# ODD06 - PART I CHAPTER 6 Applications of Integration...

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Unformatted text preview: PART I CHAPTER 6 Applications of Integration Section 6.1 Section 6.2 Section 6.3 Section 6.4 Section 6.5 Section 6.6 Section 6.7 Area of a Region Between Two Curves . . . . . . . . . . .2 Volume: The Disk Method . . . . . . . . . . . . . . . . . . 9 Volume: The Shell Method . . . . . . . . . . . . . . . . . 17 Arc Length and Surfaces of Revolution . . . . . . . . . . . 22 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Moments, Centers of Mass, and Centroids . . . . . . . . . 30 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 37 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 CHAPTER 6 Applications of Integration Section 6.1 6 Area of a Region Between Two Curves 6 Solutions to Odd-Numbered Exercises 1. A 0 0 x2 6x dx 0 x2 6x d x 3 3 3. A 0 0 x2 2x 3 x2 0 4x 3 dx 0 2x 2 1 6x d x 5. A 2 1 3 x3 x dx 6 1 x3 x dx or 6 0 x3 x dx 4 7. 0 y 5 4 3 x 1 x dx 2 6 9. 0 y 6 5 42 x3 x dx 6 3 11. 3 2 y 3 sec x d x 3 2 1 2 1 2π 3 π 3 π −1 3 2π 3 x x 1 2 3 4 5 1 2 3 4 5 6 x 2 13. f x gx A 4 x x 1 1 2 15. A 0 2 0 13 x 2 13 x 2 x2 2 4 2 2 x x 0 x 1 dx 2 1 dx Matches (d) y 3 2 x4 8 (3, 4) 16 8 y 6 5 2 0 2 (2, 6) (0, 1) 4 x 1 2 3 3 (0, 2) 1 −2 (2, 3) (0, 1) x 1 3 4 2 S ection 6.1 17. The points of intersection are given by: x2 xx 4 Area of a Region Between Two Curves 3 19. The points of intersection are given by: x2 2x 2x 2 4x 4 gx 0 0 when x f x dx 4x d x 0, 4 A x 1 1 3x 3 1, 2 0 when x A 0 gx 1 2 f x dx 3 x x2 x2 dx x3 3 2 1 4 x2 0 3x 1 2 2x 1 dx x3 3 32 3 y 4 2x2 0 1 2 2x x2 2 y x 9 2 (0, 0) 1 −1 −2 −3 −4 −5 2 3 (4, 0) 5 10 8 6 4 (2, 9) ( 1, 0) x 4 3 2 1 2 21. The points of intersection are given by: x x 1 23. The points of intersection are given by: x x 1 2 1 2 0 x and x x y 0 and 2 0 2y 0 2 3x 1 3x x 1 0, 3 x when x g x dx 1 x dx x2 2 3 0 3 A y dy y2 0 1 A 0 3 fx 3x 0 3 Note that if we integrate with respect to x, we need two integrals. Also, note that the region is a triangle. y x 1 dx 3x 3 12 0 2 2 3x 9 (1, 1) (2, 0) x y 32 3 2 1 (0, 0) 2 3 5 4 3 2 (3, 4) (0, 1) −2 x 1 2 3 4 25. The points of intersection are given by: y2 y 2 y y 2 1, 2 3 2 (4, 2) 2y gy 1 2 1 0 when y 1 x 1 2 3 4 5 A f y dy 2 y2 dy y3 3 2 1 1 (1, 1) y 1 3 2y y2 2 9 2 4 Chapter 6 2 Applications of Integration 10 ⇒x x 10 27. A 1 2 fy y2 1 g y dy 1 2 29. y A 10 y 0 dy 6 2 10 dy y 10 2 y3 3 y y 1 10 ln y 10 ln 10 10 ln 5 ln 2 16.0944 3 (0, 2) 1 (5, 2) y x 12 2 3 4 5 6 (0, 10) 8 6 4 (1, 10) (0, −1) 2 (2, −1) (0, 2) −4 −2 (5, 2) x 2 4 6 8 31. The points of intersection are given by: x3 xx 1 3x 2 1x fx 3x 3 x2 0 when x 3 11 0, 1, 3 f x dx 3 −6 (3, 9) A 0 1 g x dx 1 gx x2 dx (0, 0) −1 (1, 1) 12 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 32 x 2 1 0 x4 4 43 x 3 2.667 37 12 Numerical Approximation: 0.417 3.083 33. The points of intersection are given by: x 2 9 4x 2x x 4 3 4 4x 3 4x x 2 (0, 3) (4, 3) 12 0 when x x2 8x d x 4 0, 4 4x 3 dx −6 −3 A 0 4 3 2x 2 0 x2 2x 3 3 4x 2 0 64 3 Numerical Approximation: 21.333 S ection 6.1 35. f x x4 4x 2, g x x2 4 Area of a Region Between Two Curves 5 2 The points of intersection are given by: x4 x4 x2 5x 2 4 x2 4x 2 4 1 x2 0 0 when x ± 2, ± 1 −4 (− 2, 0) (2, 0) 4 4 (−1, − 3) −5 (1, − 3) By symmetry, 1 2 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 8.0 5x 3 3 5 3 4 4x 0 2 2 32 5 4x 1 1 5 5 3 4 8. Numerical Approximation: 5.067 2.933 37. The points of intersection are given by: 1 1 x4 x2 x2 x2 2 1 x 1 39. 3 1 x3 ≤ 1 x 2 2 on 0, 2 x2 2 0 0 ±1 −3 1 2 Numerical approximation: 1.759 2 (−1, ( ( 1, 1 ( 2 3 A 0 1 x 2 5 2 1 x3 dx 1.759 2 x2 −1 (0, 2) (2, 3) (0, 1) 5 A 2 0 1 fx 1 1 g x dx x2 dx 2 x3 6 2 1 0 −4 −1 2 0 x2 2 arctan x 2 1 6 4 1 3 1.237 Numerical Approximation: 1.237 3 41. A 2 0 3 fx 2 sin x 0 g x dx 4 y g 2 2 21 tan x d x 3 3 2 1 π , 3 3 f 2 cos x ln 2 ln cos x 0 π 2 (0, 0) π 2 x 0.614 3 4 π , 3 3 6 Chapter 6 2 Applications of Integration 1 43. A 0 2 2 2 0 cos x cos x dx 2 cos x dx 45. A 0 xe 1 e 2 x2 0 dx 1 1 x2 0 1 1 2 1 e 0.316 2x y 3 sin x 0 4 12.566 y 1 (0, 1) 2 g 1 (1, e ) (2π, 1) (0, 0) 1 x f −1 π 2 π 2π x 3 47. A 0 2 sin x 2 cos x 3 sin 2x 1 cos 2x 2 0 dx 4.0 0 49. A 1 1 1x e x2 3 0 dx e e1 3 e1 x 1 4 1.323 0 0 (0, 0) ( π, 0) 0 0 6 51. (a) y 3 x3 4 x 4 ,y 0, x d x, 3 (b) A 0 x3 x No, it cannot be evaluated by hand. (c) 4.7721 6 −1 −1 4 x 53. 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 6.1 2 2 2 1 Area of a Region Between Two Curves 7 55. F 1 cos 2 d sin sin 2 2 2 y (a) F 1 0 y (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 θ 1 −2 c 57. A 0 c 0 b c a y c a2 y 2c a y a b y dy c y a dy c y= cx b (b, c) y= c b − a (x − a ) ay 0 x (0, 0) ac 2 59. f x fx x3 3x 2 ac ac 2 1 base height 2 (a, 0) y 8 6 At 1, 1 , f 1 Tangent line: y 1 3x 3. −4 −3 −2 4 2 1 y = 3x − 2 (1, 1) x 2 3 4 1 or y 3x 2 x 3 at x x4 4 2. 3x 2 2 1 f (x) = x3 −6 The tangent line intersects f x 1 (− 2, − 8) −8 A 2 x3 3x 2 dx 2x 2 27 4 61. The variable is y. 63. x 4 A 1 1 2x 2 1 1≤1 1 x2 x2 x4 dx x 2 on x4 1, 1 2x 2 1 dx 2 y (0, 1) 1 x x3 3 x5 5 1 1 4 15 2x 2 1≤1 x 2 on 1, 1 . ( 1, 0) (1, 0) You can use a single integral because x 4 65. Offer 2 is better because the accumulated salary (area under the curve) is larger. 8 Chapter 6 3 Applications of Integration 67. 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 n 69. lim →0 i xi 1 x i2 x 0.6 y where xi 1 i and x n x2 dx 1 is the same as n x2 2 x3 3 1 0 0.4 0.2 f (x) x x2 (1, 0) x x 0 1 . 6 5 0.2 0.4 0.6 0.8 1.0 (0, 0) 5 71. 0 7.21 0.58t 7.21 0.45t d t 0 0.13t d t 0.13 t 2 2 5 \$1.625 billion 0 73. (a) y1 460 275.0675 1.0537 t 275.0675 e0.0523t] (b) y2 460 239.9407 1.0417 t 239.9407 e0.0408t 0 240 10 0 240 10 15 (c) 10 y1 y2 dt 649.5 billion dollars (d) No, model y1 > y2 forever because 1.0537 > 1.0417. No, these models are not accurate. According to news reports, E > R eventually. S ection 6.2 Volume: The Disk Method y 9 75. 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 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 2 2 1 0 x2 4 4± 2 16 2 1 x2 4 2 0 16 2±2 2 ⇒ x 2 22 Total area 8 0 x dx 2 2 8x x3 12 1 3 x2 2 16 42 3 5.5 5 8 0.4379 3.503 5 77. (a) A 2 0 1 5 32 x dx 5 5 5.5 1 0 dx 10 5 9 6.031 m 2 5000 12.062 60,310 pounds 2 (b) V 79. True 2A x 2 5 9 2 6.031 x x 0 5 25 5.5 5 12.062 m 3 (c) 5000 V 81. False. Let f x b x and g x 2 2x x 0 x2. f and g intersect at 1, 1 , the midpoint of 0, 2 . But 2x x2 dx 2 3 0. fx a g x dx Section 6.2 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 4 4 3. V 1 1 x 2 dx 1 x dx 1 x2 2 x4 0 4 1 15 2 x5 5 x7 7 1 0 5. V 0 x2 x2 ⇒ x 4 2 x3 y 2 dx x6 dx 2 35 x2 3 7. y V 9. y 4 ⇒x 1 y3 2 1 y 0 2 dy 0 y dy V 0 y3 2 2 dy 0 y3 dy y4 4 1 0 4 y2 2 4 8 0 10 11. y Chapter 6 x, y Applications of Integration 0, x x, r x 4 0 2 (a) R x V 0 4 (b) R y V 4, r y 2 y2 y4 dy 15 y 5 2 0 4 x x dx 0 y dx 4 16 0 2 x2 0 8 y 3 2 1 x 16y 128 5 3 2 1 1 −1 2 3 4 −1 x 1 2 3 4 (c) R y V 4 2 y 2, r y 4 y 22 0 dy y4 dy 15 y 5 2 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 16 0 8y 2 83 y 3 2 32 0 12y 2 4y 3 16y y 256 15 y 32y 192 5 3 4 2 1 1 x 1 −1 2 3 −1 −2 1 2 3 4 5 x 3 2 13. y x2, y 4x 4x 2 x 2 intersect at 0, 0 and 2, 4 . x2 r x 4x x2 2 (a) R x V x2 x4 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 S ection 6.2 Volume: The Disk Method 1 1 1 1 1 1 x 1 1 1 32.485 x x 2 11 15. R x V 4 3 x, r x 4 x 2 1 1 2 17. R x dx 4, r x 3 4 x 2 0 3 V 0 3 0 42 8 1 x 4 dx dx 3 0 x2 0 8x 4x2 15 d x 3 x3 3 y 5 15x 0 18 8 ln 1 8 ln 4 8 ln 4 y x 1 4 3 4 3 2 1 x 1 2 3 4 −1 3 2 1 −1 −1 x 1 2 3 4 19. R y V 6 4 y, r y 6 y 2 dy 12y 6y 2 0 5 4 3 2 y 0 4 y2 0 36 d y 4 1 x y3 3 208 3 1 −1 2 3 4 5 36y 0 21. R y V 6 2 y 2, r y 6 y2 2 2 2 2 23. R x dy 3 1 x x 1 x 1 1 1 , rx 2 0 dx 2 V 0 3 1 dx 3 2 2 0 y4 y5 5 12y 2 4y 3 32 dy 2 0 2 384 5 32y 0 y ln x 1 0 ln 4 2 1 x 1 −1 2 3 12 Chapter 6 1 , rx x 4 Applications of Integration 25. R x V 0 dx 27. R x V e x, r x 1 0 dx 1 1 x 1 x 2 e 0 1 x2 4 1 0 e e 2x dx 1 3 4 y 2 y 1 2 x 1 −1 2 3 1 2x 0 2 2 2 1 e 1.358 x 1 2 2 3 29. V 0 2 5 4x3 0 2x x2 8x2 2 x2 20x 1 2 dx 2 3 x2 4x3 2 1 8x2 10x2 2 5 20x 24x 2 2x 24 dx 3 x2 2 dx 10 8 6 y 24 dx 2 (2, 5) x4 152 3 83 x 3 125 3 10x2 277 3 1 6 3 2 24x 0 x3 83 x 3 2 −1 x −2 1 2 3 4 31. y V 6 6 0 6 3x ⇒ x 1 6 3 36 0 y 33. V 0 y sin x 2 d x 4.9348 y 12y dy 3 9 9 9 8 y 6 5 4 3 2 1 y2 dy y3 3 6 0 2 1 36y 216 6y 2 216 x 1 2 3 216 3 x 1 3 4 5 6 S ection 6.2 2 2 Volume: The Disk Method 13 35. V 0 e x2 2 dx 1.9686 37. V 1 ex 2 e x22 dx 49.0218 39. A 3 41. Disk Method: b d Matches (a) y V a Rx 2 dx or V c Ry 2 dy 2 Washer Method: b 1 V a d x 1 2 Rx Ry c 2 rx ry 2 dx dy or V 2 2 43. y 4 3 2 1 x 1 2 3 2 1 y −2 −1 x 1 2 3 4 The volumes are the same because the solid has been translated horizontally. 1 x, r x 2 6 45. R x V 0 47. R x r r2 r2 r r x 2, r x x2 dx x2 dx 13 x 3 13 r 3 r 0 0 0 12 x dx 4 6 V 2 0 12 Note: V x3 0 18 12 rh 3 1 3 18 32 6 r2 r 2x r3 y 2 2 43 r 3 y 4 3 2 1 x 1 −1 −2 2 3 4 5 6 y = r2 − x2 (−r, 0) x (r, 0) 14 Chapter 6 r y H h Applications of Integration y ,Ry H 2 49. x V r r1 y H r1 h y ,ry H 2 y H 12 y H h2 H h H 0 H y r1 0 dy r2 0 1 12 y dy H2 13 y 3H 2 h3 3H 2 h2 3H 2 2x 5 5 x6 6 2 0 h 0 h r2 y r2 h r 2h 1 −r r x 2 51. V 0 12 x 8 3 5 2 2 2 x dx 64 x4 2 0 x dx 64 30 5 3 25 9 25 9 y 53. (a) R x V 9 25 25 5 x2, r x x2 dx x2 dx x3 3 5 0 (b) R y V 9 3 y 2, r y y2 dy y3 3 3 0, x ≥ 0 25 5 5 9 0 18 25 25 0 9y 50 0 18 25x 25 y 60 0 6 4 8 6 4 x 2 −6 −4 −2 −2 −4 x 2 4 6 2 4 6 2 55. Total volume: V 4 50 3 3 500,000 3 ft 3 y0 y 60 40 Volume of water in the tank: y0 20 2500 50 2 y2 dy 50 2500 y3 3 y 2 dy y0 50 − 60 − 20 x 20 40 60 − 40 − 60 2500 y 2500 y0 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 y03 3 250,000 3 y03 3 y03 250,00 3 250,000 When the tank is three-fourths of its capacity the depth is 100 32.64 67.36 feet. S ection 6.2 h b Volume: The Disk Method r 15 57. (a) 0 r 2 d x (ii) is the volume of a right circular cylinder with radius r and height h. y (b) b a 1 x2 b2 2 d x (iv) (c) r r2 x2 2 d x (iii) 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 (−b, 0) x (−r, 0) (r, 0) (b, 0) h (d) 0 rx h 2 r d x (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 59. 4 3 2 y 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 2x 3 81 10 1 2 + x − x2 2 + x − x2 2 + x − x2 16 61. 1 3 4 1 2 1 4 Chapter 6 y Applications of Integration (a) A y V 0 b2 1 1 3 3 y 2 1 1 y 2 dy y2 3 53 y 5 3 1− 3 y 1 0 x 1 4 1 2 3 4 2y 1 3 43 y 2 3 dy 1 0 1− 3 y 1 y 1 1 2 3 3 1 10 Base of Cross Section (b) A y V 1 8 12 r 2 1 y 3 1 2 1 y y 2 1 8 1 3 y 2 2 dy 0 1 8 10 3 2 1 3 80 y 1− 3 1− 3 y (c) A y 1 bh 2 3 1 4 1 1 2 3 3 y y 1− 3 y y 3 2 V (d) A y 3 4 1 1 0 y 2 dy 3 31 4 10 y 1 2 3 3 40 1− 3 y 1 ab 2 2 1 1 3 2 y 21 2 y a V 2 1 0 3 y 2 dy 1 2 10 b 20 1− 3 y 63. Let A1 x and A 2 x equal the areas of the cross sections A2 x , of the two solids for a ≤ x ≤ b. Since A1 x we have b b 65. 4 3 25 25 r2 r2 25 32 14 23 125 2 125 2 r2 r2 51 125 32 V1 a A1 x dx a A2 x d x V2 r2 25 22 3 2 23 23 Thus, the volumes are the same. 25 25 1 r 67. (a) Since the cross sections are isosceles right triangles: Ax V (b) A x V 1 bh 2 1 2 r 2 23 3.0415 1 2 r2 r r2 y 2 dy y2 r r2 r2 0 y2 12 r 2 r 2y r2 y2 y3 3 y2 r2y y3 3 r 0 r 0 y2 dy tan 2 y2 dy 23 r 3 x y 1 bh 2 tan 2 r 1 2 r2 r r2 y2 r2 tan y 2 tan r y2 dy . r2 0 tan 23 r tan 3 As → 90 , V → Section 6.3 Volume: The Shell Method 17 Section 6.3 1. p x hx V 2 0 Volume: The Shell Method 3. p x hx x x 4 x x 2 x x dx 2 x3 3 2 0 2 x3 3 2 0 16 3 V 2 0 4 x x dx x3 2 dx 0 16 3 2 45 x 5 5. p x hx V 2 0 4 4 2 0 128 5 x x 2 2 7. p x hx x3 dx 2 y x 4x 2 x2 x 4x x2 2x 2 d x 4x 2x 2 y 4 V 2 0 2 2 x4 0 8 3 4 0 2x 2 23 x 3 x3 14 x 4 2 0 dx 16 3 −1 3 2 2 1 x 1 2 3 4 1 x 1 2 3 −1 9. p x hx V 2 x 4 2 11. p x 4x x3 0 x 1 e 2 1 x2 2 x2 4x 2 43 x 3 x2 4x d x 2 4x 4 hx V 2 x 0 1 1 e 2 e x2 2 x2 2 dx 2 y 4 3 x4 4 2x 2 0 8 3 2 0 x dx 1 2e y 1 3 4 1 2 1 4 x2 2 0 2 1 1 e 0.986 2 1 x 1 2 3 −1 x 1 4 1 2 3 4 1 13. p y hy V 2 y 2 2 y y2 0 2 y dy y2 dy y3 3 2 0 2 0 2y y2 2 8 3 18 Chapter 6 Applications of Integration 1 . 2 1 3 4 1 2 15. p y py V 2 y and h y y and h y 12 1 if 0 ≤ y < 1 y 2 12 y 1 1 if ≤ y ≤ 1. 2 1 y dy 0 1 y y2 2 1 12 y dy 1 4 2 y2 2 4 4x 2 12 2 0 x 4 4 2 1 2 1 3 2 2 17. p x hx V 2 x x2 4 0 2 19. p x x2 x 4x x3 6x 2 2 5 4x 4 x x2 5 0 4 4x 2x 2 hx V 2 2 2x 2 d x 8x d x 16 0 x 4x 9x 2 3x 3 x2 dx 20x d x 4 2 4 y 4 3 2 0 x3 0 x4 4 2x 3 4x 2 2 y 4 3 2 1 x4 4 10x 2 0 64 2 1 x x 1 2 3 1 −1 2 3 4 21. (a) Disk Rx rx 2 (b) Shell x3 0 x6 dx 0 y 8 6 8 6 px hx x7 7 2 0 x x3 2 V 128 7 V 2 0 y x4 dx 2 x5 5 2 0 64 5 4 4 2 x 1 2 3 2 x 1 2 3 −1 −1 (c) Shell px hx 2 4 x 3 x y 8 V 2 0 2 4 4x 3 0 x x3 dx x4 2 0 6 4 2 2 x4 dx 96 5 2 x 1 2 3 4 15 x 5 S ection 6.3 23. (a) Shell py hy a Volume: The Shell Method 19 (c) Shell y a1 2 ya 0 a px y1 2 2 a a1 2 a x x1 2 x a1 2 2a 3 2 x 1 2 4 32 32 ax 3 2 hx y dy y2 dy y3 3 a 15 a V 2 2 0 2a 1 2 y 1 2 2a 1 2 y3 2 4a 1 2 5 2 y 5 4a 5 3 V 2 0 a a a2 0 x1 2 2 dx 2a 1 2 x 3 2 4 12 52 ax 5 x2 dx 13 x 3 a 0 ay a2 y 2 a 2 3 2 2 2 2 y a 2x 0 3 y 4 a3 15 a 3 3 (0, a) (0, a) (a, 0) x (a, 0) x (b) Same as part (a) by symmetry d b 25. V 2 x p y h y dy or V 2 a p x h x dx 5 5 27. 1 x 1 dx 1 x 1 2 dx 29. (a) 1.5 y = (1 − x 4/3) 3/4 This integral represents the volume of the solid generated by revolving the region bounded by y x 1, y 0, and x 5 about the x-axis by using the Disk Method. 2 − 0.25 − 0.25 1.5 2 0 y5 y2 1 dy (b) x 4 3 y4 3 y V 1, x 1 2 0 0, y x4 3 1 34 0 represents this same volume by using the Shell Method. y 4 3 2 1 x 1 −1 2 3 4 5 x1 x4 3 34 dx 1.5056 Disk Method 20 Chapter 6 7 Applications of Integration 33. y 2e x, y 7.5 0, x 0, x 2 31. (a) y= 3 (x − 2) 2 (x − 6) 2 Volume 7 −1 −1 Matches (d) y 6 (b) V 2 2 x3 x 2 2 x 6 2 dx 187.249 2 1 x 1 2 35. p x hx V 2 x 2 2 12 x 2 x2 0 12 x dx 2 2 2 0 2x 13 x dx 2 2 x2 14 x 8 2 4 0 total volume Now find x0 such that x0 2 0 2x x2 14 x 8 14 x 40 13 x dx 2 2 y x0 0 1 1 1 x 04 8x 02 4 x 02 Take x0 Diameter: 2 1 2 2 x 02 0 x 1 2 4±2 3 (Quadratic Formula) 4 4 2 3 since the other root is too large. 23 1.464 37. V 4 1 1 2 1 1 x x2 1 1 dx x1 1 x2 dx 1 39. Disk Method Ry x1 1 r2 0 r y2 8 8 4 2 4 x2 32 1 12 x2 dx ry V 2 2 2 2 1 3 2 dx 4 2 r2 r h y2 dy r r h 1 x2 r 2y y y3 3 1 h 2 3r 3 h r −r x r S ection 6.3 r Volume: The Shell Method 21 41. (a) 2 0 hx 1 x d x (ii) r r (b) 2 r R x 2 r2 x 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 x=R (0, h) y=h 1− x r ( ( y= r2 − x2 (r, 0) (r, 0) x x (−r, 0) y=− r2 − x2 r r (c) 2 0 2x r 2 x 2 d x (iii) is the (d) 2 0 h x d x (i) is the volume of a volume of a sphere with radius r. y right circular cylinder with a radius of r and a height of h. y y= r2 − x2 (r, h) (r, 0) x y=− r2 − x2 x b (e) 2 0 2ax 1 x 2 b 2 d x (iv) is the volume of an ellipsoid with axes 2a and 2b. y y =a (0, a) 2 1 − x2 b (b, 0) x (0, −a) y = −a 2 1− x b2 200 43. (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 −20 −6 225 200 (c) V 2 0 xd x dx 2 213,800 1,343,345 cubic feet (d) Number gallons V 7.48 10,048,221 gallons 22 Chapter 6 Applications of Integration Section 6.4 1. 0, 0 , 5, 12 (a) d (b) y y s 0 Arc Length and Surfaces of Revolution 3. y 0 2 2 32 x 3 1 5 12 x 5 12 5 5 12 0 2 13 y s x 1 2, 0, 1 1 1 0 x dx 1 1 12 5 2 dx 13 x 5 5 2 1 3 13 2 3 8 x 1 32 0 0 1.219 5. y y s 3 23 x 2 1 , 1, 8 x1 3 8 7. y y x4 8 13 x 2 13 x 2 b 1 4x 2 1 , 1, 2 2x 3 12 , 1, 2 2x 3 1 y 2 1 1 8 1 1 x1 3 2 dx 1 y 2 x2 3 1 dx x2 3 8 s a 2 1 dx 3 2 x2 3 1 1 1 2 dx 3x 1 3 8 13 x 2 1 4x 2 1 dx 2x 3 2 1 3 2 23 x 23 55 32 1 14 x 8 33 16 2.063 22 8.352 3 44 , cot x csc2 x 9. y y 1 y 2 ln sin x , 1 cos x sin x 1 3 cot2 x 4 s 4 csc x dx 3 4 4 ln csc x ln 2 1 cot x ln 2 1 1.763 11. (a) y 4 5 x 2, 0 ≤ x ≤ 2 (b) 1 y y 2 2x 1 2 (c) L 2 4.647 4x L 0 −1 −1 3 1 4x2 dx Section 6.4 1 ,1 ≤ x ≤ 3 x 2 Arc Length and Surfaces of Revolution 23 13. (a) y (b) 1 y y 2 1 x2 1 3 (c) L 1 x4 1 1 dx x4 2.147 −1 4 L −1 1 15. (a) y sin x, 0 ≤ x ≤ 2 (b) 1 y y 2 cos x 1 cos 2 x 1 0 (c) L 3.820 L − 2 − 0.5 cos 2x d x 3 2 17. (a) x y e y, 0 ≤ y ≤ 2 ln x 2 (b) 1 y y 2 1 x 1 1 (c) L 1 x2 1 1 dx x2 2.221 1≥x≥e 3 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 19. (a) y 2 arctan x, 0 ≤ x ≤ 1 3 (b) y L 2 1 1 x2 1 4 1 x2 2 (c) L dx 1.871 0 − 0.5 1.5 −3 24 Chapter 6 2 Applications of Integration d 5 dx x 2 1 2 y 21. 0 1 s 5 dx 5 4 3 (0, 5) y = 25 x +1 Matches (b) 2 1 −1 (2, 1) x 1 2 3 4 23. 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 (d) 64.672 4 25. (a) 5 4 3 2 1 −1 y (c) y1 y2 y1 y3 1 2 3 4 5 1, L1 0 2 dx 4 5.657 1 9x dx 16 x2 dx 4 1 5.759 5.916 6.063 y2 y4 x 3 12 x , L2 4 1 x, L 3 2 5 32 x , L4 16 4 0 y3 y4 1 0 4 0 −1 25 3 x dx 256 (b) y1, y2, y3, y4 1 32 x 3 27. y 3x 1 2 0, y y 2 3. 2 Thus, the fleeting object has traveled 3 units when it is caught. 3 x 2 2 12 When x 2 1 3 12 x 32 1 1 1x 1 2 x1 2 1 4x 1 2 1 y 2 x 1 4x x 1 2 s 0 x1 dx 2x 1 2 x1 2 0 x 12 dx 1 2 32 x 23 1 2x 1 2 0 4 3 2 2 3 The pursuer has traveled twice the distance that the fleeing object has traveled when it is caught. 29. y y 1 y 2 20 cosh sinh 1 20 x , 20 20 ≤ x ≤ 20 x 20 sinh 2 cosh x 20 x dx 20 cosh 2 2 0 x 20 20 L 20 cosh x dx 20 2 20 sinh x 20 20 0 40 sinh 1 47.008 m. Section 6.4 x3 3 Arc Length and Surfaces of Revolution 25 31. y y 1 y 2 9 x 9 9 9 2 x2 x2 x2 9 9 3 9 x x 3 2 3 x dx 2 33. y y S x 2, 0, 3 3 2 0 3 x3 3 1 1 x4 x4 x4 dx 12 s 0 2 0 6 9 9 4x3 dx 0 3 dx 2 2 0 1 32 0 3 arcsin 3 arcsin 3 arcsin 82 82 1 258.85 arcsin 0 2.1892 2 3 35. y y 1 y 2 x3 6 x2 2 x2 2 1 2x 1 2x 2 12 , 1, 2 2x 2 2 37. y y S x2 2 1 dx 2x 2 3 x 2 1 , 1, 8 3x 2 3 8 2 1 x 8 1 1 dx 9x 4 3 1 dx 12x 1 3 d x S 2 1 2 x3 6 x5 12 x2 6 1 2x x 3 2 3 18 27 27 x 1 3 9x 4 3 1 8 2 1 1 dx 4x 3 1 8x 2 2 1 9x 4 3 1 1 1 12 2 x6 72 47 16 8 9x 4 3 32 1 145 145 10 10 199.48 39. y y S sin x cos x, 0, 2 0 41. A rectifiable curve is one that has a finite arc length. sin x 1 cos 2 x d x 14.4236 43. 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 xi. 26 Chapter 6 hx r h r r2 r2 Applications of Integration 45. y y 1 y 2 47. y y 9 x 9 3 9 2 x2 x2 x2 3x dx 9 x2 2 h2 r 1 r2 r h 2 2 y 2 S 2 0 x 2 r r 2 h2 x2 2 dx r S r2 h2 2 0 r 0 3 0 2x 9 9 5 x2 0 x2 2 dx 6 6 3 14.40 See figure in Exercise 48. 1 12 x 3 1 x 6 1 2 0 13 12 49. 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 91 2 2 dx x2 3x 3 0 2 6 13 13 0 1 12 x 3 x3 2 x 12 9x 1 2 d x 3 0 1 x 33 27 ft 2 0.1164 ft2 16.8 in 2 Amount of glass needed: V 27 0.00015 ft 3 0.25 in 3 51. (a) y fx 0.0000001953x4 400 0.0001804x3 0.0496x2 4.8323x 536.9270 (b) Area 0 f x dx 131,734.5 square feet 3.0 acres (Answers will vary.) 400 (c) L 0 1 f x 2 dx 794.9 feet (Answers will vary.) b 53. (a) V 1 y 1 dx x2 b x 1 1 1 b b (b) S 2 1 b 1 x 1 x 1 1 x2 1 dx x4 2 dx 2 y= 1 x 2 1 b 1 1 2 1 x 1 b x4 1 dx x3 —CONTINUED— Section 6.5 53. —CONTINUED— (d) Since x4 1 > x3 we have b 1 Work 27 (c) lim V b→ b→ lim 1 1 b x4 x3 1 > 0 on 1, b x x4 1 dx > x3 b→ b 1 1 dx x b ln x 1 ln b and lim ln b → b b→ . Thus, x4 1 dx x3 . lim 2 1 55. (a) Area of circle with radius L: A L2 Area of sector with central angle (in radians) 12 S A L2 L 2 2 2 (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 rL (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 r 1L 1 L r1 L1 L1 ⇒ L r1 r1 L1 r2 r1 By similar triangles, Hence, S r2 L L r1 L 1 r2 r1 L 1 r2 r2 . r1 r2 L L r1 Section 6.5 1. W Fd Work 1000 ft lb FD. 3. W Fd 112 4 448 joules (newton-meters) 100 10 5. Work equals force times distance, W 7. Since the work equals the area under the force function, you have c < d < a < b . 11. F x 250 W kx k 30 ⇒ k 50 9. F x 5 k W 0 kx k4 5 4 7 25 3 50 F x dx 20 20 5 x dx 4 lb lb 52 x 8 7 0 25 x dx 3 25x 2 6 50 20 8750 n cm 87.5 joules or Nm 245 in 8 30.625 in 2.55 ft lb 28 Chapter 6 Applications of Integration 13 13. F x 20 k W kx k9 20 9 12 0 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 324 lbs 13 13 Note: 4 inches 20 x dx 9 10 2 x 9 12 0 1 3 foot 40 ft 3 lb 17. Assume that Earth has a radius of 4000 miles. Fx s k Fx k x2 k 4000 2 4300 4100 (a) W 4000 80,000,000 dx x2 80,000,000 x 4100 487.8 mi 4000 tons lb 5.15 (b) W 4000 109 ft 80,000,000 80,000,000 x2 80,000,000 dx x2 1395.3 mi 1.47 ton lb 1010 ft 19. Assume that the 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 10 4 mi 1011 ft 160,000,000 3.10 160,000,000 x2 26,000 (b) W 4000 160,000,000 dx x2 160,000,000 x 26,000 6,153.846 4000 33,846.154 mi 3.38 3.57 21. Weight of each layer: 62.4 20 Distance: 4 4 10 4 mi 1011 ft 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 1 4−y (b) W 624y 2 x 1 2 3 4 5 6 23. Volume of disk: 2 2 y 4 y y y 4 Weight of disk of water: 9800 4 Distance the disk of water is moved: 5 4 W 0 5 y 9800 4 dy 39,200 0 5 5y 12 y dy y2 2 4 0 39,200 39,200 470,400 newton–meters Section 6.5 2 y 3 2 y Work 29 25. Volume of disk: y 2 y 3 2 7 5 Weight of disk: 62.4 Distance: 6 W 4 62.4 9 y y 6−y 4 3 2 x 6 6 0 y y2 dy 4 62.4 9 y 2y 3 14 y 4 6 −4 −3 −2 −1 1 2 3 4 2995.2 ft 0 lb 27. Volume of disk: Weight of disk: 62.4 Distance: y 6 36 36 y2 2 y 10 y2 y 8 4 W 62.4 0 6 y 36 36y 0 y2 dy y3 dy lb lwh 42 8 42 94 94 y2 y y2 y Tractor y x −6 −4 −2 −2 2 4 6 62.4 62.4 18y 2 14 y 4 6 0 20,217.6 ft 29. Volume of layer: V Weight of layer: W 13 Distance: 2 1.5 y 8 6 4 13 2 y 94 1.5 −y x 2 −6 −4 −2 −2 −4 2 W 1.5 42 8 336 13 2 y2 13 2 y2 dy 4 6 y dy 1.5 94 1.5 94 1.5 y2 y dy 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 2457 ft lb 31. Weight of section of chain: 3 y Distance: 15 15 y 15 0 33. The lower 5 feet of chain are raised 10 feet with a constant force. W1 3 5 10 150 ft lb W 3 y dy 15 The top 10 feet of chain are raised with a variable force. Weight per section: 3 y Distance: 10 W2 3 0 3 15 2 337.5 ft y lb 2 0 y 10 10 y dy 3 10 2 150 ft lb 10 y 2 0 W W1 W2 300 ft lb 30 Chapter 6 Applications of Integration 37. Work to pull up the ball: W1 Weight per section: 1 y Distance: 15 15 2 35. Weight of section of chain: 3 y Distance: 15 7.5 500 15 7500 ft lb 2y 15 0 Work to wind up the top 15 feet of cable: force is variable 2y d y 3 15 4 3 15 4 7.5 W 3 2y 2 0 x 15 x dx lb 1 15 2 15 168.75 ft lb W2 0 x 2 0 112.5 ft 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 7500 7987.5 ft k V k 2 2000 3 39. p 1000 k W 41. F x W k 2 1 2 x 2 2 k x 2 dx k1 2x2 3k units of work 4 k 1 1 4 2 2000 dV V 3 2 3 2000 ln V 2 2000 ln 5 810.93 ft lb 5 43. W 0 1000 1.8 ln x 1 dx 3249.44 ft lb 45. W 0 100x 125 x3 dx 10,330.3 ft lb Section 6.6 1. x 6 5 6 7 12 Moments, Centers of Mass, and Centroids 31 53 35 5 6 3 8 5 1 4 6 7 12 5 3 623 30 12 1 6 3 11 x 3 5 15 5 18 5 3. x 17 18 1 1 1 12 1 1 15 11 1 18 12 5. (a) x (b) x 17 11 8 12 3 5 99 33 3 7. 50x 50x 125x x 75 L 750 750 6 feet 52 5 52 5 10 , 9 x 75x 75 10 9. x y x, y 1 1 11 1 1 9 3 3 3 3 31 4 10 9 1 9 y 2 m2 1 (− 3, 1) −3 −2 −1 −1 −2 −3 −4 1 2 m1 (2, 2) x 3 m3 (1, − 4) S ection 6.6 3 3 2 3 7 , 8 41 34 40 34 7 16 27 10 216 21 10 216 6 60 3 7 16 7 8 y Moments, Centers of Mass, and Centroids 31 11. x y x, y m2 6 (− 1, 0) m5 (− 3, 0) −4 −2 −2 m4 (0, 0) 2 4 m3 (7, 1) x 6 8 m1 (−2, −3) 4 13. m 0 4 x dx x 2 4 4 23 x 3 x dx 4 2 0 16 3 4 4 y Mx 0 x2 4 3 4 2 52 x 5 12 5 4 0 4 0 3 2 1 y My Mx m 3 16 ( 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 1 15. m 0 1 x2 x2 2 x3 dx x3 12 x2 12 35 x3 x3 3 dx x4 4 2 1 y 0 1 12 x4 x6 dx x5 25 x7 7 1 0 1 3 4 1 2 Mx 0 0 35 y My Mx m 1 35 x x2 0 ( x, y ) 1 4 1 x3 dx 0 x3 x4 dx x4 4 x5 5 1 x 0 20 1 4 1 2 3 4 1 x x, y My m 12 20 3 5 3 12 , 5 35 3 17. m 0 3 x2 x2 4x 4x 2 2 2 x x 2 dx 2 x2 x3 3 4x 3 3x 2 2 2 8x 3 3 0 9 2 6 y Mx 0 3 x 2 dx 11x 2 12x d x 5 4 ( x, y ) 2 x2 0 5x 4 x2 3 3x d x 99 5 2 x4 0 3 2 1 −1 x 1 2 3 4 5 x5 25 y My 0 2x 4 99 5 2 9 x2 27 4 2 9 11x 3 3 22 5 6x 2 0 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 32 Chapter 6 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 8 0 y 6 4 2 Mx 0 3 73 x 27 10 7 3 83 x 8 5 8 192 7 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 2 5 96 21. m My x 2 0 2 4 4 2 256 15 y2 dy y2 3 32 0. 4 2 4y y3 3 2 0 32 3 2 y 2 0 y2 dy 8 5 16y 83 y 3 y5 5 2 0 256 15 1 ( x, y ) x 1 −1 −2 2 3 My m By symmetry, M x and y x, y 8 ,0 5 3 23. m 0 3 2y 2y y2 y2 2 4y 3 y dy y 2y 3y 2 2 y2 y5 25 y4 y3 3 3 0 9 2 3 3 y My 0 3 y dy 3 2 y 0 y 2 3y y2 dy ( x, y ) 1 x 1 −1 2 x Mx My m y4 0 3y 2 d y 3 5 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 1 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 6 1 0 Mx My 0 x5 5 x4 4 1 0 11 23 1 3 1 4 1 5 1 12 1 15 x2 0 S ection 6.6 3 3 Moments, Centers of Mass, and Centroids 33 27. A 0 2x 1 2 3 3 4 dx 4 dx 2 x2 3 4x 0 9 8x 3 12 8 dx 18 21 2x 3 3 18 36 3 Mx My 2x 0 2x 2 0 4x 2 8x 0 18 36 24 78 2x 2 0 4x d x 2x 3 3 2x 2 0 5 29. m 0 5 10x 125 10x 125 2 10x 2 125 0 x3 dx x3 1033.0 5 Mx 0 5 10x 125 10 3 x3 dx 5 50 0 x 2 125 3x 2 d x x3 dx 12,500 5 9 400 3,124,375 24 3105.6 130,208 My x y My m Mx m x3 dx 125 0 x3 3.0 126.0 −1 6 − 50 Therefore, the centroid is 3.0, 126.0 . 20 31. m 20 20 5 3 400 x2 dx x2 1239.76 x2 dx 20064.27 33. A 1 A x Mx 25 2 y Mx m 5 3 400 2 20 20 5 3 400 dx 1 2a c 2 1 ac 11 ac 2 1 2ac c 0 c 0 ac b c a 2 400 20 x 2 23 y a b c a 2 y a dy 16.18 4ab y c 4ab 2 y dy c2 4ab 3 y 3c 2 a c 0 x 0 by symmetry. Therefore, the centroid is 0, 16.2 . 50 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 0 c b c 2a y c y 3c 3c 0 y a 2a dy a y y y 0 − 25 −5 0 y2 dy c 25 2 y2 c2 x, y bc , 33 c 3 In Exercise 566 of Section P.2, you found that b 3, c 3 is the point of intersection of the medians. y (b, c) y = c (x + a) b+a y= c b − a (x − a ) ( x, y ) x (− a, 0) (a, 0) 34 Chapter 6 c a 2 2 ca 2 ca 2 ca y 2 ca 1 ca b 1 3c a b 1 3a Thus, x, y b b 1 b2 b b c Applications of Integration 35. A 1 A x b x 0 b c a 3 a c2 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 c 0 2bc 2 c 0 c 0 2ac 2 6 b c 1 ca a 2 c 2b 3a a 2b c 3a b a2 dx a a 2c b c a 2 a x3 3 2 x a dx x2 b b 2a b a x c a 2c 3 ac b b c b2 b2 2a b a x 2 c 2 a2 c a2 3ab 3ac b 3a 2 2ab 2ab a 3a 2 3a 2c a2 ab b 2 3a b y a 2b c a 2 ab b 2 , . 3a b 3a b b 2c a c a x 2b a . 2 x b. (0, a) y= b−ax +a c a ( x, y ) ( c, b) (0, 0) x The one line passes through 0, a 2 and c, b 2 . It’s equation is y The other line passes through 0, b and c, a b . It’s equation is y x, y is the point of intersection of these two lines. 37. x 0 by symmetry A 1 A y 1 ab 2 b y b (c, 0) 2 ab 21 ab 2 1 b2 ab a2 a a −a a x b a a 2x 2 a2 x3 3 a x2 dx b 4a3 a3 3 4b 3 a x, y 0, 4b 3 b 39. (a) y (e) Mx b b b b2 2 x2 b 2 x4 b2 5 dx b x2 dx x5 5 b b y=b b 12 bx 2 4b2 5 bx 4 bb 3 x3 3 b b2 b −5 − 4 −3 −2 −1 x 12345 b b b A (b) x (c) M y b b b x2 dx bb 2 3 0 by symmetry b bb xb x dx 2 0 because bx x is odd b than below 2 y Mx A 3 b (d) y > since there is more area above y 2 4b2 b 5 4b b 3 3 b. 5 S ection 6.6 41. (a) x 0 by symmetry 40 Moments, Centers of Mass, and Centroids 35 A Mx y x, y (b) y (c) y x, y Mx A 2 0 40 f x dx fx 2 40 2 2 40 30 34 40 30 2 34 72160 5560 4 29 4 29 12.98 2 2 26 2 26 2 4 20 4 20 2 0 0 20 278 3 5560 3 72160 3 dx 10 7216 3 Mx A 72160 3 5560 3 0, 12.98 1.02 10 5 x4 0.0019x 2 12.85 29.28 23697.68 1843.54 0, 12.85 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 1 y x 1 −1 −2 3 ,0 45. Centroids of the given regions: 0, Area: A x y x, y 15 12 15 0 15 3 2 135 34 7 12 0 34 12 5 34 34 70 3 15 , 0, 5 , and 0, 2 2 7 6 5 4 y 0 135 34 −4 −3 −2 −1 3 2 1 7 15 2 x 1 2 3 4 0, 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 49. V 2 rA 2 5 16 160 2 1579.14 36 Chapter 6 1 44 2 11 82 y 8 3 2 4 Applications of Integration 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 2 rA 8 8 3 128 3 134.04 x 1 2 3 4 53. m My Mx x m1 m1x1 m1y1 My ,y m ... ... ... Mx m mn mn xn mn yn 55. (a) Yes. x, y (b) Yes. x, y (c) Yes. x, y (d) No. 55 , 6 18 5 6 5 , 6 2, 5 18 2 5 18 5 41 , 6 18 17 5 , 6 18 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 centroid is d S s 4 r2 r 4 r. r y This distance is also the circumference of the circle of radius y. d 2y (0, y) −r r x 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 2 1 0 1 n 1 y m Mx My n 2 xn 2 dx 0 1 x 2n 1 2n 1 xn 2 n2 1 0 1 0 2 2n n 2 1 1 y=xn (1, 1) x xn dx 0 x y My m Mx m n n n n 1 2 n 4n 1 2 1 2 x 1 n1 2 2n 1 1n , 2 4n Centroid: As n → , x, y → 1, 1 . 4 1 as n → . The graph approaches the x-axis and the line x Section 6.7 Fluid Pressure and Fluid Force 37 Section 6.7 1. F PA Fluid Pressure and Fluid Force 936 lb 3. F 62.4 h 62.4 2 6 26 62.4 h 6 62.4 5 3 748.8 lb 5. h y Ly F 3 4 y 7. h y Ly 3 2 y 3 y 1 3 3 62.4 0 3 3 3 0 y 4 dy F y dy 124.8 y2 2 3 0 2 62.4 0 3 3 3 y3 9 y y 3 1 dy 249.6 249.6 3y y y2 dy 3 3 1123.2 lb 0 124.8 3y y 748.8 lb 0 4 3 2 1 x 1 2 3 4 x −2 −1 1 2 2 1 4 9. h y Ly F 4 2y y 4 11. h y Ly 4 0 4 2 y 2 2 62.4 4 y y dy y3 2 dy F 9800 0 24 y2 y dy 2 124.8 0 4y1 2 8y 3 2 3 9800 8y 1064.96 lb y 117,600 Newtons 0 124.8 y 2y 5 2 5 4 0 3 3 x 1 x −2 −1 1 2 −2 −1 1 2 38 Chapter 6 12 6 9800 0 Applications of Integration y 15. h y Ly 2y dy 3 2y 3 9 9 13. h y Ly F 2 10 y 2y 3 9 2 F 140.7 0 2 2 2 0 y 10 d y y dy y2 2 2 12 y6 7y 2 9800 72y y 2,381,400 Newtons 0 1407 1407 2y y 2814 lb 0 9 4 6 3 x −3 3 6 9 x −6 −4 −2 −1 −2 2 4 6 3 17. h y Ly F 4 6 y 19. h y Ly 2 42 y 1 2 0 9 y 4y 2 9 4y 2 4y 2 d y 12 4 140.7 0 4 4 4 0 y 6 dy F y dy y2 4 2 0 32 844.2 844.2 4y y 5 42 8 6753.6 lb 21 4 0 9 32 8y d y 0 2 9 3 4y 2 32 32 94.5 lb y 2 3 1 1 x −3 −2 −1 −1 1 2 3 x −2 −1 −1 −2 1 2 21. h y Ly F w k y y2 y r2 r r y 2 r2 r water level k r r r2 y2 2 dy r −r r x w 2k 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 Section 6.7 23. h y Ly F w h2 Fluid Pressure and Fluid Force 39 k b h2 y 25. From Exercise 22: F k y2 2 y b dy h2 64 15 1 1 960 lb wb k y y wb hk h2 wkhb water level h 2 x −b 2 −h 2 b 2 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 29. h y Ly F 12 2 42 4 y 3 y2 332 31. (a) If the fluid force is one half of 1123.2 lb, and the height of the water is b, then hy b 4 b y 62.4 0 2 12 y 42 3 y2 332 dy Ly F 6448.73 lb y 62.4 0 b b y 4 dy b y dy y2 2 b 1 1123.2 2 2.25 2.25 10 8 6 4 0 by x 0 −6 −4 −2 2 4 6 b2 b2 2 b2 2.25 4.5 ⇒ b 2.12 ft. (b) The pressure increases with increasing depth. d 33. F Fw w c h y L y dy, see page 471. ...
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