ODD0R - 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

ODD0R - 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 ...
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