4 3 y 32. y x V 9 9 5 x 2, y y 9 0 5 0, x 2, x 3 y 12y dy y2 dy y3 6 3 0 y y dy 2 2 2 dy y 9 8 7 6 5 4 3 2 1 6 2 1 x 1 3 4 5 6 9 9 9 8 5 0 36y 216 6y 2 216 5y 25 216 3 Volume of cone y2 2 25 2 5 0 (2, 5) 12 r h, 3 25 2 x 123456789 2 33. V 0 sin x 2 dx 1 0 y 34. V 0 2 0 cos x 2 dx 1 cos 2x dx 2 2 0 2 y cos 2x dx 2 1 sin 2x 2 2 3 2 1 2 2 x 1 0 x 1 2 3 2 x 1 sin 2x 2 2 x 1 2 2 22 4 Numerical approximation: 4.9348 Numerical approximation: 2.4674 24 Chapter 7 2 Applications of Integration 2 35. V 1 2 ex e2x 1 12 dx 36. V 1 2 ex ex 1 2 e e x x22 dx 2 dx ex e2 2 dx 2 2 2 2 e2x e2 2 1 e e e x 2x 1 2 4 6 e 2 e 1 e 1 2 1 e2 e Numerical approximation: 10.0359 2 3 Numerical approximation: 49.0218 5 37. V 0 e x2 2 dx 1.9686 38. V 1 ln x 2 d x 3.2332 39. V 0 2 arctan 0.2x 15.4115 2 dx 40. x 2 x4 x3 x V 2x 2 21 2x 2 y y = x2 3 2 0 2 0 13 13 1 y= 2x 1.2599 2x 2x 2 x2 2 dx x 1 2 x 4 dx 2.9922 3 22 5 3 2 4 41. 0 sin2 x dx represents the volume of the solid generated by revolving the region bounded by y y 0, x 0, x 2 about the x-axis. y 42. 2 y 4 dy represents the volume of the solid generated by revolving the region bounded by x x 0, y 2, y 4 about the y-axis. y 4 sin x, y 2, 1 3 2 1 π 4 π 2 x x 4 8 12 16 43. A 3 2 y 44. A 3 4 1 3 4 1 2 1 4 y Matches (a) Matches (b) 1 x 1 2 1 4 1 2 3 4 x 1 S ection 7.2 45. y 4 3 2 1 x 1 2 1 2 3 4 3 2 1 x y Volume: The Disk Method 25 −2 −1 The volumes are the same because the solid has been translated horizontally. 4x 46. (a) 4 2 2 4 x −4 8 z x2 4 (c) x z 4 4 2 2 (b) 4 2 z y x 8 8 y 8 x 16 y a<c<b 47. R x V 1 x, r x 2 6 0 0 4 3 2 1 y 48. R x V r x, r x h h 0 0 y 12 x dx 4 6 r2 2 x dx h2 h 0 r y= r x h (h, r) 12 Note: V x3 0 18 1 −1 2 3 4 5 6 x r2 3 x 3h 2 r h3 3h 2 2 x h 12 rh 3 1 3 18 32 6 −2 12 rh 3 49. R x r r2 r2 r r x 2, r x x2 dx x2 dx 0 y 50. x V r2 r y2, R y r2 y2 2 r2 dy y2, r y 0 V 2 0 h y = r2 − x2 r r2 r 2x r3 r2 h y2 dy y3 3 r3 3 r 2h 3r 2h r h 2 2 13 x 3 13 r 3 r 0 (−r, 0) x r2y r3 2r 3 3 3 2r 3 (r, 0) 43 r 3 r 2h h3 3 h3 3 y r h3 h x 26 Chapter 7 r y H h Applications of Integration y ,Ry H 2 51. x V r r1 y H r1 h y ,ry H 2 y H 0 H y r1 0 dy r2 0 1 12 y dy H2 13 y 3H 2 h3 h 0 r2 y r2 h r 2h 1 12 y H h H 2 h −r r x 3H 2 h2 3H 2 4 c h H x2 2 4 4 52. (a) V 0 x 2 dx 0 x dx 8 0 (b) Set 0 x dx 8 , c2 3 8 (one third of the volume). Then 3 16 ,c 3 4 3 43 . 3 Let 0 < c < 4 and set c x dx 0 x2 c 2 0 c2 2 c2 2 4. To find the other value, set d c2 c 8 8 22 x dx 0 16 (two thirds of the volume). 3 16 , d2 3 32 ,d 3 32 3 46 . 3 Thus, when x 2 2 , the solid is divided into two parts of equal volume. Then d2 2 The x-values that divide the solid into three parts of equal 4 3 3 and x 4 6 3. volume are x 2 53. V 0 12 x2 8 2 2 x dx 64 x4 2 0 x dx 2x 5 64 5 x6 6 2 0 30 54. y V 0.1x 3 2.95, 11.5 2.2x 2 10.9x 22.2, 0 ≤ x ≤ 11.5 11.5 < x ≤ 15 8 y 15 0.1x 3 0 2.2x 2 10.9x 22.2 11.5 2 dx 11.5 2.95 2 15 dx 6 0.1x 4 4 2.2x 3 3 10.9x 2 2 4 2 x 4 8 12 16 22.2x 0 2.95 2x 11.5 1031.9016 cubic centimeters 55. (a) R x V 9 25 3 5 5 25 25 5 5 x2, r x x2 dx 0 y 8 (b) R y V 25 9 25 9 50 x 2 −2 4 6 5 3 9 3 y 2, r y 9 y2 dy y3 3 3 0 0, x ≥ 0 y 0 6 18 25 25 0 x dx x3 3 5 0 −6 −4 −2 2 6 4 2 9y 4 18 25x 25 60 2 x 2 4 6 −4 S ection 7.2 Volume: The Disk Method 27 56. (a) First find where y b x2 x V 0 4 b intersects the parabola: z 5 4 16 x2 4 4b b b 44 b 24 2 4 4 4 0 4 0 x2 4 2 2 4 b dx bx 2 2 dx 2 4 b b 4 x2 4 2 dx x 2 2 4 y x2 4 2x 2 b x4 16 2x 3 3 128 3 b2 8b 8bx 32b 512 15 16 d x 4 x5 80 64 5 (b) Graph of V b 120 bx 3 6 32 b 3 4b 2 b 2x 4b 2 64 b 3 16x 0 64 4b 2 64 b 3 512 15 8b 8 64 3 0⇒b 64 3 8 8 3 2 2 3 (c) V b Vb >0⇒b 8 is a relative minimum. 3 0 0 4 Minimum volume is 17.87 for b 4 50 3 3 2.67. 57. Total volume: V 500,000 ft 3 3 y 60 40 Volume of water in the tank: y0 2500 50 y2 2 y0 20 dy 50 2500 2500 y 2500 y0 y3 3 y 2 dy y0 50 − 60 − 20 x 20 40 60 − 40 − 60 y03 3 250,000 3 When the tank is one-fourth of its capacity: 1 500,000 4 3 125,000 y03 7500 y0 125,000 y0 Depth: 17.36 50 2500 y0 7500 y0 0 17.36 32.64 feet 32.64 67.36 feet. y03 y03 3 250,000 3 250,000 When the tank is three-fourths of its capacity the depth is 100 28 Chapter 7 10 Applications of Integration 58. (a) V 0 fx 2 dx a 2 Simpson’s Rule: b V 3 3 (b) f x 6 10 0 2 10, n 4 2.35 2 10 2 2.6 2 2.1 2 4 1.9 2 2.1 4 2.85 2 2 2.9 2 4 2.7 2 2 2.45 2 4 2.2 2 2.3 2 178.405 0.00249x 4 186.83 cm 3 10 0.0529x 3 0.3314x 2 0.4999x 2.112 (c) V 0 f x 2 dx 186.35 cm 3 0 0 10 h b 59. (a) 0 r2 dx (ii) (b) b a 1 x2 b2 2 r d x (iv) (c) r r2 x2 2 d x (iii) is the volume of a right circular cylinder with radius r and height h. y is the volume of an ellipsoid with axes 2a and 2b. y 2 y=a 1− x b2 is the volume of a sphere with radius r. y y=r (h, r) (0, a) y= r 2 − x2 x x x (−r, 0) (r, 0) (−b, 0) (b, 0) h (d) 0 rx h 2 r dx (i) (e) r R r2 x2 2 R r2 x2 2 d x (v) is the volume of a right circular cone with the radius of the base as r and height h. y is the volume of a torus with the radius of its circular cross section as r and the distance from the axis of the torus to the center of its cross section as R. y (h, r) y= r x h R+ r 2 − x2 R x R− r2− x2 −r r x 60. Let A1 x and A 2 x equal the areas of the cross sections of the two solids for a ≤ x ≤ b. Since A1 x A 2 x , we have b b V1 a A1 x dx a A2 x d x V2. Thus, the volumes are the same. S ection 7.2 61. 4 3 2 y Volume: The Disk Method 29 x 2 3 4 Base of cross section (a) A x b2 4 2 x x 3x 2 3x 2 x3 x2 1 2 x2 1 2 x x2 (b) A x bh 2 2 4x 2 x x x2 1 2x x2 2 x3 3 2 1 2x 3 2x 3 14 x 2 x4 x4 dx 15 x 5 2 1 V 1 2 x2 dx 9 2 V 1 4 4x 4x 2x2 81 10 1 2 + x − x2 2 + x − x2 2 + x − x2 62. 3 y 1 −3 −1 x 1 3 −3 Base of cross section (a) A x 2 24 x2 2 x2 (c) A x 2 4 − x2 b2 44 2 24 x2 dx x3 3 2 2 12 r 2 2 2 4 x2 2 2 4x x3 3 4 2 2 x2 16 3 V V 2 4 2 x2 dx 2 4 4x 128 3 2 4 − x2 (b) A x 1 bh 2 34 2 1 24 2 x2 x2 dx x3 3 2 2 x2 34 x2 (d) A x V 2 4 − x2 1 bh 2 2 1 24 2 x 2 dx x2 4x 4 x3 3 2 2 x2 32 3 4 x2 V 3 2 4 4 2 3 4x 32 3 3 2 4 − x2 2 4 − x2 2 4 − x2 4 − x2 2 4 − x2 30 63. 1 3 4 1 2 1 4 Chapter 7 y Applications of Integration x 1 4 1 2 3 4 1 Base of cross section (a) A y 1 1 3 3 y (b) A y 1− 3 b2 1 0 1 1 3 y 2 12 r 2 1 8 1 1 2 3 1 2 y 2 3 y 2 1 8 1 3 y 2 V y 2 dy y2 3 53 y 5 3 3 y V 1 0 dy 1 8 10 80 1 0 2y 1 3 43 y 2 3 dy 1 0 1− 3 y y (c) A y 1 10 3 1 2 3 1− 3 y 1 bh 2 3 1 4 1 1 2 3 y y (d) A y V 1 ab 2 1 2 1 21 3 3 y 1 2 1 2 10 3 y 2 20 1 3 y 2 y 3 2 V 3 4 1 1 0 y 2 dy 31 4 10 3 40 2 y 2 dy 0 1− 3 y 1− 3 y a b 1− 3 y 1− 3 y R2 r2 64. The cross sections are squares. By symmetry, we can set up an integral for an eighth of the volume and multiply by 8. Ay V 8 0 65. V R2 R2 r2 r2 R2 x2 2 r2 dx b r 2 r 2 y 22 2 0 R2 r2 x3 3 R x2 dx R2 r2 r2 y2 dy 2 13 y 3 r x y R2 R2 R2 r2 x r2 r2 32 0 2 8 r2y 16 3 r 3 0 2 4 3 r2 3 32 32 R r R2 − r 2 R S ection 7.2 4 3 14 23 125 2 125 2 r2 r2 51 2 23 23 1 Volume: The Disk Method y3 3 1 0 31 66. 25 25 25 25 25 1 r2 r2 32 125 67. V 0 y 2 dy 3 32 r2 25 22 3 2 23 r 1 3.0415 1 68. V 0 1 12 2y 0 1 y 2 dy 69. V 0 x2 x3 3 1 3 x5 5 1 5 x 4 dx 1 0 y 2 dy y3 1 3 1 3 0 y2 1 2 3 2 15 1 1 70. V 0 1 1 1 0 1 x2 2x 2 2 1 x4 x 1 2 dx 2x x 2 dx 71. V 0 1 y 1 y dy y2 2 1 2 1 0 2x 0 3x 2 x3 x5 5 x 4 dx 1 0 x2 1 5 1 2 5 1 72. V 0 1 1 1 0 y 2 dy y dy y2 1 2 0 73. V 0 y y2 2 1 2 6 y 2 dy y3 3 1 3 1 0 2y 43 y 3 4 3 2 y 1 1 2 6 32 Chapter 7 1 Applications of Integration 74. V 0 1 1 1 0 1 y 2y 2 1 y2 3y 1 y 2 dy y3 3 1 0 y 2 dy y dy 75. (a) When a When a 1: x 2: x y y 2 1 represents a square. 2 y 1 represents a circle. 2y 1 a=2 a=1 x 2y 0 43 y 3 4 3 6 2 3y2 2 1 3 −1 1 3 2 −1 (b) y A 2 1 1 x 1 1 a 1a 1 x a 1a dx 4 0 1 xa 1a dx To approximate the volume of the solid, form n slices, each of whose area is approximated by the integral above. Then sum the volumes of these n slices. 76. (a) Since the cross sections are isosceles right triangles: Ax V (b) A x V 1 bh 2 1 2 r 1 2 r2 r r2 y2 r2 dy y2 r r2 r2 0 y2 y2 y 2 tan r 12 r 2 dy y2 x r 2y tan 2 y2 dy r2 y3 3 r 0 23 r 3 y 1 bh 2 tan 2 r 1 2 r2 r y2 r2 tan y2 r2y y3 3 r 0 y2 dy . r2 0 tan 23 r tan 3 As → 90 , V → r 77. (a) x R 2 y2 x r r2 R± r2 r2 y2 2 (b) 0 r2 y2 dy is one-quarter of the area of a circle of y2 R r2 y2 2 radius r, 1 r2. 4 dy V 8R 1 4 V 2 0 r R 4R r2 0 r r2 2 22 rR 2 8R y2 dy y2 dy r2 0 y x R Section 7.3 Volume: The Shell Method 33 Section 7.3 1. p x V 2 0 Volume: The Shell Method x 2. p x V 16 3 2 0 2 0 1 x, h x 2 x, h x 1 1 x 3. p x V 2 x, h x 4 x x x dx 2 x3 3 x1 x 0 x dx x2 d x x3 3 1 0 x x dx 0 4 2 2 x2 2 2 0 x3 2 dx 45 x 5 4 2 0 3 128 5 4. p x V 2 x, h x 2 8 x2 4 4 x2 5. p x V 2 x, h x 2 x2 4 y x4 0 2 x2 dx x3 dx x4 4 2 x3 dx 0 2 3 2 0 4x 2x 2 2 8 x4 0 8 2 1 x 1 2 3 2 0 −1 6. p x V 2 x, h x 6 0 12 x 2 y 18 15 12 9 6 7. p x V 2 x, h x 2 4x x2 x2 4x y 4 2x 2 13 x dx 2 6 x 4x 0 2 2x 2 d x x3 dx 14 x 4 2 0 x4 4 324 0 4 0 x 1 2 3 4 5 6 2x 2 23 x 3 3 3 −1 −3 4 16 3 −1 2 1 x 1 2 3 8. p x V 2 x, h x 2 4 x2 y 9. p x hx x 4 x2 2 y 4x 0 x3 dx 3 4x 4x x3 0 x2 4 4x 2 43 x 3 4x d x 4 3 2 2x 2 14 x 4 2 8 0 2 1 x 1 2 2 1 V 2 2 8 3 x4 4 2 −2 −1 2x 2 0 −1 x 1 2 3 10. p x V 2 x, h x 2 4 2x 4 y x4 0 2 2x d x 3 2 0 4x 2x 2 2x 2 d x 23 x 3 2 0 2 1 2 16 3 −1 x 1 2 3 34 Chapter 7 Applications of Integration 1 e 2 1 e 2 x2 2 11. p x V 2 x, h x 1 x2 2 12. p x y 1 x, h x 2 0 sin x x 3 2 1 y x 0 1 dx 3 4 1 2 1 4 V sin x x dx x sin x d x 2 0 e 2e x2 2 x x2 2 1 0 dx 2 0 π 4 π 2 3π 4 π x 2 cos x 0 x 4 −1 2 0.986 13. p y V 2 0 2 1 1 e 1 4 1 2 3 4 1 y, h y 2 2 y 14. p y hy V 2 4 y, 2 0 p y ≥ 0 on y y2 2 y dy y2 dy 0 2 2, 0 y y2 2y 0 y dy y2 dy y3 3 2 0 2 2 y2 2 0 8 3 2 2 2y y2 y3 3 2 8 3 15. p y py V 2 y and h y y and h y 12 1 if 0 ≤ y < 1 y 2 12 1 . 2 1 3 4 1 2 y 1 1 if ≤ y ≤ 1. 2 1 y dy 0 1 y y2 2 y2 1 12 y dy 1 4 2 y2 2 12 x 2 0 4 4 2 17. p y 1 2 1 3 2 2 16. p y V 2 y, h y 4 16 y2 dy y, h y 8 3 y 8 y y 16 0 4 y 4 3 V 2 0 8 y 3 y dy y 4 3 dy 0 6 4 2 2 0 16y 8y2 128 y 4 y3 dy 44 0 2 1 x −1 −2 −3 −4 4 8 12 2 2 2 2 37 y 7 8 3 0 −2 x 2 4 6 64 128 6 27 7 y 768 7 18. p y V 2 y, h y 9 y 9 y y dy 0 2 0 y3 2 dy 9 6 2 25 y 5 52 9 2 0 3 4 9 5 4 243 5 972 5 x 3 6 9 S ection 7.3 19. p y V 2 0 2 Volume: The Shell Method y y y2 y2 y 2 dy 3 35 y, h y 2 4 2y dy y y 4 y 4 2y 20. p y V 2 y, h y 2 2 2 y y y2 y4 4y 0 y2 0 2 2 2 2 2y 2 dy 23 y 3 16 3 2 0 −1 3 2 1 x 1 −1 2 3 4 (2, 2) 2 0 2y y2 4 8 3 y3 dy y4 4 2 0 −2 −1 (2, 2) 2 1 2y 8 2 2 2 y3 3 4 x 1 −1 2 16 3 4x x2 16 3 4x x2 x2 4x 2x 2 21. p x V 2 4 2 x, h x 4 0 2 x2 4x 2x 2 22. p x V 2 2 2 x, h x 2 0 2 x 4x x3 6x 2 2x 2 d x 8x d x 2 x 4x 8x 2 83 x 3 2x 2 d x 2x 3 d x 14 x 2 2 0 2 4 y 4 3 2 0 2 0 8x 4x2 y x4 4 2x 3 4x 2 0 16 2 16 3 4 3 2 1 x 1 2 3 2 1 x 1 3 −1 23. p x V 2 5 4 x, h x 5 0 4 4x x2 24. p x V 2 6 4 x, h x 6 0 4 x x dx x3 2 x 4x 9x 2 3x 3 x2 dx 20x d x 4 x 2 0 x3 x4 4 2 0 6x 1 2 4x3 2 dx 4 2 y 4 3 2 10x 2 0 64 y 4 3 2 1 2 25 x 5 2 0 192 5 1 x x 1 −1 2 3 4 1 −1 −2 2 3 4 5 4 25. The shell method would be easier: V 4 2 0 4 4 x 2 y 2 2 2 y dy shells 4 x 2 Using the disk method: V 0 2 dx Note: V 128 3 36 Chapter 7 Applications of Integration ln 4 26. The shell method is easier: V Using the disk method, x ln 4 2 0 x4 y and V e x dx 3 ln 4 0 y 2 dy. Note: V 8 ln 2 2 8 ln 2 3 27. (a) Disk Rx V 0 y 8 6 (b) Shell x 3, r x 2 0 x7 7 2 0 px 128 7 V 2 x, h x 2 x3 2 x5 5 2 0 x6 dx x 4 dx 0 y 64 5 8 6 4 4 2 x 1 2 3 2 x 1 2 3 −1 −1 (c) Shell px V 2 0 2 4 2 x, h x 4 4x 3 0 x3 8 y x x3 dx 6 2 2 x4 dx 15 x 5 2 0 4 2 x4 96 5 x 1 2 3 4 28. (a) Disk Rx V 1 y (b) Shell Rx V x 10 , rx x2 5 10 0 dx −1 8 6 4 2 1 2 3 4 5 x, r x 5 0 10 x2 5 2 2 1 x 5 10 dx x2 1 dx x 5 100 1 x x 4 dx −2 20 1 35 20 ln x 1 20 ln 5 100 100 3 (c) Disk Rx V 1 3 1 1 125 1 496 15 10, r x 5 10 10 200 x 5 1 10 x2 10 x2 2 102 100 3x3 dx 1904 15 S ection 7.3 29. (a) Shell py V 2 0 a Volume: The Shell Method 37 (c) Shell y, h y a a1 2 y1 2 2 px V 2 a a x, h x a 0 a a1 2 x1 2 2 ya ay 0 2a 1 2 y 1 2 2a 1 2 y3 2 4a 1 2 5 2 y 5 4a3 5 a3 3 y dy y2 dy y3 3 15 a x a1 2 2a 3 2 x 1 2 4 32 32 ax 3 x1 2 2 dx 2a 1 2 x 3 2 4 12 52 ax 5 x2 dx 13 x 3 a 0 2 2 2 y 2 0 a2 a 2x a2 y 2 a3 2 2 0 y 4 a3 15 a3 (0, a) (0, a) (a, 0) x (a, 0) x (b) Same as part (a) by symmetry y 30. (a) Disk Rx V a a (b) Same as part (a) by symmetry y a2 3 a x2 3 3 2, rx 0 (0, a) (0, a) a2 3 2 0 x2 3 3 dx 3a 4 3x 2 3 9 43 53 ax 5 93 a 5 93 a 7 3a 2 3x 4 3 9 23 73 ax 7 13 a 3 x2 dx 13 x 3 32 a 3 105 a (a, 0) a2 a 2x a3 (− a, 0) (a, 0) x x 2 2 (0, − a) 0 31. Answers will vary. (a) The rectangles would be vertical. (b) The rectangles would be horizontal. 32. (a) 2 z (b) 2 z (c) 2 x 5 z 10 y 2 x −2 5 y x 5 5 y a<c<b 38 Chapter 7 5 Applications of Integration 5 4 33. 1 x 1 dx 1 x 1 2 dx 34. 2 0 x x dx 2 This integral represents the volume of the solid generated x 1, y 0, by revolving the region bounded by y and x 5 about the x-axis by using the disk method. 2 represents the volume of the solid generated by revolving the region bounded by y x 2, y 0, and x 4 about the y-axis by using the shell method. 2 2 2 0 y5 y2 1 dy 0 16 2y 2 dy 0 4 2 2y 2 dy represents this same volume by using the shell method. y 4 3 2 1 x 1 −1 2 3 4 5 represents this same volume by using the disk method. y 4 3 2 1 x 1 −1 2 3 4 5 Disk method 35. (a) Disk method 36. (a) y = (1 − x ) 4/3 3/4 1.5 y 1 3 4 − 0.25 − 0.25 1.5 1 2 1 4 (b) x 4 3 y V 1 2 y4 3 x4 3 1 1, x 34 0, y 0 x 1 4 1 2 3 4 1 1 x1 0 x4 3 34 dx 1.5056 (b) V 2 0 x1 x3 dx 2.3222 37. (a) 7 38. (a) y= 3 y 4 3 (x − 2) 2 (x − 6) 2 −1 −1 7 2 1 6 (b) V 2 2 x3 x 2 2 x 6 2 dx 187.249 (b) V x 1 2 3 4 3 2 1 1 2x dx e1 x 19.0162 39. y 2e x, y 7.5 0, x 0, x 2 y 40. y tan x, y 1 0, x 0, x 4 y Volume Volume 2 Matches (d) Matches (e) 2 1 1 x 1 2 π 4 π 2 x S ection 7.3 12 x 2 2 Volume: The Shell Method 39 41. p x V 2 x, h x 2 2 x2 0 12 x dx 2 2 0 2x 13 x dx 2 2 x2 14 x 8 2 4 0 total volume Now find x0 such that: y 2 1 1 x 04 8x 02 4 x 02 Take x0 Diameter: 2 4 4 2 x2 2 x 02 0 x0 0 2x 14 x 8 14 x 40 13 x dx 2 x0 0 2 1 x 1 2 4±2 3 23 23 (Quadratic Formula) 0.73205, since the other root is too large. 1.464 3 3 42. Total volume of the hemisphere is 1 4 r 3 2 23 3 px x, h x 9 x 2. Find x0 such that: x0 18 . By the Shell Method, 6 6 2 0 x9 x0 x2 dx 12 y 9 0 x2 2x d x x0 0 2 1 x 2 9 3 9 x 02 32 x2 3 2 18 2 9 3 x0 2 32 −3 −2 −1 −1 −2 −3 1 2 3 18 9 18 2 3 1.460 x0 Diameter: 2 9 1 18 2 3 2.920 r 43. V 4 1 1 2 1 1 x 1 x2 dx 1 x2 dx 1 44. V x1 1 4 r r R r2 r x r2 x2 dx 2 x2 dx r 8 8 4 2 4 x2 x2 32 x2 dx 4R 4R 2 22 4 r x r2 r x 2 dx 2 2 2 1 x1 2 1 3 12 2 dx 4 1 2 r2 2 22 r 3 x2 32 r 1 rR 45. (a) d sin x dx Hence, x cos x x sin x dx C sin x cos x x sin x C. cos x x sin x x cos x —CONTINUED— 40 Chapter 7 Applications of Integration 45. —CONTINUED— (b) (i) p x V 2 0 2 x, h x 2 sin x (ii) p x V 2 x, h x 2 sin x sin x y 2 3 sin x x sin x dx sin x 1 0 x cos x 0 x 3 sin x dx 0 2 2 y 1.0 0.5 6 0 x sin x dx sin x 6 x cos x 0 2 −π 4 1 x 0 2 6 6 −1 −2 π 2 π 5π 4 −π 4 − 1.0 π 4 π 2 3π 4 π x 46. (a) d cos x dx Hence, x sin x x cos x dx cos x ⇒ x 0.8241 C cos x sin x x sin x sin x C. x cos x x cos x (b) (i) x 2 V ± 0.8241 (ii) 4 cos x V 2 0 x 1.511 2 2 ⇒x x 0, 1.5110 2 2 22 0 x cos x cos x x sin x x 2 dx x4 4 0.8241 x 4 cos x 1.511 dx x 3 y 4 3 2 1 4 2 0 0 4 cos x 4x sin x 2 3 1.511 0 2.1205 y 2 6.2993 −1 −1 x 1 −2 −1 1 2 x 2 2 1 1 47. 2 0 x3 dx 2 0 x x 2 dx x 2, y 0, x 0, x 2 48. 2 0 y y3 2 dy 2 0 y1 y dy y, x 1, y 0 (a) Plane region bounded by y (b) Revolved about the y-axis y 4 3 2 1 −1 x 1 2 3 4 (a) Plane region bounded by x (b) Revolved about the x-axis y 1 x= y (1, 1) x 1 Other answers possible Other answers possible S ection 7.3 6 1 Volume: The Shell Method 41 49. 2 0 y 2 6 y dy 6 2 y, x 0, y 0 50. 2 0 4 x e x dx e x, y 4 0, x 0, x 1 (a) Plane region bounded by x (b) Revolved around line y y 6 5 4 3 2 1 −3 −2 −1 −2 x 1 2 3 4 5 (a) Plane region bounded by y (b) Revolved about the line x y x= 6−y y = ex 3 2 1 x 1 2 3 4 Other answers possible 51. Disk Method Ry ry V r h 52. y2 x2 a2 y2 b2 y2 b2 1 1 ±b y r2 0 r x2 a2 1 b xb 1 1 x2 a2 x2 a2 x2 dx a2 b x r2 r 2y y y2 r r h dy 12 h 3r 3 h px V y a y3 3 x, h x a 22 0 r 4b a x r a a2 0 x2 x dx x2 3 32 a 0 −r 4b a 4b3 a 3a Note: If a a2 42 ab 3 b, then volume is that of a sphere. b 53. (a) Area region 0 abn ab nx ab n 1 axn dx xn 1 a n1 a 1 1 n bn n 1 n 1 n 1 1 b 0 (c) Disk Method: b V 2 0 x abn b axn dx xn 1 2a 0 xbn bn 2 x 2 bn 2 ab n 2 dx b 0 1 abn 1 abn R1 n abn 1 1 n n 1 2a 2a R2 n (d) lim R2 n n→ n→ xn 2 n2 bn 2 n2 2 nn abn b lim n n abn 2 n 2 n n 2 (b) lim R1 n n→ n→ n→ 1 1 nn b2 abn n n n 2 lim abn b n→ lim 2 1 lim b2 abn , the graph approaches the line x 1. (e) As n → 42 Chapter 7 r Applications of Integration x dx r r 54. (a) 2 0 hx 1 (ii) (b) 2 r R x 2 r2 x2 dx (v) is the volume of a right circular cone with the radius of the base as r and height h. y is the volume of a torus with the radius of its circular cross section as r and the distance from the axis of the torus to the center of its cross section as R. y x=R y=h 1− x r (0, h) ( ( (−r, 0) x y= r2 − x2 (r, 0) (r, 0) x y=− r r2 − x2 b r (c) 2 0 2x r 2 x2 dx (iii) (d) 2 0 h x d x (i) (e) 2 0 2ax 1 x2 b2 dx (iv) is the volume of a sphere with radius r. y is the volume of a right circular cylinder with a radius of r and a height of h. y is the volume of an ellipsoid with axes 2a and 2b. y y= r2 − x2 y =a (r, h) (0, a) 2 1 − x2 b (r, 0) x (b, 0) x y=− r2 − x2 x (0, −a) y = −a 2 1− x b2 4 55. (a) V 2 0 x f x dx 40 0 34 5800 50 4 10 45 2 20 40 4 30 20 0 2 20 3 121,475 cubic feet 40 20 40 1 x 2 12 x 2 25x 2 0 (b) Top line: y 50 x 0 0 40 40 x 20 0 20 40 1 x⇒y 2 2x x 20 40 1 x 2 20 ⇒ y 50 2x 80 Bottom line: y 20 V 2 0 20 x 50 d x 50x dx 20 2 2 20 2x 2x 2 40 80 d x 80x dx 2 0 2 2 x3 6 26,000 3 2 32,000 3 2x 3 3 40x 2 20 2 121,475 cubic feet (Note that Simpson’s Rule is exact for this problem.) S ection 7.3 200 Volume: The Shell Method 43 56. (a) V 2 0 x f x dx 200 0 38 4 25 19 2 50 19 4 75 17 2 100 15 4 125 14 2 150 10 4 175 6 0 2 1,366,593 cubic feet (b) d 24 0.000561x 2 0.0189x 19.39 (c) V 2 200 xd x d x 0 2 213,800 1,343,345 cubic feet −20 −6 (d) Number gallons 225 V 7.48 10,048,221 gallons 57. y 2 y1 y2 (a) V x4 x4 x 2, x 2 0≤x≤4 4 2 y 4 3 2 x 4 x x x x4 4 x x4 1 x −1 −2 −3 −4 1 2 3 4 5 6 x dx 8x 2 8x3 3 x 2 2 0 4 x3 0 16x dx 4 x4 4 4 8x 2 0 64 3 4 (b) V 4 0 x4 4 x dx x5 2 (c) V 4 0 4 4 x4 x 8x3 2 x dx x5 2 4 0 4x3 85 x 5 2 dx 4 4 0 16 x 32 3 x 3 2 dx 4 2 0 4 27 x 7 2 0 2048 35 4 16 5 x 5 2 27 x 7 8192 105 58. y 2 y1 y2 (a) V x2 x x2 x 5, 5 5 x2 x 5 5≤x≤0 xx 5 5 −6 −4 −2 4 y x2 x 0 xx 5 dx x 2 −4 x4 4 0 5x3 3 xx x 5 0 5 625 12 0 (b) V 4 x 5 5 dx dx. (c) V 4 5 5 x 5 xx x dx. 5 5u3 2u5 5 dx Let u V 4 5, du u 5 2 2 Let u V 25u1 50 3 u 3 2 0 2 5, du uu u du 2 4 0 5 u du 2 0 5 4 0 u5 27 u 7 10u3 4u5 2 du 1600 5 21 4 0 u5 27 u 7 2 du 5 5 4 2 4 2 2 0 400 5 7 44 Chapter 7 c Applications of Integration 1 x 2x 14 c 14 c 59. V1 14 c 1 dx x2 x 1 dx x 1 c 4c 1 2 2 c 1 c 2 c 4 1 4 4c c 1 y V2 V1 2 14 V2 ⇒ 4c 2 c 1 4 1 4 1 4 c x 2c c 0 0 2 c 4c2 4c 9c 1c 1 yields no volume. 4 Section 7.4 1. 0, 0 , 5, 12 (a) d 13 (b) y y s 0 Arc Length and Surfaces of Revolution 2. 1, 2 , 7, 10 0 2 3. y 7 1 2 2 32 x 3 x 1 2, 1 1 0≤x≤1 5 12 0 2 (a) d 10 (b) y y 4 x 3 4 3 7 10 2 2 y s 12 x 5 12 5 5 2 3 1 0 x dx 1 2 1 3 4 3 10 2 x 1 32 0 1 13 x 5 5 12 5 13 2 dx s 1 1 5 x 3 3 23 x 2 1 , x1 3 8 7 1 dx 2 3 8 1.219 0 4. y y s 2x3 2 3 0≤x≤9 5. y y 6. 1≤x≤8 1 1 x1 3 2 y y x4 8 13 x 2 13 x 2 b 1 4x 2 1 , 2x 3 1≤x≤2 1, 2 dx 3x1 2, 9 1 0 9x dx 9 2 1 27 2 823 27 2 s 1 8 1 dx 1 y 2 9x 32 0 12 , 2x 3 1 y 2 1 54.929 x2 3 1 dx x2 3 8 s a 2 1 3 2 x2 3 1 1 1 2 dx 3x 1 3 8 13 x 2 1 4x 2 2.063 1 dx 2x 3 2 1 3 2 23 x 23 55 32 1 14 x 8 33 16 22 8.352 S ection 7.4 x5 10 14 x 2 2 Arc Length and Surfaces of Revolution 32 x 2 x 45 7. y y 1 y 1 6x 3 1 2x 4 1 , 2x 4 1 y 14 x 2 14 x 2 2 2 8. y y s 1≤x≤2 dx 1 2x 4 2 3 4 1 ≤ x ≤ 27 1 1 x1 3 3 2 1 3, 27 1 27 1 14 x 2 b dx x2 1 3 s a 2 1 2 1 x2 27 dx 2 3x1 32 1 dx 3 2 x2 1 3 1 1 3 27 dx 1 dx 2x 4 1 6x 3 2 1 3 2 103 2 3.2458 22 x 3 3 23 2 28.794 15 x 10 779 240 9. y y 1 y 2 ln sin x , 1 cos x sin x 1 3 3 , 44 cot x csc2 x 10. y y 1 y 2 ln cos x , sin x cos x 1 tan2 x 3 0≤x≤ tan x sec2 x 3 cot2 x 4 s 4 csc x dx 3 4 4 s 0 3 sec2 x dx sec x dx 0 3 ln csc x ln 2 1 cot x ln 2 1 1.763 ln sec x ln 2 tan x 0 3 ex ex 1 1 1 ex 4e2x 2e2x 1.3170 11. y y 1 y 2 1x e 2 1x e 2 1x e 2 2 e e x 12. , 2 x y dy dx 1 dy dx 2 ln ln e x ex 1 e4x 1 1 2 1 2e x 1 ln e x 1 x 0, 2 , e x ex ex 1 1 b e2x 2e x 1 e2x e 0, 2 2 x 1 s 0 1x e 2 2 dx s 2e2x e4x 1 e2x 2 1 dy dx e e x x e2x e2x 2 1 2 ex 0 ln 3 e e x dx 12 e 2 1 e2 3.627 dx ln 2 ln 3 a 2 0 ln 3 ln 2 1 e2x dx e2x 1 1x e 2 ex ex dx ln 2 coth x dx ln 4 3 2 ln 4 3 ln 3 4 0.57536 ln 3 ln sinh x ln 2 ln 43 34 ln 16 9 46 Chapter 7 12 y 3 y y2 4 Applications of Integration 1 3 13. x dx dy s 2 2 1 32 , 0≤y≤4 14. x x yy 2 3, 3y1 1 y 2 2 1≤y≤4 12 13 y 3 11 y 2 1 1 y 4 2 y2 y2 2y 2 1 dy 4 2 dy 1 dy 1 0 4 dx dy dx dy 2 12 y4 0 4 1 y 4 2 y 2 1 y 4 y 1 1 2 1 4 y 1 y 2 y2 0 1 y 3 3 y 0 64 3 4 76 3 4 s 1 1 2 1 y 2y1 2 dy 4 1 123 y 23 1 16 23 15. (a) y 4 5 4 12 23 4.647 2 10 3 x 2, 0 ≤ x ≤ 2 (b) 1 y y 2 2x 1 2 (c) L 4x2 1 4x2 dx L 0 −1 −1 3 16. (a) y x2 x 5 2, 2≤x≤1 (b) 1 y y 2 2x 1 1 1 4x2 2 2 (c) L 4x 4x 1 4x2 dx 5.653 −3 2 L −5 17. (a) y 1 , x 2 1≤x≤3 (b) 1 y y 2 1 x2 1 3 (c) L 1 x4 1 1 dx x4 2.147 −1 4 L 1 −1 18. (a) y 1 1 5 x , 0≤x≤1 (b) 1 y y 2 1 1 1 1 x 1 1 1 2 (c) L 1.132 y= 1 x+1 −1 −2 2 x 4 L 0 1 1 x 4 dx S ection 7.4 19. (a) y sin x, 2 Arc Length and Surfaces of Revolution (c) L 3.820 47 0≤x≤ (b) 1 y y 2 cos x 1 cos 2 x 1 0 L − 2 − 0.5 cos 2 x d x 3 2 20. (a) y cos x, 2 2 ≤x≤ 2 (b) 1 y y 2 sin x 1 sin 2 x 2 (c) 3.820 y = cos x L −2 2 1 2 sin 2 x d x −2 21. (a) x y e y, ln x 0≤y≤2 (b) 1 y y 2 1 x 1 1 (c) L 1 x2 1 1 dx x2 2.221 1≥x≥e 3 2 0.135 L e 2 −1 −1 3 Alternatively, you can do all the computations with respect to y. (a) x e y, 0≤y≤2 (b) 1 dx dy dx dy 2 e 1 2 y (c) L 2y 2.221 e 1 L 0 e 2y dy 22. (a) y ln x, 2 1≤x≤5 (b) 1 y y 2 1 x 1 5 (c) L 1 x2 1 1 dx x2 4.367 −1 6 L −6 1 23. (a) y 2 arctan x, 3 0≤x≤1 (b) y L 2 1 1 x2 1 0 (c) L 4 1 x2 2 1.871 dx − 0.5 1.5 −3 48 Chapter 7 Applications of Integration dx dy 1 36 2 y 36 3 24. (a) x y 36 36 y2 , x, 10 0≤y≤3 3 3≤x≤6 (b) y2 12 2y (c) L 3.142 ! 2 y2 1 y2 36 y2 dy y2 dy L −10 10 0 3 −5 0 6 36 Alternatively, you can convert to a function of x. y y L 3 3 36 dy dx 6 x2 x 36 1 x2 x2 36 6 dx x2 3 3 6 36 x2 dx 3.142. Although this integral is undefined at x 2 0, a graphing utility still gives L 4 25. 0 1 s 5 d 5 dx x 2 1 2 dx 5 4 3 2 1 −1 y 26. (0, 5) y = 25 x +1 0 1 s 1 d tan x dx 2 dx 2 y Matches (b) y = tan x Matches (e) (2, 1) x 1 2 3 4 1 ( π , 1( 4 π 4 3π 8 x (0, 0) 27. y x 3, 0, 4 4 1 64.525 4 4 (a) d (b) d 0 0 2 2 64 1 0 0 2 2 64.125 2 1 2 8 1 2 3 2 2 27 8 2 4 3 2 64 27 2 (c) s 0 1 3x 2 2 d x 0 1 9x 4 d x 64.666 (Simpson’s Rule, n 10) (d) 64.672 28. f x (a) d (b) d x2 4 1 4 2, 0 0 2 2 0, 4 144 9 16 16 2 2 128.062 2 1 2 0 9 2 3 2 2 25 0 2 4 3 2 144 25 2 160.151 4 (c) s 0 1 4x x 2 4 2 dx 159.087 (d) 160.287 S ection 7.4 29. (a) f x 10 Arc Length and Surfaces of Revolution 49 x2 3 (b) No, f 0 is not defined. −2 −2 10 (c) 1 fx fx 2 2 x 3 1 13 4 9x2 3 9x2 3 9x2 4 3 Divide 1, 8 into two intervals. 0 1, 0 : s1 1 9x2 3 9x2 0 4 3 8 dx 4 1 x1 12 3 dx, 0, 8 : s2 0 9x2 3 9x2 8 4 3 dx x≥0 1 3 1 18 9x2 1 0 3 x≤0 dx 1 3 9x2 0 3 3 4 4 1 dx, x1 3 8 9x2 1 3 3 4 4 32 6 x1 3 1 9x2 27 1 403 27 1 403 27 32 0 1 9x2 27 13 4 27 1 8 27 s1 s2 1 403 27 1 403 27 30. x2 3 2 2 0 1 2 43 8 2 133 133 2 2 2 9.0734 1.4397 133 16 2 8 133 8 2 2 10.5131 y2 y2 3 3 4 4 4 3 4 2 1 8 0 8 31. (a) x2 x2 3 332 5 4 y y y 1 y 2 , 0≤x≤8 2 x 3 4 x2 3 13 3 2 y2 y1 y3 x 1 2 3 4 5 x2 3 1 2 4 x2 4 x2 x 23 3 4 x1 x 2312 −1 y4 1 3 −1 x 23 3 (b) y1, y2, y3, y4 4 1 s 4 dx 8 (c) y1 6x1 3 0 1, L1 0 2 dx 4 5.657 1 9x dx 16 x2 dx 4 1 5.759 5.916 6.063 2 0 dx 48 12 y2 y3 y4 3 12 x , L2 4 1 x, L 3 2 5 32 x , L4 16 4 0 Total length: s 4 12 1 0 4 0 25 3 x dx 256 50 Chapter 7 Applications of Integration 1 and L1 x dx dy e 32. Let y ln x, 1 ≤ x ≤ e, y 1 1 1 d x. x2 1 1 Equivalently, x e y, 0 ≤ y ≤ 1, e y, and L 2 0 1 e 2y d y 0 1 e 2x d x. Numerically, both integrals yield L 1 32 x 3 2.0035. 33. y 3x 1 2 2 34. y y 1 y 2 31 10 e x 20 20 e x 20 x 20 2 When x 0, y 3. Thus, the fleeing object has traveled 2 3 units when it is caught. 1x e 2 1 e 10 y 1 y 2 1 3 12 x 32 1 1 3 x 2 2 12 1x 1 2 x1 2 1 4x 1 2 1x e 4 1x e 2 20 20 2 e x 20 e 2 x 10 x 1 4x x 1 2 s 0 x1 dx 2x 1 2 x1 2 0 1 x 2 2 3 12 dx s 20 1x e 2 ex 20 20 20 2 20 e x 20 x 20 dx 1 2 32 x 23 2x 1 2 0 4 3 1 2 20 e e x 20 dx 20 The pursuer has traveled twice the distance that the fleeing object has traveled when it is caught. 10 e x 20 e 20 1 e 47 ft 4700 square feet of roofing Thus, there are 100 47 on the barn. x , 20 35. y y 1 y 2 20 cosh sinh 1 20 20 ≤ x ≤ 20 36. y y s 693.8597 68.7672 cosh 0.0100333x x 20 x sinh 2 20 cosh x dx 20 x 20 20 0.6899619478 sinh 0.0100333x 299.2239 1 299.2239 0.6899619478 sinh 0.0100333x 2 d x x cosh 2 20 20 1480 x dx 20 47.008 m (Use Simpson’s Rule with n 100 or a graphing utility.) L 20 2 0 cosh 2 20 sinh 40 sinh 1 0 37. y y 1 y 2 9 x 9 9 9 2 x2 x2 x2 9 9 x 3 x 2 2 dx 0 38. y y 1 y 2 25 x 25 25 25 4 x2 x2 x2 25 25 x 5 4 4 s 0 3 9 2 3 dx x2 arcsin 0 s 3 x 2 dx 3 5 25 4 5 x2 dx 3 5 2 3 arcsin 3 arcsin 2 3 3 arcsin 0 5 arcsin 7.8540 1 2 4 5 7.8540 5 arcsin 3 arcsin 2.1892 s S ection 7.4 x3 3 x 2, 2 0 3 Arc Length and Surfaces of Revolution 51 39. y y S 40. y 0, 3 3 2x 1 x 2 4 9 y 1 x4 dx 4x3 dx S , 9 4, 9 2x x 1 1 dx 9 32 4 2 x3 3 1 1 dx x 6 9 9 x4 1 2 3 0 4 4 1 x4 32 0 8 3 258.85 x 1 82 82 1 8 103 3 53 2 171.258 41. y y 1 y 2 x3 6 x2 2 x2 2 1 2x 1 2x 2 12 , 2x 2 2 42. y y x 2 1 2 5 , 4 2 0 1, 2 1 2x x 3 x2 2 1 dx 2x 2 1 y 2 0, 6 6 S 2 1 2 x3 6 x5 12 x2 6 S x 2 5 6 5 dx 4 x2 0 2 1 1 dx 4x 3 1 8x 2 2 1 2 8 95 2 x6 72 47 16 44. y 9 2x 3 43. y y S 3 x 2 1, 8 x 1 8 x2, 0, 3 1 , 3x 2 3 8 y 1 dx 9x 4 3 1 dx 12x 1 3 d x S 2 2 2 3 18 27 27 1 x1 0 3 4x2 dx 12 x 1 3 9x 4 3 1 8 4 6 6 199.48 1 0 4x2 4x2 2 8x dx 3 9x 4 3 1 1 1 12 1 373 32 0 8 9x 4 3 32 1 1 117.319 145 145 10 10 45. y y S sin x cos x, 2 0 46. 0, sin x 1 cos 2 x d x 1 y y y 2 ln x 1 x x2 x2 e 1 , 1, e x2 x2 1 e 14.4236 S 2 x 1 dx 2 1 x2 1 dx 22.943 52 Chapter 7 Applications of Integration 48. The precalculus formula is the distance formula between two points. The representative element is xi 2 47. A rectifiable curve is one that has a finite arc length. yi 2 1 yi xi 2 xi. 49. The precalculus formula is the surface area formula for the lateral surface of the frustum of a right circular cone. The representative element is 2 f di xi2 yi2 2 f di 1 yi xi 2 50. The surface of revolution given by f1 will be larger. r x is larger for f1. xi. 51. y y 1 y 2 hx r h r r2 r2 r 52. y y r2 x r r2 2 x2 x2 x2 r h2 r2 r2 h2 x2 2 x2 x 9 3 9 2 1 h2 y 2 r2 r2 r2 r S 2 0 x 2 r2 r 9 dx r S 2 r r r2 r dx r x2 2 rx x2 dx r 0 r2 h2 2 4 r2 r 53. y y 1 y 2 54. From Exercise 53 we have: a y x2 x2 3x dx 9 x2 2 S 2 0 a rx r2 x2 r dx h r−h x r 0 2x d x r2 x2 a S 2 0 2r 2r 2 r2 2r x2 0 x2 + y2= r2 3 0 2x 9 9 5 x2 x r r2 2 a2 a2 dx 2 2r r 2 6 6 3 2 r h where h is the height of the zone 0 14.40 See figure in Exercise 54. 1 12 x 3 1 x 6 1 2 0 13 12 55. y y 1 y 2 x3 2 3 12 x 2 1 1 x 6 12 9x 1 2 1 x 36 1 x 36 12 1 x 36 13 18 81 x x3 2 9x 2 d x 0.015 12 9x 1 2 2 S 1 12 x 3 1 3 2x 12 9 x1 2 2 dx 13 2 6 13 0 11 x 3 2 x3 2 x 12 9x1 2 dx 3 0 1 x 33 x2 3x 3 0 27 ft 2 0.1164 ft2 16.8 in. 2 Amount of glass needed: V 27 0.00015 ft 3 0.25 in. 3 S ection 7.4 x2 9 y2 4 Arc Length and Surfaces of Revolution x2 , 9 x2 9 53 56. (a) 1 2 2 1 1 x2 9 x2 9 −5 4 (b) y 5 2 2 2 1 2 1 1 0≤x≤3 12 Ellipse: y1 y2 y 2x 9 2x x2 −4 x2 + y = 1 94 91 3 2x x2 9 1 39 L 0 4x2 dx 81 9x2 (c) You cannot evaluate this definite integral, since the integrand is not defined at x 3. Simpson’s Rule will not work for the same reason. Also, the integrand does not have an elementary antiderivative. 57. (a) We approximate the volume by summing six disks of thickness 3 and circumference Ci equal to the average of the given circumferences: 6 6 V i 1 ri 2 3 i 1 2 Ci 2 2 3 65.5 2 3 4 70 6 Ci 2 i 2 1 3 4 50 2 65.5 70 2 66 2 66 2 58 2 58 2 51 2 51 2 48 2 3 57.75 2 4 67.75 2 68 2 62 2 54.5 2 49.5 2 3 21813.625 4 5207.62 cubic inches r . For the first frustum: r s (b) The lateral surface area of a frustum of a right circular cone is s R S1 32 50 2 65.5 2 65.5 9 50 2 12 50 2 50 65.5 2 2 12 65.5 2 . h Adding the six frustums together: S 50 2 70 2 58 2 224.30 1168.64 18 R 65.5 66 51 9 9 9 15.5 2 4 2 7 2 2 12 65.5 2 66 2 51 2 202.06 58 48 70 9 8 2 3 2 4.5 2 2 12 2 12 2 12 9 9 2 12 2 12 208.96 208.54 174.41 150.37 (c) r 20 0.00401 y 3 0.1416 y 2 1.232 y 7.943 (d) V 0 18 r2 dy 2 ry 0 5275.9 cubic inches 1 r y 2 dy S 1179.5 square inches −1 −1 19 54 Chapter 7 fx Applications of Integration 0.0000001953x4 400 58. (a) y 0.0001804x3 0.0496x2 3.0 acres 4.8323x 1 acre 536.9270 43,560 square feet (b) Area 0 f x dx 131,734.5 square feet (Answers will vary.) 400 (c) L 0 1 f x 2 dx 794.9 feet (Answers will vary.) b 59. (a) V 1 y 1 dx x2 b x 1 1 1 b (c) lim V b→ b→ lim 1 1 b (d) Since y= 1 x 2 x4 1 > x3 we have b x x4 x 3 1 > 0 on 1, b x 1 1 b 1 x4 1 dx > x3 b→ b 1 1 dx x b ln x 1 ln b b (b) S 2 1 b 1 x 1 x 1 1 1 x2 1 dx x4 2 and lim ln b → dx b→ b . Thus, 1 dx . lim 2 1 x4 x3 2 1 b 2 1 x4 1 dx x3 L2 (in radians): (b) Let s be the arc length of the sector, which is the circumference of the base of the cone. Here, sL 2 r, and you have S 12 L 2 1 2s L 2 L 1 Ls 2 1 L2 r 2 L L1 r2 r1 60. (a) Area of circle with radius L: A Area of sector with central angle S 2 A 2 L2 12 L 2 r L. (c) The lateral surface area of the frustum is the difference of the large cone and the small one. S r2 L r2 L L1 L 1 r2 L r2 r1 r1L 1 r1 L1 L1 ⇒ L r1 r1 r2 L L r1 L 1 r2 r1 . Hence, By similar triangles, S r2 L L r1 61. Individual project L 1 r2 r2 . 62. Essay S ection 7.4 Arc Length and Surfaces of Revolution 1 x4 12 4 4 55 63. x2 3 y2 3 4 4 4 3 4 2 1 2 0 8 64. x 23 332 y2 y x 2, 0≤x≤4 y2 3 y y 1 y 2 x2 , 0≤x≤8 2 x 3 4 x2 3 4 dx x2 3 dx 8 352 0 13 xx 12 3x 3 12 x 4 3x 48x 16 4 2 x2 4 8 312 4 x2 x1 3 312 y 1 y 2 x2 x2 3 4 4 3 1 48x S x2 332 24x 48x xx 12 x4 24 8x 4x 2 9x2 4 3x 4 3x 2 , 48x x 0 4 0 x2 3 x1 3 4 x2 32 S 192 5 2 0 4 4 4 3x dx 48x dx 12 5 2 0 4 [Surface area of portion above the x-axis] 12 12 12 w 16 0 3x 2 dx 4 16x 64 x3 0 64 64 16 3 65. y 1 h kx 2, y y 2 2kx 1 4k2x 2 h ⇒1 w2 w 66. C 2 0 700 1 1 0 4h2 2 x dx w4 4 155 2x2 dx 7004 kw2 ⇒ k y 1 1 4h2 w4 2 x2 1444.5 meters By symmetry, C y 2 0 4h2 2 x dx. w4 (w, h) h −w x w 67. Let x0, y0 be the point on the graph of y 2 makes an angle of 45 with the x-axis. y y x0 L 0 x3 where the tangent line y x3 2 (x 0 , y 0 ) 3 12 2x 4 9 49 1 y2 = x3 45 x 1 9 4x dx 8 27 (0, 0) 22 1 56 Chapter 7 Applications of Integration Section 7.5 1. W Fd 1000 ft 4. W Fd b Work 2. W Fd 2800 4 lb 3. W Fd 112 4 100 10 lb 9 2000 1 2 11,200 ft 5280 47,520,000 ft lb 448 joules (newton-meters) 5. Work equals force times distance, W FD. 6. W a F x dx is the work done by a force F moving an a to x b. object along a straight line from x 9 7. Since the work equals the area under the force function, you have c < d < a < b . 8. (a) W 0 7 6 dx 20 dx 0 54 ft 9 lbs 10x 90 dx 140 20 9. F x 5 k kx k4 5 4 7 (b) W 7 160 ft 9 lbs x3 81 9 (c) W 0 9 12 x dx 27 x dx 9 ft 0 9 0 lbs 18 ft lbs W 0 5 x dx 4 lb 52 x 8 7 0 (d) W 0 2 32 x 3 2 27 3 245 in. 8 30.625 in. 9 lb 2.55 ft lb 10. W 5 5 x dx 4 lb 52 x 8 9 11. F x 5 kx k 30 ⇒ k 50 12. F x 25 3 800 W kx k 70 ⇒ k 70 35 in. 250 W 80 7 F x dx 20 50 20 F x dx 0 70 0 25 x dx 3 cm 25x2 6 50 20 80 x dx 7 cm 40x 2 7 70 0 8750 n 28,000 n 280 Nm 87.5 joules or Nm 13. F x 20 k W 0 kx k9 20 9 12 14. F x 15 W 20 x dx 9 13 kx k1 4 k 4 2 0 15x d x 15x 2 0 240 ft lb 10 2 x 9 12 160 in. lb 0 40 ft lb 3 15. W W 18 0 7 12 kx dx 324 x d x kx2 2 162x 2 13 0 7 12 k ⇒k 18 37.125 ft 16 324 lbs 16. W W 7.5 0 5 24 kx dx 540x d x kx 2 2 270x 2 16 0 5 24 k ⇒k 72 4.21875 ft 540 lbs 13 13 16 16 Note: 4 inches 1 3 foot S ection 7.5 17. Assume that Earth has a radius of 4000 miles. Fx 5 k Fx k x2 k 4000 2 4100 Work 57 (a) W 4000 80,000,000 dx x2 tons 5.15 80,000,000 x 10 ft 9 4100 4000 487.8 mi 4300 lb 80,000,000 80,000,000 x2 80,000,000 dx x2 4000 20,000 mi ton h (b) W 4000 80,000,000 dx x2 ton 1.47 1010 ft ton 1395.3 mi 18. W h→ 80,000,000 x 2.1 1011 ft h 4000 80,000,000 h 20,000 lim W lb 19. Assume that Earth has a radius of 4000 miles. Fx 10 k Fx k x2 k 4000 2 15,000 (a) W 4000 160,000,000 dx x2 160,000,000 x 15,000 10,666.667 4000 40,000 ton ton lb 40,000 ton ton lb 29,333.333 mi 2.93 3.10 26,000 10 4 mi 160,000,000 160,000,000 x2 (b) W 4000 1011 ft 160,000,000 dx x2 160,000,000 x 26,000 6,153.846 4000 33,846.154 mi 3.38 3.57 20. Weight on surface of moon: 1 6 10 4 mi 1011 ft 12 2 tons Weight varies inversely as the square of distance from the center of the moon. Therefore: Fx 2 k W 1100 k x2 k 1100 2.42 1150 2 10 6 10 6 x2 dx 2.42 x 10 6 1150 2.42 2.42 1100 10 6 1 1100 ton 1 1150 1.01 10 9 ft lb 95.652 mi 21. Weight of each layer: 62.4 20 Distance: 4 4 y 6 5 y y 4 4 (a) W 2 4 62.4 20 4 62.4 20 4 0 y dy y dy 4992y 4992y 624y 2 2 4 2496 ft 9984 ft 0 lb lb 3 2 4−y (b) W 624y 2 1 x 1 2 3 4 5 6 58 Chapter 7 Applications of Integration 22. The bottom half had to be pumped a greater distance than the top half. 23. Volume of disk: 2 2 y 4 y y y 4 24. Volume of disk: 4 y y Weight of disk of water: 9800 4 Weight of disk: 9800 4 Distance the disk of water is moved: 5 4 Distance the disk of water is moved: y 12 W 0 5 39,200 39,200 y 9800 4 dy 5y 12 y2 4 2 0 39,200 0 5 y dy W 10 y 9800 4 dy 39,200 39,200 y2 2 22 12 10 862,400 newton–meters 470,400 newton–meters 2 y 3 2 25. Volume of disk: y 7 y 26. Volume of disk: 6−y 2 y 3 2 y 7 y Weight of disk: 62.4 Distance: 6 W 4 62.4 9 4 62.4 9 2995.2 ft y 6 2 y 3 2 y 5 4 3 2 Weight of disk: 62.4 Distance: y 2 y 3 2 y 5 4 3 2 y x x 6 0 y y2 dy 14 y 4 6 0 −4 −3 −2 −1 1 2 3 4 (a) W 4 62.4 9 4 62.4 9 2 y3 dy 0 −4 −3 −2 −1 1 2 3 4 2y 3 lb 14 y 4 6 2 110.9 ft 0 lb (b) W 4 62.4 9 4 62.4 9 y3 dy 4 14 y 4 6 7210.7 ft 4 lb 27. Volume of disk: Weight of disk: 62.4 Distance: y 6 36 36 y2 2 y y 28. Volume of each layer: y 3 3 3 3 y y y 3 y y2 Weight of each layer: 53.1 y Distance: 6 y yy 3y 3y2 2 3 dy y 2 dy y3 3 3 0 W 62.4 0 6 y 36 36y 0 y2 dy 62.4 18y 2 14 y 4 6 0 3 W 0 53.1 6 3 62.4 y3 dy lb 53.1 0 18 20,217.6 ft y 10 8 53.1 18y 53.1 117 2 4 y x −6 −4 −2 −2 2 4 6 3106.35 ft lb 4 3 2 1 y (2, 3) y = 3x − 3 x 1 2 3 4 S ection 7.5 29. Volume of layer: V Weight of layer: W 13 Distance: 2 1.5 Work y 59 lwh 42 8 42 94 94 y2 y y2 y Tractor 8 6 4 y 9 4 y 2 2 13 2 −y x W 1.5 42 8 13 2 y dy 13 336 2 1.5 1.5 9 4 1.5 y dy 1.5 2 9 4 y y dy 2 −6 −4 −2 −2 −4 2 4 6 The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius 3. Thus, the work is 2 W 336 13 2 3 2 2 1 2 12 2 2457 ft lb. 30. Volume of layer: V Weight of layer: W Distance: 2.5 25 4 25 4 y2 y y2 y Ground level 12 y 42 24 19 2 6 y 25 4 25 4 y2 19 2 y2 dy 2.5 3 −9 −6 −3 −3 −6 3 19 2 −y x 6 9 W 2.5 42 24 1008 19 2 2.5 2.5 y dy 2.5 25 4 y2 y dy The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. 5 The first integral represents the area of a semicircle of radius 2. Thus, the work is W 1008 19 2 5 2 2 1 2 29,925 ft lb 94,012.16 ft lb. 31. Weight of section of chain: 3 y Distance: 15 15 y y dy 15 32. The lower 10 feet of chain are raised 5 feet with a constant force. W1 3 10 5 150 ft lb W 3 0 15 3 15 2 y lb The top 5 feet will be raised with variable force. Weight of section: 3 y Distance: 5 5 2 0 y 5 y dy 150 3 5 2 75 2 5 337.5 ft W2 W 3 0 y 2 0 75 ft 2 lb lb W1 W2 375 ft 2 33. The lower 5 feet of chain are raised 10 feet with a constant force. W1 3 5 10 150 ft lb 34. The work required to lift the chain is 337.5 ft lb (from Exercise 31). The work required to lift the 500-pound load is W 500 15 7500. The work required to lift the chain with a 100-pound load attached is W 337.5 7500 7837.5 ft lbs. The top 10 feet of chain are raised with a variable force. Weight per section: 3 y Distance: 10 10 y 10 y dy 300 ft lb 3 10 2 10 W2 W 3 0 y 2 0 150 ft lb W1 W2 60 Chapter 7 Applications of Integration 6 35. Weight of section of chain: 3 y Distance: 15 7.5 36. W 3 0 12 2 2y d y 108 ft lb 3 12 4 6 2y 2 0 2y 2y d y 3 15 4 3 15 4 2 7.5 W 3 0 15 2y 2 0 3 12 4 168.75 ft lb 37. Work to pull up the ball: W1 500 15 7500 ft lb 38. Work to pull up the ball: W1 500 40 20,000 ft lb Work to wind up the top 15 feet of cable: force is variable Weight per section: 1 y Distance: 15 15 Work to pull up the cable: force is variable Weight per section: 1 y Distance: 40 x x dx lb W2 20,000 800 20,800 ft lb 1 40 2 40 x x dx lb 1 15 2 15 40 W2 0 15 112.5 ft x 2 0 W2 0 40 800 ft x 2 0 Work to lift the lower 25 feet of cable with a constant force: W3 W 1 25 15 W1 W2 375 ft W3 lb lb 112.5 375 W W1 7500 7987.5 ft 39. p 1000 k W k V k 2 2000 3 2 40. p 2500 W k V k ⇒k 1 3 1 41. F x 2500 W k 2 1 2 x 2 k 2 k x 1 2 dx k1 1 4 2000 dV V 3 2 2500 dV V 3 2 x 2 2500 ln V 1 3k units of work 4 2000 ln V 2000 ln 3 2 8000 810.93 ft lb 2500 ln 3 2746.53 ft lb 42. (a) W (b) W (c) F x 25,000 FD 2 16,000 ft 2 22,000 lbs 4 15,000 157,738.64x 2 2 10,000 104,386.36x 4 5000 32.4675 0 24,88.889 ft lb 20 0 36 4 20,000 16,261.36x 4 85,295.45x 3 0 0 2 (d) F x (e) W 0 0 when x 2 0.524 feet. F x is a maximum when x 25,180.5 ft lbs 0.524 feet. F x dx S ection 7.6 5 Moments, Centers of Mass, and Centroids 4 61 43. W 0 1000 1.8 ln x 1 dx 3249.44 ft lb 44. W 0 ex 1 dx 100 2 11,494 ft lb 5 2 45. W 0 100x 125 x3 dx 10,330.3 ft lb 46. W 0 1000 sinh x d x 2762.2 ft lb Section 7.6 1. x 6 5 6 Moments, Centers of Mass, and Centroids 31 53 35 6 7 1 15 11 5 1 4 12 5 1 18 2. x 7 3 7 12 6 4 4 2 3 35 8 6 6 2 3 86 17 11 30 11 11 8 3. x 17 18 1 7 12 5 6 1 1 12 1 8 12 4. x 14 12 1 0 5. (a) x (b) x 5 15 5 18 3 5 17 11 8 12 3 5 99 33 3 3 3 623 30 12 1 6 3 11 6. The center of mass is translated k units as well. 7. 50x 50x 125x x 75 L 750 750 6 feet 52 5 52 5 10 , 9 1 1 11 1 1 9 2 x 75x 75 10 x 8. 200x 200x 750x x 550 5 2750 2750 32 feet 3 10 1 10 10 0, 0 1 10 x (Person on left) 550x 9. x y x, y 3 3 3 3 31 4 10 9 1 9 y 10. x y x, y 25 2 5 5 4 0 0 y 8 6 25 50 25 m2 1 (− 3, 1) −3 −2 −1 −1 −2 −3 −4 1 2 m1 (2, 2) x 3 4 m2 (5, 5) m3 (− 4, 0) −4 −2 2 x m3 (1, − 4) −2 m1 4 (1, −1) 6 11. x y x, y 3 3 2 3 45 27 342 45 34 10 1 6 6 60 3 5 8 13 16 m5 (− 3, 0) y 6 m2 (5, 5) m4 (0, 0) 2 4 21 10 216 m3 (7, 1) x 6 8 5 13 , 8 16 −4 −2 −2 m1 (−2, −3) 62 Chapter 7 12 2 12 3 Applications of Integration 6 12 1 6 15 2 6 15 2 15 2 15 15 2 8 15 2 15 15 2 93 40.5 96 40.5 62 27 8 12. x y x, y y 65 12 6 64 27 m3 (6, 8) 6 m2 (− 1, 5) 2 −2 −2 62 64 , 27 27 m1 (2, 3) x 2 m4 (2, − 2) 6 8 4 13. m 0 4 x dx x 2 4 4 23 x 3 x dx 4 2 0 16 3 4 y Mx 0 x2 4 3 4 2 52 x 5 12 5 4 0 4 4 0 3 2 y My Mx m 3 16 1 ( x, y ) x 1 2 3 4 x x dx 0 64 5 x x, y My m 64 5 3 16 12 3 , 54 2 14. m 0 2 12 x dx 2 112 x 22 45 43 x3 6 2 0 4 3 2 2 y Mx 0 12 x dx 2 3 5 1 2 3 2 2 8 x 4 dx 0 40 x5 0 32 40 4 5 2 1 y My Mx m 2 0 ( 3, 3) 25 x 1 2 1 x x2 dx 2 2 43 2 x3 dx 0 8 x4 0 2 x x, y My m 33 , 25 1 15. m 0 1 x2 x2 2 x3 dx x3 12 x2 12 35 x3 3 x3 dx x4 4 2 1 y 0 1 12 1 (1, 1) Mx 0 x4 0 x6 dx x5 25 x7 7 1 0 35 3 4 1 2 y My Mx m 1 ( x, y ) 1 4 35 x x2 0 1 x3 dx 0 x3 x4 dx x4 4 x5 5 1 0 x 20 1 4 1 2 3 4 1 x x, y My m 12 20 3 5 3 12 , 5 35 S ection 7.6 1 Moments, Centers of Mass, and Centroids 63 16. m 0 1 x x 2 6 12 1 x dx x 1 2 x 2 32 x 3 x dx 2 x2 2 1 1 y 0 6 1 Mx 0 x 0 x dx 2 x2 22 x3 3 1 0 12 3 4 1 2 y My Mx m x 0 ( x, y ) 1 4 1 x 6 15 x dx 0 x3 2 x2 dx 2 52 x 5 x3 3 1 0 x 15 1 4 1 2 3 4 1 x x, y My m 21 , 52 3 2 5 17. m 0 3 x2 Mx 0 3 4x x2 2 4x 2 2 x 2 dx x 2 x2 x3 3 4x 3 3x 2 2 2 3 0 9 2 x 2 dx 11x 2 12x d x y 6 5 4 (3, 5) ( x, y ) 3 2 x2 0 5x 4 x2 3 3x d x 99 5 2 x4 0 8x 3 2 1 x5 25 y My 0 2x 4 99 5 2 9 x2 27 4 2 9 11x 3 3 22 5 6x 2 0 −1 x 1 2 3 4 5 Mx m 3 3 x My m 4x 3 2 2 x 2 dx 0 x3 3x 2 dx x4 4 3 x3 0 27 4 x x, y 3 22 , 25 9 18. m 0 9 x x 1 1 1 x 3 1 3x 2 13 x 3 9 2 0 9 1 1 dx 0 x 1 2x 27 9 1 x dx 3 9 23 x 3 x 0 2 x2 6 1 x 3 9 18 0 27 2 1 x dx 3 9 2 Mx 0 9 x 1 x 3 1 dx 2 9 2 x 2 x 0 13 x 3 x3 27 x 81 5 92 2 2 12 x 9 2 27 2 2 x dx 3 36 45 4 2 0 1 x 3 12 x 9 2 x dx x2 26 9 43 x 3 1 My 0 x My m 18 5 , 52 1 x 3 18 ;y 5 1 dx 0 x3 45 4 92 2 12 x dx 3 5 2 25 x 5 2 13 x 9 9 0 y 5 486 5 81 81 5 x x, y Mx m 4 3 2 1 −2 (9, 4) (18, 5 ) 52 (0, 1) x 2 4 6 8 10 −1 64 Chapter 7 8 Applications of Integration 3 53 x 5 8 0 19. m 0 8 x 2 3 dx x2 3 2 3 x dx 2 192 7 8 96 5 6 y Mx 0 3 73 x 27 10 7 3 83 x 8 5 8 8 0 192 7 4 2 y My Mx m 5 96 ( x, y ) x 2 4 6 8 x x2 3 dx 0 96 0 −2 x x, y My m 5, 96 10 7 5 96 8 20. m 2 0 4 x2 3 dx 2 0. 4x 3 53 x 5 8 0 128 5 12 8 y By symmetry, My and x 8 Mx y x, y 2 0 4 5 128 x2 3 4 2 20 7 x2 3 dx 16x 3 73 x 7 8 0 512 7 −8 −4 −4 ( x, y ) 512 7 0, 20 7 x 4 8 2 21. m My x 2 0 2 4 4 2 256 15 y2 dy y2 3 32 4 2 y2 8 5 0. 4y dy y3 3 2 0 32 3 2 y 2 0 16y 83 y 3 y5 5 2 0 256 15 1 ( x, y ) x 1 −1 −2 2 3 My m By symmetry, Mx and y x, y 8 ,0 5 2 22. m 0 2 2y 2y 2 y2 dy y2 2y 2 5 y2 dy 1 y2 y2 dy y3 3 2 0 4 3 y4 y5 5 2 0 y My 0 4y 3 23 8 15 2 ( x, y ) 1 x Mx My m 2 83 15 4 y 2y 0 2y 3 3 y4 4 2 0 4 3 x 1 2 y x, y Mx m 2 ,1 5 43 34 S ection 7.6 3 Moments, Centers of Mass, and Centroids 65 23. m 0 3 2y 2y y2 y2 2 4y 3 y dy y 2y 3y 2 2 y2 y4 y3 3 3 0 9 2 3 y (−3,3) 3 My 0 3 y dy 3 2 y 0 y2 3y y2 dy ( x, y ) 1 x 1 −1 2 x Mx My m y4 0 3y 2 d y 3 5 y5 25 y3 0 27 10 −3 −2 −1 27 2 10 9 3 3 y 2y 0 y2 2 9 3 2 y dy 0 3y 2 y3 dy y3 y4 4 3 0 27 4 y x, y Mx m 27 4 33 , 52 2 24. m 1 2 y y 2 2 2 2 2 9 2 y2 2 y2 dy y2 y 2 y2 2 y2 y 2 2y dy 2 3 3 y3 3 2 1 9 2 3 2 1 y My 1 2 2 x Mx My m y 1 y4 dy 8 5 y2 dy y3 dy 1 2 y5 5 2 1 36 5 ( x, y ) x 1 2 3 4 36 5 2 −1 yy 1 2 2y 1 y2 y3 3 y4 4 2 1 9 4 y x, y Mx m 81 , 52 1 92 49 25. A 0 x 1 2 1 1 x2 dx x2 x4 dx x3 dx 12 x 2 1 x3 23 x3 3 x3 3 1 0 1 0 1 6 11 23 1 3 1 4 1 5 1 12 1 15 4 26. A 1 1 dx x 4 1 4 ln x 1 ln 4 1 x 4 4 1 Mx My 0 x5 5 x4 4 1 0 Mx My 1 2 4 1 dx x2 1 dx x 1 2 x 1 1 8 1 2 3 8 x2 0 x 1 3 3 3 27. A 0 2x 1 2 3 4 dx 4 2 dx x2 3 4x 0 9 8x 36 3 12 8 dx 24 18 78 21 Mx 2x 0 2x 2 0 3 2x 3 3 3 4x 2 8x 0 18 2x 3 3 My 0 2x 2 4x d x 2x 2 0 18 36 66 Chapter 7 2 Applications of Integration 2 28. A 2 x2 1 2 2 4 dx 44 8x 3 3 2 0 4 x2 dx 1 2 32 5 2 8x x4 8x 2 32 2x 3 3 2 16 0 16 3 32 3 Mx x2 2 x 2 dx 2 16 d x 256 15 2 1 x5 25 My 16x 2 64 3 0 by symmetry. 5 29. m 0 5 10x 125 10x 125 2 5 x3 dx x3 1033.0 10x 125 3,124,375 24 10 3 5 400 Mx 0 x3 dx −1 6 − 50 50 0 5 x 2 125 10x 2 125 0 x3 dx x3 dx 130,208 125 x3 3x 2 dx My x y My m Mx m 0 12,500 5 9 3105.6 3.0 126.0 Therefore, the centroid is 3.0, 126.0 . 4 20 30. m 0 4 xe xe x2 dx xe 2.3760 x2 31. m 20 20 5 3 400 x2 dx x2 1239.76 x2 dx 20064.27 50 x2 Mx 0 4 2 x 2e x dx Mx 25 2 y 3 5 3 400 2 20 20 5 3 400 23 2 My dx dx 0.7619 5.1732 400 20 x2 dx 0 4 x 2e 0 x2 Mx m 16.18 x y My m Mx m 2.2 0.3 −1 x 0 by symmetry. Therefore, the centroid is 0, 16.2 . − 25 −5 5 25 Therefore, the centroid is 2.2, 0.3 . 2 −1 32. m 2 2 8 x2 4 dx 6.2832 8 x2 4 2 3 Mx 2 8 1 2 x2 4 0.8 dx 32 2 1 x2 4 2 dx 5.14149 −3 −1 3 y x Mx m 0 by symmetry. Therefore, the centroid is 0, 0.8 . S ection 7.6 1 2a c 2 1 ac 11 ac 2 1 2ac c 0 c 0 Moments, Centers of Mass, and Centroids 67 33. A 1 A x ac 34. A 1 A bh 1 ac 11 ac 2 1 2ac ac c 0 c 0 b c a 2 y a b c a 2 y a dy x b y c 2ab y c 2 a a2 dy c b y c 2 dy 4ab y c 4ab 2 y dy c2 4ab 3 y 3c 2 a c 0 1 2ab 2 y 2ac c y 1 ac 1 ac c 12 abc 2ac 3 b c 2 c c b 3 a dy y x, y 1 ab 2 y 2ac c 1 abc 2ac 1 ac b 2 c a 2y 0 y 0 c b c 2a y c y3 c 3c 0 y a 2a dy a y y a 2c b y c a 1 b 2 a b y c dy 1 y2 c2 c 0 y 0 0 y2 dy c y 0 c 2 2 c2 x, y y2 c 3 y ac , 2 bc , 33 y= c b + a (x + a) This is the point of intersection of the diagonals. y (b, c) y= c b − a (x − a ) y= c x b (b, c) ( x, y ) x From elementary geometry, b 3, c 3 is the point of intersection of the medians. (a + b, c) ( x, y ) (a, 0) (0, 0) (a, 0) y = c (x − a ) b x (− a, 0) 35. A 1 A x c a 2 2 ca 2 ca 2 ca b b c b b 2 x 0 b c a x a dx ac 2 2 2 ca 2 ca b 1 ca b a 2x b c 0 b c a x2 ax dx 3ac 2 2 ca a b b b a x3 c3 ax 2 2 c 0 b 3 c 0 a c2 b c a c 2 2bc 2 c 0 c 0 2ac 2 6 b c 1 ca 2 c 2b 3a a 2b c 3a b a2 dx a a 2c y y ca 1 ca 1 b2 b a x3 3 2 x a dx a 2 x2 b b 2a b a x c a 2c 3 ac b b 2a b a x 2 c 2 ac a2 a2 3ab ab 3a b 2 1 3c a b 1 3a Thus, x, y b b b2 2 2ab 2ab 3ac b 3a 2 b2 . a 3a 2 3a c a2 ab b 2 3a b (0, a) y= b−ax +a c a ( x, y ) ( c, b) (0, 0) x a 2b c , 3a b b (c, 0) The one line passes through 0, a 2 and c, b 2 . It’s equation is y passes through 0, b and c, a b . It’s equation is y a c 2b x b 2c a x a . The other line 2 b. x, y is the point of intersection of these two lines. 68 36. x Chapter 7 Applications of Integration 37. x 0 by symmetry. A 1 A r 0 by symmetry. A 1 A y 12 r 2 2 r2 21 r2 2 r2 r 1 ab 2 2 ab 21 ab 2 1 b2 ab a2 a a x2 2 dx 4r 3 y y b a a 2x 2 a2 x3 3 a x2 dx b 4a3 a3 3 y 1 r 2x r2 x, y 0, 4r 3 x3 3 r r 1 4r3 r2 3 a 4b 3 x, y 0, 4b 3 r b r x −a a x 1 38. A 0 1 3 1 2x x2 dx 1 3 1 A x y 1 3 0 1 x1 1 2x 2x 2 4x 2 x2 dx x2 1 3 0 x 2x 2 x3 dx 3 2 1 3 1 0 x2 2 2x 1 0 23 x 3 x2 2 x4 4 dx 1 0 1 4 3 0 2x x4 dx x2 dx 3 x 2 43 x 3 3 2 x, y 1 1 0 4x 3 x4 x5 5 7 10 17 , 4 10 b 39. (a) y (e) Mx b b b b2 2 x2 b 2 x4 dx x2 dx x5 5 b b y=b b 12 bx 2 4b2 b 5 bx 4 3 b 5 bb 3 x3 3 b2 b −5 − 4 −3 −2 −1 x 12345 b2 b 5 b x2 dx bb 2 3 4b2 b 5 4b b 3 b b b A (b) x (c) M y b 0 by symmetry. b b xb x2 dx 0 because bx x 3 is odd. b than below. 2 y bb Mx A b (d) y > since there is more area above y 2 S ection 7.6 Moments, Centers of Mass, and Centroids b 2 69 40. (a) M y My 0 by symmetry. 2n (b) y > 0 (d) b xb 2n x 2n d x 1 b because there is more area above y 2 than below. n y 1 3 5b b because bx 2n x 2n b is an odd function. 2n 2 5 9b 3 7 13 b 4 9 17 b b b (c) M x 2n x2n b 2 x 4n 1 4n 1 b 4n 4n x2n 2n dx 12 b 2n b2 b x 4n dx (e) lim y n→ n→ 12 bx 2 b 2 b 1 2n 2n b 2n lim 2n 4n 1 b 1 1 b 2 b (f) As n → 1 2n , the figure gets narrower. y 1 2n 1 4n b 4n 4n 1 2 bx 1 2n y = x 2n b A 2n b b x 2n d x b 2n 2n 1 2n 1 2n x 2n 1 2n 1 2n b 0 y=b 2b y Mx A b 1 2n 4n b 4n 4n b 24n 1 4n 2n 4n b 2n 2n 1 1 1 2n 4n 1 2n x − 2n b 2n b 1 b 1 41. (a) x 0 by symmetry. 40 A Mx 2 0 40 40 f x dx fx 2 2 2 40 30 34 40 30 2 34 72,160 5560 4 29 4 29 12.98 2 2 26 2 26 2 4 20 4 20 0 2 20 278 3 0 5560 3 72,160 3 dx 10 7216 3 y x, y (b) y (c) y x, y Mx A Mx A 72,160 3 5560 3 0, 12.98 1.02 10 5 x4 0.0019 x 2 12.85 29.28 (Use nine data points.) 23,697.68 1843.54 0, 12.85 42. Let f x be the top curve, given by l x f d 0 2.0 0.50 0.5 1.93 0.48 1.0 1.73 0.43 1.5 1.32 0.33 d. The bottom curve is d x . 2.0 0 0 —CONTINUED— 70 Chapter 7 Applications of Integration 42. —CONTINUED— 2 (a) Area 2 0 fx 2 1.50 34 d x dx 4 1.45 4.62 dx fx 2 (b) f x dx 2 1.30 4 .99 0 (c) y x, y d x dx Mx A 0.1061x 4 0.02648x 4 4.9133 4.59998 0, 1.068 0.06126x 2 0.01497x 2 1.068 1.9527 .4862 2 1 13.86 3 2 Mx 2 2 fx 2 fx 2 dx dx 3 0 2 3.75 34 1 29.878 6 y x, y Mx A 4 3.4945 4.9797 1.078 2 2.808 4 1.6335 0 d −2 0 f 2 4.9797 4.62 0, 1.078 y 2 1 x 1 −1 3 43. Centroids of the given regions: 1, 0 and 3, 0 Area: A x y x, y 4 41 4 40 4 4 4 3 3 0 4 4 0 1.88, 0 3 −2 ,0 44. Centroids of the given regions: Area: A x y x, y 3 2 2 22 7 21 2 7 7 13 1 7 , , 2, , and , 1 22 2 2 4 3 y 31 2 33 2 25 15 , 14 14 27 2 21 25 2 7 15 2 7 25 14 15 14 2 1 m1 m2 m3 x 1 2 3 4 45. Centroids of the given regions: 0, Area: A x y x, y 15 15 0 15 3 2 135 34 12 7 12 0 34 12 5 34 0, 3.97 34 70 3 15 , 0, 5 , and 0, 2 2 7 6 5 y 0 135 34 4 3 2 1 −4 −3 −2 −1 x 1 2 3 4 7 15 2 0, S ection 7.6 7 2 8 7 6 8 7 ,P 41 7 8 7 16 0, 55 16 Moments, Centers of Mass, and Centroids 71 46. m1 m2 0, y 287 ,P 64 2 0. 5 4 3 m2 By symmetry, x y x, y 2 7 4 7 16 74 0, 789 304 287 64 55 16 287 64 0, 2.595 16,569 6384 789 304 m1 3 −2 −1 1 −2 x 1 2 1 3 2 47. Centroids of the given regions: 1, 0 and 3, 0 Mass: 4 x y x, y 2 41 4 0 2 2 3 ,0 2.22, 0 2 2 3 2 2 3 48. Centroids of the given regions: 3, 0 and 1, 0 Mass: 8 y x x, y 0 81 8 8 8 2 rA 3 3 8 8 3 ,0 1.56, 0 49. r A V 5 is distance between center of circle and y-axis. 4 2 16 is area of circle. Hence, 2 rA 1 44 2 11 82 y 8 3 2 8 8 3 128 3 4 x 3 6 4 50. V 2 34 24 2 2 5 16 160 2 1579.14. 51. A y r V 8 4 y 4 0 x4 x dx 1 16x 16 x3 3 4 0 8 3 3 ( x, y ) 2 1 x 1 2 3 4 2 rA 134.04 6 6 52. A 2 6 2x x2 x 2 2 dx 2 dx u u du 2 32 2 32 3 6 y My Let u My 2 0 2 2 xx dx: 4 2 dx 4 (6, 4) x 4 2, x u 2 32 3 2, du 2 0 2 u3 2 2u1 2 du 2 25 u 5 2 43 u 3 4 2 0 2 ( x, y ) x 4 6 2 x r V My A x 64 5 704 15 22 5 704 15 32 3 22 5 2 22 5 2 rA 32 3 1408 15 294.89 72 Chapter 7 m1 m1x1 m1 y1 My ,y m ... Applications of Integration mn mn xn mn yn 54. A planar lamina is a thin flat plate of constant density. The center of mass x, y is the balancing point on the lamina. 53. m My Mx x ... ... Mx m 55 6 , 18 5 6 5 6, 55. (a) Yes. x, y (b) Yes. x, y (c) Yes. x, y (d) No 2 2, 5 18 5 18 5 41 6 , 18 17 5 6 , 18 56. Let R be a region in a plane and let L be a line such that L does not intersect the interior of R. If r is the distance between the centroid of R and L, then the volume V of the solid of revolution formed by revolving R about L is V 2 rA where A is the area of R. 57. The surface area of the sphere is S 4 r 2. The arc length of C is s r. The distance traveled by the y centroid is d S s 4 r2 r 4 r. 58. The centroid of the circle is 1, 0 . The distance traveled by the centroid is 2 . The arc length of the circle is also 2 . Therefore, S 22 4 2. y 2 This distance is also the circumference of the circle of radius y. d 2y r 1 C d x 1 3 (0, y) −r r x −1 −1 −2 Thus, 2 y 4 r and we have y 2r . Therefore, the centroid of the semicircle y r 2 x 2 is 0, 2 r . 1 59. A 0 xn dx A 1 xn 1 n1 1 dx 2 1 0 1 n 1 m Mx My n xn 2 0 1 60. Let T be the shaded triangle with vertices 1, 4 , 1, 4 , and 0, 3 . Let U be the large triangle with vertices 4, 4 , 4, 4 , and 0, 0 . V consists of the region U minus the region T. 1 0 2 x 2n 1 2n 1 xn 2 n2 1 0 Centroid of T: 0, 131 ; Area 2 2n n 2 1 8 Centroid of U: 0, 3 ; Area 1 16 x xn dx 0 Area: V x 15 y 16 1 15 0 by symmetry. 1 11 3 x y My m Mx m n n n n 1 2 n 4n 1 2 1 4 16 117 3 13 5 8 3 15 y n1 2 2n 1 1n , 2 4n 1 2 y x, y y 7 6 Centroid: 0, 153 As n → , x, y → 1, . The graph approaches the x-axis and the line x 1 as n → . y (− 1, 4) V T (1, 4) (4, 4) (0, 3) (− 4, 4) 1 2 x y=xn (1, 1) −4 −3 −2 −1 (0, 0) 3 4 x 1 Section 7.7 Fluid Pressure and Fluid Force 73 Section 7.7 1. F 3. F PA 62.4 h 62.4 2 6 5. h y Ly F 3 4 3 Fluid Pressure and Fluid Force 936 lb 62.4 h 6 2. F 4. F PA 62.4 h 62.4 5 16 4 48 4992 lb 62.4 h 48 62.4 5 3 26 748.8 lb 6. h y Ly 3 62.4 4 48 3 4 y 3 3 11,980.8 lb y 4 y 4 y y 62.4 0 3 3 3 0 y 4 dy y dy 2 F 1 x 1 2 3 4 62.4 0 3 3 y 3y 4 y dy 3 y2 dy y3 3 3 −2 −1 2 1 x 1 2 249.6 249.6 3y 1123.2 lb y2 2 3 0 4 62.4 3 0 4 3y 2 62.4 3 2 374.4 lb 0 Force is one-third that of Exercise 5. 7. h y Ly F 3 2 y 3 3 y 1 3 0 3 8. h y Ly y y 3 y2 3 y3 3 9 0 2 −1 y 24 0 y2 y2 4 0 F 1 dy 62.4 2 y2 dy 332.8 lb 2 2 62.4 124.8 0 62.4 dy 4 y y 1 3 2 4 3 y2 32 124.8 3y 748.8 lb x 1 1 x −2 −1 1 2 −3 9. h y Ly F 4 2y y 10. h y Ly 4 y 4 3 62.4 3 9 0 y2 y 4 3 9 3 0 2 62.4 0 4 4 4y1 2 0 y y dy F y3 2 dy y 9 y2 y2 dy 12 124.8 62.4 2y 5 2 5 4 3 8y 3 2 124.8 3 1064.96 lb 2 3 0 2y d y y 0 62.4 1 x −2 −1 1 2 4 9 9 y2 32 3 x 748.8 lb −1 −1 −2 1 −4 74 Chapter 7 4 2 2 Applications of Integration y 11. h y Ly F y 3 9800 0 24 y2 y dy 2 9800 8y 117,600 newtons 0 −2 −1 1 2 x 12. h y L1 y L2 y F 1 2y 32 y 1+ 3 2 32 32 2 3 2 y lower part y 3 22 23 2 2 9800 0 upper part 1 32 y3 3 9 3 0 3 y y dy 3 22 22 1 3 2y 18y 32 y3 3 y3 2 62 2 1 y y dy 3 3 2 −3 −2 −1 x 1 2 3 19,600 19,600 y2 2 3 2y 1 3 22 92 2 4 2 4 1 44,100 3 2 13. h y Ly F 12 6 9800 0 2 newtons 14. h y Ly 2y dy 3 2y 3 9 9 0 y y 2y 3 9 6 1 y 6 y 5 F 9800 0 16 y2 2 y dy 5 0 x −3 −2 −1 1 2 3 12 y6 7y 2 9800 6y 9800 72y 171,500 newtons 2,381,400 newtons 9 6 3 x −3 3 6 9 15. h y Ly F 2 10 y 4 3 y 2 140.7 0 2 2 2 0 y 10 d y x 1407 1407 2y y dy y2 2 2 −6 −4 −2 −1 −2 2 4 6 2814 lb 0 S ection 7.7 16. h y Ly F 2 140.7 3 Fluid Pressure and Fluid Force 75 y 4 3 0 2 y 9 y2 −2 2 x y2 0 4 3 9 9 y2 y2 32 y2 dy 2y d y 0 3 −4 −6 140.7 4 3 140.7 4 3 3376.8 lb 17. h y Ly F 4 6 4 3 2 9 3 y 5 y 18. h y Ly 5 y 5 y 3 0 3 140.7 0 4 4 4 0 y 6 dy 1 F x 1 2 3 140.7 3 0 y5 5y 3 5 y dy 3 52 y dy 3 1 y 844.2 844.2 4y y dy y2 2 4 −3 −2 −1 −1 140.7 140.7 6753.6 lb 0 52 y 2 15 53 y 9 0 3 −1 −2 2 3 4 6 x 45 140.7 2 1055.25 lb 19. h y Ly F 42 32 −3 −4 −5 (5, −3) y 2 1 2 0 2 y 9 y 4y 2 1 x 9 4y 2 4y 2 d y 12 −2 −1 −1 −2 1 2 42 8 21 4 3 2 2 42 0 9 32 8y d y 0 2 9 3 4y 2 32 32 94.5 lb 20. h y Ly F y 1 2 32 32 9 3 2 4y 2 32 y 9 4y 2 d y 63 32 9 4y 2 d y 21 4 32 9 32 4y 2 8y d y The second integral is zero since it is an odd function and the limits of integration are symmetric to the origin. 3 The first integral is twice the area of a semicircle of radius 2. 9 4y 2 2 94 9 4 y2 141.75 445.32 lb. Thus, the force is 63 76 Chapter 7 k y Applications of Integration y 21. h y Ly F w 2 r2 r y2 y r2 r2 y2 2 dy −r water level r k r r r x r w 2k r y2 dy r r2 y2 2y d y −r The second integral is zero since its integrand is odd and the limits of integration are symmetric to the origin. The first integral is the area of a semicircle with radius r. F w 2k r2 2 0 wk r 2 22. (a) F (b) F 23. h y Ly F w k wk r2 wk r y 2 62.4 7 62.4 5 22 3 2 1747.2 lbs 2808 lbs y b h2 water level k h 2 k h2 y b dy y2 2 h2 −b 2 b 2 −h 2 x wb k y wb hk h2 wkhb 24. (a) F wkhb 62.4 11 2 25. From Exercise 23: 35 5148 lbs F 64 15 1 1 960 lb 26. From Exercise 21: F 64 15 12 2 753.98 lb (b) F wkhb 62.4 17 2 5 10 26,520 lbs 27. h y F 4 4 y 4 0 62.4 y L y dy 8 we have: 23 5 4 2.5 8 22 9 4 1.5 10 2 1 10.25 4 0.5 10.5 0 Using Simpson’s Rule with n F 62.4 40 0 38 4 3.5 3 3010.8 lb 28. h y 3 y 5x 2 x 2 y. 2 y 5 4 Solving y x Ly F 4y 5 2 62.4 2 4 for x, you obtain 4y 5 3 1 y 3 y y 4y 5 y 5 y y dy dy 546.265 lb x −3 −2 −1 −1 1 2 3 0 3 2 124.8 0 3 R eview Exercises for Chapter 7 29. h y Ly F 12 2 42 4 77 y 3 y y2 332 10 8 62.4 0 2 12 y 42 3 y2 332 dy 6 4 6448.73 lb x −6 −4 −2 2 4 6 30. h y Ly F 12 2 62.4 0 y 7 16 2 4 y y2 7 16 7 16 y 16 y2 10 8 6 12 4 y 12 y2 dy x 62.4 7 0 y2 dy 21373.7 lb −6 −4 −2 2 4 6 31. (a) If the fluid force is one-half of 1123.2 lb, and the height of the water is b, then hy Ly F b 4 b (b) The pressure increases with increasing depth. y 62.4 0 b b y 4 dy b y dy y2 2 b 1 1123.2 2 2.25 2.25 0 by b2 0 b2 2 b2 2.25 4.5 ⇒ b 2.12 ft. d 32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body. 33. F Fw w c h y L y dy, see page 508. 34. The left window experiences the greater fluid force because its centroid is lower. Review Exercises for Chapter 7 5 1. A 1 y 1 dx x2 1 x 5 1 4 5 5 2. A 12 4 4x 1 x 1 dx x2 5 12 y 6 5 1 ( 2 , 4) 81 5 (5, 4) 1 (1, 1) 3 2 5, , 2 3 4 1 25 x 1 1 2 3 4 1 ( 5, 25 ) 6 x (1, 0) (5, 0) 78 Chapter 7 1 Applications of Integration 1 3. A 1 1 x2 1 1 dx y 4. A 0 1 y2 y2 0 1 2y 2y 1 2 dy 1 dy 1 dy y arctan x 1 2 4 4 2 1, 1 2 1 1, 1 (−1, 0) 1 2 x y 0 3 2 1 (1, 0) y 3 1 31 0 1 3 −2 (−1, 1) 1 1 2 3 −2 1 −2 x 1 2 5. A 2 0 x 12 x 2 x3 dx 14 x 4 1 0 1 y 6. A 1 2 y 2 1 3 y 12 y 2 y2 y2 dy 13 y 3 2 1 1 dy 2 1 2 (1, 1) x 1 (0, 0) 1 1 2y 9 2 ( 1, 1) y 3 (5, 2) 2 1 x 2 −1 −2 3 4 5 (2, −1) 2 2 7. A 0 e2 xe 2 e2 y ex dx 2 8. A 2 6 2 csc x d x 2 ex 0 2 2x 2 2 y 3 ln csc x 0 cot x 6 1 3 3 ln 2 3 1.555 (0, e 2) 6 4 (2, e 2) 2 3 ln 2 (0, 1) x 1 1 2 3 (π , 2) 6 1 ( 56π, 2) π 6 π 3 π 2 2π 3 5π 6 x R eview Exercises for Chapter 7 5 4 5 3 3 5 4 4 7 3 79 9. A 4 sin x cos x 1 2 4 2 cos x d x sin x 10. A y 2 1 2 sin y cos y d y 5 5 3 3 3 cos y y 2 7 5 3 3 1 dy 2 sin y 1 2 22 1 2 1 2 3 23 y 3 y 2 2 )π ) 4 , 1 2 π 4 ( 21 , 73π ) ( 21 , 53π ) 1 ( 2, π ) 3 x −2 −1 x 2 ) 8 5π , − 1 4 2 ) 11. A 0 8 3 16x 0 8x x2 x2 8x 3 dx 12. Point of intersection is given by: x3 x2 0.783 4x 3 3 4x 0⇒x x2 0.783. 2x 2 d x 23 x 3 8 0 A 512 3 170.667 0 x3 dx 14 x 4 0.783 0 8x 2 20 3x (8, 3) 2x 2 13 x 3 1.189 −4 (0, 3) 10 4 − 16 −1 − 2 (.7832, .4804) 3 13. y A 1 1 x 1 2 2 14. A x dx 2 2 0 2 2x 2 4x 2 0 x4 x4 dx 15 x 5 2 0 2x 2 d x 0 1 2 x dx 2 1 6 0.1667 (−2, 8) 1 0 2x 1 4 32 x 3 2 43 x 3 10 128 15 8.5333 x 2 12 x 2 1 0 (2, 8) (0, 1) −1 −4 4 −2 (1, 0) −1 2 (0, 0) 80 15. x A Chapter 7 y2 0 Applications of Integration 1 1 y 1 1 2 2y ⇒ x 1 1 0 ⇒y x 1± x 1 16. y y A 0 x x 2 2 1⇒x 1 ⇒x 1 1 1 32 y2 2y y2 x 2 1 x 4 1 1 1 x 1 dx y2 y2 dy 2 0 1 dx 2x 1 2 2y 5 1 dy dx 5 A 0 2 0 2y 0 2y dy 3 y x 1 (0, 2) 1 y 2 x 3 1 2 1 4 3 y2 4 3 13 y 3 (0, 0) −2 −1 −1 x 2 3 (5, 2) 2 1 x (1, 0) 2 −1 −2 3 4 5 2 17. A 0 2 0 1 x dx 2 1 3 x 2 3 2 3 y dx 2 1 x 2 dx 3 2 x dx 2y (0, 1) (3, 1) y y A 1 x 1 x ⇒x 2 2⇒x y 2 2 y 2 32 y 2 5 1 0 (2, 0) x 2, y 1 1 3 2y d y 3 2 y 0 1 3y d y 0 1 18. A 0 2 dx 1 2 x 1 dx 5 4 3 x A y2 2 1 y2 1 dy (5, 2) 0 1 13 y 3 2 y 0 14 3 (1, 0) x 2 3 4 5 19. Job 1 is better. The salary for Job 1 is greater than the salary for Job 2 for all the years except the first and 10th years. 20. (a) y 80 6.8335 1.2235 t 6.8335 e0.2017t (b) R2 5 6.83e0.2 t 20 Difference: 15 R2 y dt 12.06 billion dollars 5 0 12 R eview Exercises for Chapter 7 21. (a) Disk 4 81 (b) Shell x2 dx 0 V y 4 3 x3 4 3 0 64 3 4 V y 4 3 2 0 x2 dx 23 x 3 4 0 128 3 2 1 x 1 2 3 4 2 1 x 1 2 3 4 (c) Shell 4 (d) Shell 4 V 2 0 4 4 4x 0 x x dx x2 dx x3 3 4 0 V 2 0 4 6 6x 0 x x dx x2 dx 13 x 3 4 0 2 2 y 2 64 3 y 5 2x 2 2 3x 2 160 3 4 4 3 3 2 2 1 1 1 x 1 2 3 −1 2 3 4 5 x 22. (a) Shell 2 2 (c) Disk 2 2 V y 4 3 2 0 y3 dy 2 y4 0 8 V 0 y 4 3 y4 dy 5 y5 0 32 5 2 1 x 1 2 3 4 2 1 x 1 2 3 4 (b) Shell 2 (d) Disk 2 V 2 0 2 2 2y 2 0 y y2 dy 4 y V 0 2 y2 y4 0 1 2 12 d y 5 4 3 y 2 2 8 3 y3 dy 14 y 4 2 0 3 2y 2 d y 23 y 3 2 23 y 3 1 x 1 2 3 4 15 y 5 176 15 2 0 1 x 1 2 3 4 82 Chapter 7 Applications of Integration (b) Disk 4 23. (a) Shell V 4 0 x 3 y 4 3 4 4 16 2 16 3 x2 dx 4 V 64 0 2 0 3 4 2 16 x3 3 4 x2 48 0 dx 1 2 x2 32 9 8 16x y 4 2 1 −3 −2 −1 −2 −4 x 1 2 3 −3 −2 −1 2 1 x 1 2 3 −2 −4 24. (a) Shell a (b) Disk 0 V 4 2b a b x a a a2 a2 x2 x2 x2 dx 12 a V 2x d x 2 0 b2 2 a a2 x2 dx 13 x 3 a 0 0 a 32 0 2 b2 2 ax a2 4 ab2 3 4b 2 a 3a 42 ab 3 y 2 x2 + y =1 a2 b2 (0, b) y 2 x2 + y =1 a2 b2 (0, b) x (a, 0) x (a, 0) 25. Shell 1 y 26. Disk 1 y V 2 0 1 0 x x4 2x x2 2 1 dx dx 1 V 2 0 1 1 1 0 2 3 x2 dx 2 1 1 2 arctan x x 1 −2 −1 −1 x 1 2 arctan x 2 0 2 2 2 4 0 4 0 4 2 R eview Exercises for Chapter 7 6 83 27. Shell: V u x dx V u2 2u du 6 2 2 x 1 x 2 y dx 3 2 1 x 2 2 x 1 −1 2 3 4 5 6 2 2 2 x 1 u3 1 x 2u du u 12 u 2 2 2 −2 dx 4 4 0 2 u2 1 u 2 2u du u 3 3 1 4 20 3 u du 9 ln 3 42.359 −3 4 0 u2 0 4 13 u 3 3u 3 ln 1 u 0 0 28. Disk 1 29. Since y ≤ 0, A 1 xx 1 d x. V 0 1 e e 0 x2 dx 1 u x dx A e 2x 0 x u du 1 1 2x dx 1 2 2 2e 2 y 2 1 e2 1 1 u 0 1 u du 0 u3 2 4 15 u1 2 du 2 52 u 5 y 2 32 u 3 1 0 1 −1 x 1 x −1 30. (a) Disk 0 (b) Shell x2 x 1 0 V 1 dx x dx 0 1 y u x u2 x 1 1 −1 y x x 1 3 2 dx V 2u du 0 x 4 4 x3 3 2 1 1 x2 x u2 0 1 1 dx −1 12 4 x 1 2 u 2du 2u 4 25 u 5 u2 du 13 u 3 1 0 −1 4 0 u6 17 u 7 4 −1 32 105 84 Chapter 7 Applications of Integration V 1 V 4 y0 31. From Exercise 23(a) we have: 64 ft 3 16 16 1 9 Disk: 1 9 16 9 39 y0 y2 dy y2 dy 13 y 3 27 y0 3 9 3 9y 9y0 13 y 30 y03 9 27 9 0 1.042 1.958 ft. 4 54 x 5 x1 4 1 1 u 2u 4 27y0 By Newton’s Method, y0 1 bh 2 3 a2 a 1.042 and the depth of the gasoline is 3 32. A x 1 2 a2 2 x2 a2 a x2 3 a2 x2 33. fx fx V 3 3 4a3 3 x2 dx 3 a 2x x3 3 a a 1 fx 2 x x 1 2 u x dx 1 du 3 Since 4 3 a 3 3 a 3 10, we have a 3 1.630 meters. 5 3 2. Thus, s 53 2 1 0 3 x dx 2 1 uu 1 du 2 1 u3 2 2 52 u 5 u1 2 du 2 32 u 3 3 1 2 a 2− x 2 2 a 2− x 2 2 4 32 u 3u 15 3 5 1 2 a 2− x 2 8 1 15 63 6.076 34. y y 1 y 2 x3 6 12 x 2 12 x 2 3 1 2x 1 2x 2 1 2x 2 2 35. y y s 300 cosh x 2000 280, 2000 ≤ x ≤ 2000 x 3 sinh 20 2000 2000 1 2000 3 x sinh 20 2000 400 9 sinh2 2 dx s 1 12 x 2 1 dx 2x 2 13 x 6 1 2x 3 1 14 3 1 20 2000 2000 x dx 2000 4018.2 ft by Simpson’s Rule or graphing utility 36. Since f x tan x has f x sec 2 x, this integral represents the length of the graph of tan x from x This length is a little over 1 unit. Answers (b). 0 to x 4. R eview Exercises for Chapter 7 3 x 4 y 1 y 2 85 37. y 38. y 3 4 25 16 4 2x y 1 x 1 2 0 1 y 2 1 x 3 x x 2x 1 x x 3 S y 4 2 0 3 x 4 25 dx 16 15 8 x2 2 4 S 15 1 3 dx 56 3 4 0 x 1 dx 0 4 (4, 3) 2 x 3 1 32 0 3 2 1 x 1 2 3 4 39. F 4 F W kx k1 4x 5 5 40. F 50 F W kx k9 ⇒ k 50 x 9 9 50 9 4x d x 0 2x 2 0 50 in. lb 4.167 ft lb 0 50 x dx 9 lb 25 2 x 9 9 0 225 in. 1 3 2 18.75 ft lb 41. Volume of disk: Weight of disk: 62.4 Distance: 175 W 62.4 9 150 y 1 3 2 42. We know that y dV dt V 4 gal min 12 gal min 7.481 gal ft3 r 2h dh 9 dt 9 dV dt 9 3.064 t 8 7.481 150 49 minutes 3.064 ft min. 1 h 9 8 ft3 min 7.481 y 175 y dy 163.4 ft 62.4 9 ton 175y y2 2 150 0 0 dV dt dh dt 104,000 ft lb Depth of water: Time to drain well: t 49 12 150 3.064 588 gallons pumped Volume of water pumped in Exercise 41: 391.7 gallons 391.7 52 x Work 588 x 588 52 391.7 78 ft 78 ton 86 Chapter 7 Applications of Integration 43. Weight of section of chain: 5 x Distance moved: 10 10 x x dx 5 10 2 x 2 10 W 5 0 10 250 ft 0 lb 44. (a) Weight of section of cable: 4 x Distance: 200 200 x 200 W 4 0 200 x dx 2 200 x 2 0 80,000 ft lb 40 ft lb ton 30 ft ton (b) Work to move 300 pounds 200 feet vertically: 200 300 Total work work for drawing up the cable 40 ft b 60,000 ft work of lifting the load ton b ton 30 ft ton 70 ft 45. W a 4 F x dx ax2 dx 0 46. ax3 3 4 0 W a F x dx 2 9x 4 3x 9 80 a 64 a 3 Fx W 0 6, 16, 6 dx 0≤x≤9 9 ≤ x ≤ 12 12 9 3 80 64 15 4 3.75 2 x 9 12 x 9 9 54 lbs a 0 4 x 3 16x 9 16 d x 12 9 6x 0 22 x 3 96 192 54 144 51 ft a a 47. A 0 a 1 A x y 6 a2 6 a2 a x 2 dx 0 a 2 ax 1 2 x dx ax 4 3 ax 3 2 12 x 2 a2 6 x 0 a a a 0 x 2 dx 4 6 a2 dx 6ax a ax 0 2 ax 3 2 x2 dx a 5 a y 61 a2 2 3 a2 a x ( x, y ) a2 0 4a 3 2x 1 2 8 32 32 ax 3 4a 1 2x 3 2 8 12 52 ax 5 x2 dx 13 x 3 a 0 a x 32 ax a2 x, y aa , 55 3ax 2 a 5 R eview Exercises for Chapter 7 3 87 48. A 1 2x 3 32 3 32 3 3 x2 dx x2 3x 13 x 3 3 1 32 3 1 A x y x 2x 1 3 3 2x x 2 dx 3 43 x 3 2 3 32 3 3x 1 2x 2 3 x 3 dx 12 x 4x2 3 32 x 32 2 x 4 dx 23 x 3 14 x 4 3 1 1 31 32 2 3 9x 64 x4 dx 15 x 5 3 1 1 3 64 17 5 9 1 6x 2 x, y y 1, 17 5 9 6 ( x, y ) 3 −3 x 3 6 8 49. By symmetry, x 1 0. x2 dx 2 a 2x x3 3 a 0 50. 4a 3 3 A 0 x2 3 5 16 5 16 8 1 x dx 2 3 53 x 5 12 x 4 8 0 16 5 A 1 A y 2 0 a2 1 A x 3 4a 3 31 4a 3 2 6 8a 3 a a x x2 3 0 1 x dx 2 13 x 6 8 0 a2 a x2 2 dx x4 dx 15 x 5 15 a 5 a 0 5 3 83 x 16 8 y 51 16 2 15 2 16 x, y y 6 10 3 12 x dx 4 a4 0 2a 2x 2 2a 2 3 x 3 25 a 3 8 x4 3 0 6 a 4x 8a 3 6 a5 8a 3 x, y 0, y 3 73 x 7 13 x 12 8 0 40 21 2a 2 5 10 40 , 3 21 2a 2 5 4 2 a2 ( x, y ) x 2 4 6 8 ( x, y ) −2 −a a x 88 51. y Chapter 7 Applications of Integration y 4 1 y= 6x+1 0 by symmetry. 3 2 1 For the trapezoid: m My 0 6 0 (6, 2) 46 6 16 1 x 6 12 x 3 1 2x d x 18 1 x 6 x3 9 1 x2 0 x x dx 6 −1 −2 −3 −4 1 2 3 4 5 7 (6, −2) 60 For the semicircle: m My 6 1 2 8 2 x 2 2 8 4 x u 6 2 4 3 4 6 2 4 x 6 2 dx 6, u u2 du 16 3 2 6 x4 x 8, u 6 2 dx 2. Let u My x 2 6, then x 2 6 and dx u2 du 2 du. When x 2 0. When x 2 u 0 2 0 u4 2 4 2 12 0 4 44 3 u2 du 9 2 Thus, we have: x 18 2 1 2 u2 32 0 12 12 60 x 180 44 3 44 3 29 3 9 9 29 49 ,0 . 9 53. Let D surface of liquid; d 1 29 3 49 9 The centroid of the blade is 52. Wall at shallow end: 5 weight per cubic volume. g y dy d F 62.4 0 y 20 d y 1248 y2 2 5 15,600 lb 0 F c d D y fy Wall at deep end: 10 10 Dfy c g y dy c yfy d g y dy F 62.4 0 y 20 d y 624 y 2 0 62,400 lb d Side wall: 5 5 yfy fy g y dy D c d g y dy g y dy c F1 F2 F y 20 15 10 5 62.4 0 5 y 40 d y 10 0 1248 y 2 0 5 31,200 lb Area D 80y 0 fy c y 62.4 F1 F2 y 8y d y 72,800 lb 62.4 8y 2 d y y Area depth of centroid D d g c f ( x, y ) x −5 x 5 10 15 20 25 30 35 40 45 P roblem Solving for Chapter 7 54. F 62.4 16 5 4992 lb 89 Problem Solving for Chapter 7 1. T R 0 12 cc 2 c 13 c 2 x2 dx cx2 2 x3 3 c 0 1 2. R 0 x1 x dx x2 2 x3 3 1 0 1 2 1 3 1 6 cx c3 2 c3 3 c3 6 Let c, mc be the intersection of the line and the parabola. Then, mc 11 26 1 12 c1 1 0 c ⇒m m 1 c or c 1 m. T lim c→0 R 13 c 2 lim c→0 1 c3 6 3 x x2 2 1 2 1 61 1 1 1 2 1 1 m m m m m 1 2 13 2 2 2 x2 m x2 2 mx dx 1 0 m x3 3 m 2 1 3 41 41 2m m 3 m m 3 1 2 6m 1 6m m 2 m 2 6 2 m 3 1 2 13 m 1 0.2063 3. (a) 1 V 2 1 2 0 1 1 41 y2 2 2 1 y2 1 4 y2 2 dy y2 1 y2 dy 4 0 1 y2 41 8 0 1 2 y2 dy 2 Integral represents 1 4 area of circle 4 R± 2 2 8 (b) x 1 V 2 R 2 r 4 y2 R ⇒V r2 ⇒ x r2 y2 r2 R y2 r2 y2 2 dy 0 r 4R r2 0 y2 dy 2 r2 R 4R V 2 12 r 4 2 r2R 90 Chapter 7 Applications of Integration r 4. 8y2 y ± x2 1 x y 0.5 0.25 x2 1 x2 5. V 22 r2 h2 4 x r2 22 r 3 h3 8 x2 + y2 = r2 r x2 dx 22 2 4 3 x x2 32 r2 h2 4 h3 which does not depend on r! 6 −1.5 − 0.5 − 0.25 − 0.5 0.5 1.5 r h 2 For x > 0, y 1 1 2x2 2 2 1 x2 x 1 1 2x2 2 2 1 x2 2 2 r2 − h 4 r S 22 0 dx 52 3 6. By the Theorem of Pappus, V 2 rA 2 d 1 2 w2 l2 lw 7. (a) Tangent at A: y y x3, y 1 y 3x 3x 3x2 1 2 x3 3x 0 0⇒B 2, 8 2 (b) Tangent at A a, a3 : y a3 y 3a2 x 3a2x 2a3 2a 2a, 0 0 8a3 2a 16a3 16a3 x 4a a 2a 3 To find point B: x3 x 3a2x a 2 To find point B: x3 x Tangent at B: y y 1 x3, y 8 y To find point C: x3 x 1 2 x 3x x 2 2 ⇒B Tangent at B: y 8a3 y 12a2 x 12a2x 12a2x 2a 2 3x2 12 x 12x 2 16 x3 12x 16 4 2 dx x3 dx area S area R 12x 0 0⇒C 27 4 108 16 4, 64 Area of S 2a To find point C: x3 x 16 a 0 0 ⇒C Area of R 2a 4a 4a, 64a3 3a2x 16a3 2a3 dx x3 dx 27 4 a 4 108a4 x3 12a2x 2 2 x Area of R 2 4 x3 12x 2 3x 16 Area of S 16 area of R Area of S Area of S 16 area of R P roblem Solving for Chapter 7 8. f x 2 91 ex ex 2 2 fx fx f0 fx 2e x C 2 0⇒C 2e x x 2 2 9. s x (a) s x (b) ds ds (c) s x 2 1 ds dx 1 1 x f t 2 dt 1 fx 2 f x 2 dx fx 1 2 dx 31 t 2 2 1 2 dy dx x 2 dx 1 2 dx 2 dy 2 2 dt 1 1 2 9 t dt 4 22 27 1 to x 22 2. 13 27 13 2.0858 (d) s 2 1 1 9 t dt 4 8 1 27 9 t 4 32 2 1 This is the length of the curve y x3 2 from x 10. Let f be the density of the fluid and iceberg. The buoyant force is 0 0 the density of the 11. (a) y My 0 by symmetry 6 x 1 6 F fg h A y dy m x y 3 1 x3 1 dx x3 12 7 1 x3 1 x2 dx 6 1 62 1 x2 dx 2 1 x 6 1 5 3 where A y is a typical cross section and g is the acceleration due to gravity. The weight of the object is L h 2 1 35 36 12 ,0 7 W F fg h 0 f 0g h A y dy. 53 35 36 x, y W 0 L h 2 A y dy 0g h A y dy 0.92 1.03 103 103 0.893 or 89.3% 1 x −1 −2 −3 2 3 4 5 6 submerged volume total volume b (b) m My x (c) lim x b→ 2 1 6 1 dx x3 1 dx x2 1b 1 b2 b2 b2 2b b 1 1 2b b 1 2 2b b 2, 0 1 2 1 2b b2 b→ x, y x, y ,0 lim 2b b 1 92 Chapter 7 Applications of Integration y 12. (a) y My m x (b) My m x 0 by symmetry 6 2 1 6 x 1 dx x4 6 2 1 1 dx x3 35 36 −1 3 2 1 y= 1 x4 2 1 1 dx x4 215 324 63 43 b2 b2 2 b3 1 3b3 3b b 1 2 b2 b 1 3 ,0 2 6 1 160 0 28 160 0 38 0.5x 80, p 12 2 1 2 2 50 4 50 7 1 2 2 54 2 54 x, y 1 x, y 63 ,0 43 x −1 −2 −3 2 3 4 5 35 36 215 324 b y=−1 x4 2 1 b 1 dx x3 1 dx x4 2 1 b2 2 b3 3 2 1 b2 1 3b3 x, y 3b b 1 ,0 2 b2 b 1 b→ lim x 13. (a) W (b) W area area 2 3 4 1 14. (a) Trapezoidal: Area (b) Simpson’s: Area 2 82 4 82 2 82 2 82 2 73 4 73 2 75 2 75 2 80 4 80 0 0 9920 sq ft 10,4131 sq ft 3 15. Point of equilibrium: 50 x P0, x0 10, 80 80 0.125x 10 Consumer surplus 0 80 50 10 0 0.5x 0.125x dx 10 dx 400 1600 Producer surplus 16. Point of equilibrium: 1000 x P0, x0 840, 20 20 0.4x2 840 42x 20, p Consumer surplus 0 20 1000 840 0 0.4x2 42x dx 840 dx 8400 2133.33 Producer surplus P roblem Solving for Chapter 7 17. We use Exercise 23, Section 7.7, which gives F Wall at shallow end From Exercise 23: F Wall at deep end From Exercise 23: F Side wall From Exercise 23: F1 4 93 wkhb for a rectangle plate. 62.4 2 4 20 9984 lb 62.4 4 8 20 39,936 lb 20 15 y 62.4 2 4 40 8 y 10y d y y2 dy 19,968 lb y=8 10 5 x = 40 F2 62.4 0 4 624 0 8y 624 4y 2 y3 3 4 5 10 15 20 25 40 45 1 y = 10 x x 0 26,624 lb Total force: F1 18. (a) Answers will vary. f1 x f2 x 6x 2 sin x2 x F2 46,592 lb (b) f1 arc length f2 arc length 3.2490 3.3655 (c) See the article by Professor Larson Riddle at http://ecademy.agnesscott.edu/lriddle/arc/contest.htm One such function is f3 x 8 x x2 arc length 2.9195 ...
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This note was uploaded on 11/13/2010 for the course MATH MAT 231 taught by Professor Thurber during the Spring '08 term at Thomas Edison State.

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