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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online