CHAPTER-7-TEXTBOOK-ANSWERS - C H A P T E R 7 Applications of Integration Section 7.1 Area of a Region Between Two Curves 313 Section 7.2 Volume The

CHAPTER-7-TEXTBOOK-ANSWERS - C H A P T E R 7...

This preview shows page 1 out of 50 pages.

Unformatted text preview: C H A P T E R 7 Applications of Integration Section 7.1 Area of a Region Between Two Curves . . . . . . . . . . 313 Section 7.2 Volume: The Disk Method . . . . . . . . . . . . . . . . . 321 Section 7.3 Volume: The Shell Method Section 7.4 Arc Length and Surfaces of Revolution . . . . . . . . . . 335 Section 7.5 Work Section 7.6 Moments, Centers of Mass, and Centroids . . . . . . . . . 344 Section 7.7 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 350 Review Exercises . . . . . . . . . . . . . . . . 328 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 C H A P T E R 7 Applications of Integration Section 7.1  Area of a Region Between Two Curves 6 1. A   6 0  x 2  6x dx   0   3 x 2  6x dx 3. A  0 x 2  2x  3  x 2  4x  3 dx 0 3  2x 2  6x dx 0   0 5. A  2  0 1 3x 3  x dx  6 1 x 3  x dx 1 or 6 x 3  x dx 0  4 7. x  1  0  6  x dx 2 9. 4 2x3  0 y  3  x dx 6 11. 3 2  sec x dx y y 6 5 3 5 4 3 3 2 2 1 2π 3 1 x 1 2 3 4 x 5 1 2 3 4 5 x (0, 2) x −6 −4 6 (− 5, − 3)  y  3 y  2  0 −6 Intersection points: 0, 2 and 5, 3   0  4 x  2  4  x dx  2 4  x dx  0 2 3 2π 3 4 y2  y  6  0 (b) A  −1 3 6 4  y2  y  2 5 π y xy2 A 3 6 x  4  y2 13. (a) π 4  y2   y  2 dy  15. f x  x  1 61 32 125   6 3 6 125 6 y gx  x  1 2 3 A 4 2 (3, 4) Matches (d) (0, 1) x 1 2 3 313 314 Chapter 7    2 17. A  0 2    1 3 x  2  x  1 dx 2 y 5 4 3  x 4 x2  x 8 2  16 4  2 02 8 2 2 (2, 3) (0, 2) 0 1 (0, 1) x −2 1 19. The points of intersection are given by: x2 (2, 6) 6 1 3 x  x  1 dx 2 0  Applications of Integration 3 21. The points of intersection are given by:  4x  0 x 2  2x  1  3x  3 xx  4  0 when x  0, 4  4 x  2x  1  0 when x  1, 2 4 A 0  (0, 0)  (4, 0) 1 x 2  4x dx  x3  2x  3 2 3 x  −2 4 −3 0 −4  gx  f x dx 1 3x  3  x 2  2x  1 dx 2  2 −5 32 3 1 2 5 −1 0  A y 4     2  gx  f x dx 1 2  x  x 2 dx   2x  x2 x3  2 3  2 1 y  9 2 10 (2, 9) 8 6 4 ( 1, 0) x 4 23. The points of intersection are given by: 3 x1 2  x0 A x2   2  y   y dy  2y  y 2 0 1 0 1 (0, 0)  y  2 y  1  0 when y  1, 2 3 A 0   3x  1  x  1 dx  3x12  x dx  3 3  2 0 1  5   g y  f  y dy y 3  y  2  y 2 dy y2 y3  2y   2 3 y 0 2 x2  3x 32  9 2 1 2 3    2  f x  gx dx 0 3 y2  y  2 3x  x when x  0, 3  x 2 27. The points of intersection are given by: 3x  1  x  1 3 2 (2, 0) 25. The points of intersection are given by:     1 (1, 1) 1 Note that if we integrate with respect to x, we need two integrals. Also, note that the region is a triangle. A 2 y x  2  x and x  0 and 2  x  0 1 3 1  2 1 9  2 x 1 (1, (3, 4) 3 3 2 (0, 1) x 1 2 3 4 2 1 4 −2 (4, 2) 2 1) 3 4 5 Section 7.1   Area of a Region Between Two Curves 2 29. A  1  f  y  g y dy 3 (0, 2) 2  1 31. y  y  y 2  1  0 dy (5, 2)   1 6 10 dy y   2 x y3 y  3  (0, −1) 3 4 5 6 (2, −1) (0, 10) (1, 10) 8 6 10  10 ln y 2 y 12 10 A 1 2 2 10 10 ⇒ x x y 315 4 (0, 2) 2  10ln 10  ln 2 −4 −2 (5, 2) x 2 4 6 8  10 ln 5 16.0944 33. (a) (b) The points of intersection are given by: 11 x 3  3x 2  3x  x 2 (3, 9) −6 (0, 0) xx  1x  3  0 when x  0, 1, 3 (1, 1) 12 −1 (c) Numerical approximation: 0.417  2.667 3.083     1 A 3  f x  gx dx  0 1  1  x 2  x 3  3x 2  3x dx 1  3 x 3  4x 2  3x dx  0   3 x 3  3x 2  3x  x 2 dx  0 35. (a)  gx  f x dx 1 x 3  4x 2  3x dx 1  x4  34 x 4 3     x4 1 3 2 x 2 4 0 4 3  x3  x2 3 2  3  1 5 8 37   12 3 12 (b) The points of intersection are given by: 9 x 2  4x  3  3  4x  x 2 (0, 3) (4, 3) −6 12 2xx  4  0 when x  0, 4   4 −3 A 3  4x  x 2  x 2  4x  3 dx 0 (c) Numerical approximation: 21.333 4  2x 2  8x dx 0    37. (a) f x  x 4  4x 2, gx  x 2  4 (−2, 0) 4 0  64 3 (b) The points of intersection are given by: x 4  4x 2  x 2  4 2 −4  2x 3  4x 2 3 (2, 0) 4 x 4  5x 2  4  0 x 2  4x 2  1  0 when x  ± 2, ± 1 (−1, −3) (1, −3) −5 By symmetry:   1 (c) Numerical approximation: 5.067  2.933  8.0 A2 0 1 2  2 x 4  4x 2  x 2  4 dx  2  x 2  4  x 4  4x 2 dx 1 2 x 4  5x 2  4 dx  2 0 x 4  5x 2  4 dx 1 2  x5  5x3 2  5  35  4  2  5 5 1 3   4x 1 0  2  32  x 5 5x 3   4x 5 3  2 1 1 5 40 8    4 3 5 3   8 316 Chapter 7 Applications of Integration 39. (a) 41. (a) 3 5 (−1, 12 ( ( 1, 12 ( (2, 3) (0, 2) −3 3 (0, 1) −4 5 −1 −1 (b) The points of intersection are given by: (b) and (c) 1  x3 ≤ x2 1  1  x2 2 x4  x2 You must use numerical integration because y  1  x3 does not have an elementary antiderivative. 20 x2  2x2  1  0   1 x  2 on 0, 2 2  2 A x  ±1 0  1 x  2  1  x3 dx 1.759 2 1  f x  gx dx A2 0 1 2 0  x2 1  dx 1  x2 2   2 arctan x  2 x3 6  1 0 4  61  2  31 1.237 (c) Numerical approximation: 1.237 43. A  2   3  f x  gx dx 45. A  y 0 2 3 4 2 0  0 π 2  21  ln 2 0.614 1  cos x dx 0 f (0, 0) 2  cos x  cos x dx 2 2 3 1 3  2 2 cos x  ln cos x π , 3 3 2 0 g 2 sin x  tan x dx    2 2   2 x  sin x x π 0  4 12.566 y 3 π , 3 3 3 4 2 (0, 1) g (2π, 1) f −1  1 47. A  xex  0 dx 2 y 0   1 2   ex 2 1 0  1 1 1 0.316 2 e 1 (1, e1 ) (0, 0) x 1 π 2 π 2π x Section 7.1 49. (a) (b) A  3   Area of a Region Between Two Curves 2 sin x  sin 2x dx 317 (c) Numerical approximation: 4.0 0   2 cos x   0 0 (b) A  4  1 (1, e)  (3, 0.155) 0 53. (a) (c) Numerical approximation: 1.323  e 6 1 1x e dx x2 3  e1x 0 1 e13 (c) A 4.7721 (b) The integral 6  3 A 0 −1 4 x3 dx 4x does not have an elementary antiderivative. −1 55. (a) 0 1 1  2  4 2 2  2 3 51. (a)   1 cos 2x 2 (b) The intersection points are difficult to determine by hand. 5  c −3 (c) Area  3 c 1.201538. −1 57. Fx   x 0 4 cos x  x 2 dx 6.3043 where c   1 t2 t  1 dt  t 2 4 x  0 x2 x 4 (b) F2  (a) F0  0 22 23 4 (c) F6  62  6  15 4 y y y 6 5 4 3 2 6 5 5 4 4 3 3 2 2 t −1 −1 59. F   6 1  2 3  cos 1 4 5 6 −1 −1  2  sin d  2  2  (a) F 1  0   1  t 1 2 3 4 5 6 −1 −1 t 1 2 3 4 5 6 2  2 sin   2  (b) F 0  2 0.6366  (c) F 12  2  2 1.0868 y y y 3 2 1 2 − 21 − 21 1 2 1 3 2 3 2 1 2 1 2 θ − 21 − 21 1 2 1 θ − 21 − 21 1 2 1 θ 318 Chapter 7    4 61. A  2 4    9 x  12  x  5 dx  2 7 x  7 dx  2 2  Applications of Integration  6 4 4 4  5  x  16  x  5 dx 2  6 4 y = 29 x − 12 y 7   x 2  21x 4 2 6 (4, 6) 6 7  x  21 dx 2   7 2 x  7x 4   y = − 25 x + 16 4 2  7  7  14 (6, 1) x 6 2 8 10 −2 (2, −3) −4 63. Left boundary line: y  x  2 ⇔ x  y  2 y 4 Right boundary line: y  x  2 ⇔ x  y  2   3  y  2   y  2 dy 2 2 (4, 2) 1 x −4 2  (0, 2) 2 2 A y=x−5  4 dy  4y 2 2 −2 −1 (− 4, − 2)  8  8  16 −3 2 3 4 (0, − 2) −4 65. Answers will vary. If you let x  6 and n  10, b  a  106  60. (a) Area 60 0  214  214  212  212  215  220  223  225  226  0 210  3322  966 sq ft (b) Area 60 0  414  214  412  212  415  220  423  225  426  0 310  2502  1004 sq ft 67. f x  x 3 y 8 f  x  3x 2 6 4 At 1, 1, f  1  3. Tangent line: y  1  3x  1 or y  3x  2 A   x3 2 69. f x   (1, 1) x −4 −3 −2 The tangent line intersects f x  x 3 at x  2. 1 y = 3x − 2 2 1 2 3 4 f (x) = x3 −6  x 4 3x 2  3x  2 dx    2x 4 2 1 2 (− 2, − 8) 27  4 −8 1 x2  1 y (0, 1) 2x f  x   2 x  1 2 1 f ( x) = 2 x +1 3 4 (1, 21 ) 1 2 1 1 At 1, , f  1   . 2 2 1 4 y=− 1x+1 2 Tangent line: y  1 1 1   x  1 or y   x  1 2 2 2 The tangent line intersects f x   1 A 0 1 1   x1 x2  1 2 x 1 2 1 at x  0. x2  1  dx  arctan x  x4  x 2 1 0  3 0.0354 4 1 3 2 2 Section 7.1 71. x 4  2x 2  1 ≤ 1  x 2 on 1, 1   Area of a Region Between Two Curves 319 y 1 A 1 1  x 2  x 4  2x 2  1 dx 2 (0, 1) 1  1 x2  x 4 dx  x3 x5   3 5  x 1 ( 1, 0) 4  15 1 (1, 0) You can use a single integral because x 4  2x 2  1 ≤ 1  x 2 on 1, 1. 73. Offer 2 is better because the accumulated salary (area under the curve) is larger.  3 A 75.   9  x 2 dx  36 3  9  4   b dx  18 x2 9  b  x 2 dx  9 9  bx  x3  3 9b ( 9 0 2 9  a 0  4a  a  8a  8  0 a  4 ± 2 2 x 6 x2 2 2 2 0  4  x dx  4x  0 6 4 9b  a 10 9b  9b 77. Area of triangle OAB is 12 44  8. y 2 b, b) 6 Since 0 < a < 4, select a  4  2 2 1.172. b, b) 9 ( y 2 9  b 32  9 3 B A 3 27  2 9  b 32 9b b9 n  x 79. lim  →0 i1 9 2 1 9 3 4 x  x 2 dx  0  4 y  2  3 1 x x  2 3 0 f (x) 0.4 0.2 x2 (1, 0) 0.2 0.4 0.6 0.8 1.0 (0, 0)  5 7.21  0.58t  7.21  0.45t dt  0 x x 1  . 6 5 0.13t dt  0 83. (a) y1  270.31511.0586t  270.3151e0.05695t  0.13t 2  2 5 0  $1.625 billion (b) y2  239.97041.0416t  239.9704e0.04074t R E 600 500 400 300 200 100 t 2 4 6 8 10 12 Time (in years)  17 (c) Surplus   y1  y2  dt 926.4 billion dollars 12 (Answers will vary.) Expenditures (in billions) Receipts (in billions) 3 0.6 1 81. 2 3.330 3 4 i 1 where xi  and x  is the same as n n  x 1  x i2 x i a O 600 500 400 300 200 100 t 2 4 6 8 10 12 Time (in years) (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. a2 2 320 Chapter 7 Applications of Integration 85. 5%: P1  893,000e 0.05t 312%: P2  893,000e 0.035t  5 Difference in profits over 5 years: 893,000e 0.05t  893,000e 0.035t dt  893,000 0  0.05  0.035 e 0.05t e 0.035t 5 0 893,00025.6805  34.0356  20  28.5714 893,0000.2163 $ 193,156 Note: Using a graphing utility, you obtain $193,183. 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  6.25 0.16   6.25 (b) Area  2 (a) The value of k is given by  y2  y1 dx 0 y1  y 2 6.25 2 6.25  0.086.25  k 2 0.08x 2  3.125  x dx 0 k  3.125. 2  0.08x 3 3  3.125x  x2 2  6.25 0  26.510417 13.02083  161   2  18  5.908 2 89. (a) A 6.031  2  2 91. True (b) V  2A 25.908 11.816 m 3 (c) 5000V 500011.816  59,082 pounds 93. Line: y  A  76 0 3 x 7  95. We want to find c such that:   3x2  cos x  14  3 2   b 3x sin x  dx 7 2x  3x3  c dx  0 0 x 76  0 7 1 24 2   34 x 4  cx b 0 b2  34 b 4  cb  0 But, c  2b  3b3 because b, c is on the graph. b2  34 b 4  2b  3b3b  0 2.7823 4  3b2  8  12b2  0 y 9b2  4 1 b  23 1 2 (0, 0) π 6 −1 0 4π 3 c  49 x ( 76π , − 12 ( y y = 2x − 3x 3 c (b, c) x Section 7.2 Section 7.2  Volume: The Disk Method  1 1. V   1 x  1 2 dx   0 4 x 1  x 2  2x  1 dx   0  3. V    2 dx    4 x dx   1 1 5. V     x2  2 4 1  7. y  x 2 ⇒ x  y  4 y 0   y2  2 x 4  x 6 dx   0 4  2 dy    3 x x3  x 2 1 0  3  15 2 1 x 2 2  x 3 2 dx   0 V Volume: The Disk Method  5  7 x5 x7 1 0  2 35 9. y  x 2 3 ⇒ x  y 3 2  4 1 V y dy 0  1  y 3 2 2 dy   0 y 3 dy   0  8 0 11. y  x, y  0, x  4 (a) Rx  x, rx  0   4 V x 0 x dx  0  2  2 dx 4  (b) R y  4, r y  y 2 V 16  y 4 dy 0  2 x  4 2 0   1   16y  y 5 5  8 y 2 0  128 5 y 3 3 2 2 1 1 x 1 2 3 x 4 1 −1 2 3 4 −1 (c) R y  4  y 2, r y  0   (d) R y  6  y 2, r y  2 2 V V 0 6  y 2 2  4 dy 0 2    2 4  y 2 2 dy 2 16  8y 2  y 4 dy  0 32  12y 2  y 4 dy 0   8 1   16y  y 3  y 5 3 5 2 0  256 15   1   32y  4y 3  y 5 5 y y 4 3 3 2 2 1 1 x x 1 −1 2 3 1 −1 −2 2 3 4 5 2 0  192 5 4 y4 1 0   4 321 322 Chapter 7 Applications of Integration 13. y  x2, y  4x  x 2 intersect at 0, 0 and 2, 4. (b) Rx  6  x2, rx  6  4x  x 2 (a) Rx  4x  x 2, rx  x 2   4x  x 2 2  x 4 dx V 2 2 16x 2  8x 3 dx x 3  5x 2  6x dx  8 0 0  6  x 2 2  6  4x  x 2 2 dx 0 0    2 2 V 3x 16 3  2x 4  2 0 32 3   8  4  3x x4 5 3   3x 2 2 0 64 3  y y 4 5 3 4 3 2 2 1 1 x −1 1 2 3 −2 1 3 V 4  x 2  1 2 dx 3   3 V 0  x2  8x  15 dx 3 y 0  5  4 42  4  0 3  2 17. Rx  4, rx  4  15. Rx  4  x, rx  1   x −1  0 3 x3   4x2  15x 3 0  x −1 1 2 3 4 2  8 ln 4  1 1x  3 0  1 1 4   8 ln 4  1  dx   2 1 1x 8 1  dx 1  x 1  x2   8 ln1  x  3  18 1 1x y 3  4 3 2 32.485 1 x −1 1 2 3 4 −1 19. R y  6  y, r y  0   21. R y  6  y 2, r y  2 4 V V 0 4   2 6  y 2 2  2 2 dy 2  y 2  12y  36 dy  2 0    2 6  y 2 dy  y 4  12y 2  32 dy 0  y3  6y 3 208 3 2   36y 4  2 y 0 5 4  3  y5  4y 5 3   32y 2 y 0 4 384 5 3 2 1 2 x −1 1 x 1 −1 2 3 4 5 −2 −3 1 2 3 5 Section 7.2 23. Rx  1 x  1   3 V 0 3  1 25. Rx  , rx  0 x , rx  0 1 x  1 dx V    1  x  0   ln x  1 1 x 1 1 dx x1 0  4 2 3   1 V 2 dx ex  2 dx 0 1  4 e2x dx 0 1    e2x 2   y y 323 27. Rx  ex, rx  0 3 4    ln 4 4.355 Volume: The Disk Method  1 0  1  e2 1.358 2 2 y 2 1 2 x 1 1 2 3 4 −1 x 1 2 1 −2 3 −1 x 1 2 x 2  1  x 2  2x  5 29. 2x 2  2x  4  0 x2  x  2  0 x  2x  1  0 The curves intersect at 1, 2 and 2, 5.   2 V 0 2    8x2  20 x  24 d x      4x 3 10  8x 2 8  20x  24 dx  2 0   8   x 4  x 3  10x 2  24x 3 1 31. y  6  3x ⇒ x  6  y 3 3 2 2 −1   6 0  2 dy  9 36  12y  y 2 dy  y3  36y  6y 2  9 3 1  8   r 2 h, 3 4  0  Volume of cone  0  2 6 216  216  216  9 3   1 0  5 3 6  x −2 1 2 3 sin x2 dx 4 y 0 6 1 6  y 3   33. V   y (2, 5) 6 152 125 277   3 3 3   y 2 8   x 4  x3  10x 2  24x 3  x2  12  5  2x  x22 dx 2 3 4 x3 0 V  3 5  2 x  x 22  x 2  12 d x   x 1 3 4 5 6 1  cos 2x dx 2  1 x  sin 2x 2 2    3 2 1 0  2    2 2 Numerical approximation: 4.9348 x 1 2 3 324 Chapter 7   Applications of Integration 2  2 ex12 dx 35. V   37. V   1 39. V   2 0 2 arctan0.2x 2 dx 0 2   5 ex  2 dx 1.9686 15.4115 e2x2 dx 1   e2x2 2   2 1  2 e  1 2 Numerical approximation: 10.0359 41.    2 43. A 3 sin2 x dx represents the volume of the solid y 0 generated by revolving the region bounded by y  sin x, y  0, x  0, x   2 about the x-axis. Matches (a) 2 y 1 1 x 1 π 4 45. 2 x π 2 y y 4 −2 3 3 2 2 1 1 x −1 1 x 2 1 2 3 4 The volumes are the same because the solid has been translated horizontally. 4x  x 2  4  x  22 1 47. Rx  x, rx  0 2  6 V 0 1 2 x dx 4 4   y r V 3 2 r y = r2 − x2 r 2  x 2 dx r  3  x 12  49. Rx  r 2  x 2, rx  0 y  6 0 1  18 1 Note: V   r 2h 3 1  326 3  18 x 1 −1 −2 2 3 4 5 6  2 r 2  x 2 dx 0 x   1  2 r 2x  x 3 3 r 0 1 4  2 r 3  r 3  r 3 3 3 (−r, 0) (r, 0) Section 7.2 r y y yr 1 , R y  r 1  , r y  0 H H H 51. x  r    h V r 1 0 y H  2  h dy   r 2 1 0  h2   r2 h  H   r 2h 1   2 0 1 2 x 2  x 8 dx  64  h3  3H 2 0 h2 h  H 3H 2   9 25   x 4 2  x dx  0  2x 5 x 6  64 5 6  25  x 2 dx  8 5 6 25  x 2 dx 18 25x  25 3    4 0 −6 0  30 −4 x −2 2 4 25 9  3 9  y 2 dy 25 y3 9y  9 3   6 3 4 0 2 6 −2 x 2 −4 4 503 500,000  3  ft 3 3 2500  y 2  2 dy   50 y 60 40  2500  y 2 dy  y3 3 y0 20 −60 50   2500y    2500y0   y0 −60 y03 250,000  3 3 y 3 250,000 1 500,000   2500y0  0  4 3 3 3 −20 −40 50 When the tank is one-fourth of its capacity: y 0  50 2 x3 5 Volume of water in the tank:  0 V y 5 57. Total volume: V   2 5 (b) R y  9  y 2, r y  0, x ≥ 0 3  60 y0  5 18 25 x r 3 55. (a) Rx  25  x 2 , rx  0 5 V h h −r 2 2 H  325 y 2 1 y  2 y 2 dy H H 1 2 1 3 y  y H 3H 2   r2 y  53. V   Volume: The Disk Method 125,000  7500y0  y03  250,000 y03  7500y0  125,000  0 y0 17.36 Depth: 17.36  50  32.64 feet When the tank is three-fourths of its capacity the depth is 100  32.64  67.36 feet. x 20 40 60 4 6 326 Chapter 7  Applications of Integration   b h 59. (a)  r 2 dx (b)  (ii) 0 1 a b x2 b2 (c)  dx (iv) y r r 2  x 2 (0, a) (h, r) (iii) y y y=r  2 dx is the volume of a sphere with radius r. is the volume of an ellipsoid with axes 2a and 2b. is the volume of a right circular cylinder with radius r and height h.  r 2 2 y=a 1− x b2 y= r 2 − x2 x  h (d)  0 rx h (−b, 0) (e)  (i) is the volume of a right circular cone with the radius of the base as r and height h. y (r, 0) (b, 0)   r 2 dx (−r, 0) x x r R  r 2  x 2   R  r 2  x 2   dx (v) 2 2 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) r 2 − x2 R+ y= r x h R x R− r2− x2 −r x r y 61. 4 3 2 x 2 3 4 Base of cross section  x  1  x 2  1  2  x  x 2 (a) Ax  b 2  2  x  x 2 2 (b) Ax  bh  2  x  x 21  4  4x  3x 2  2x 3  x 4  2 V 1 4  4x  3x 2  2x 3  x 4 dx   1 1  4x  2x2  x 3  x 4  x 5 2 5 2 + x − x2 2 + x − x2  2 V 2 1  81 10 1  2  x  x 2 dx  2x  1 2 + x − x2 x2 x3  2 3  2 1  9 2 Section 7.2 63. Volume: The Disk Method 327 y 1 3 4 1 2 1 4 x 1 4 1 2 1 3 4 3 y Base of cross section  1   3 (a) A y  b 2  1   y   1 V 0 3 1  y  dy 2 1− 3 y  1 1 V  8 0 1 1  2y 1 3  y 2 3 dy   81  1  3 y 1 1 1 (b) A y   r 2   2 2 2 2 1− 3 3 1  y  dy  2 2 3  y 2  1   8 10 80 y 0  3 3  y  y 4 3  y 5 3 2 5  1 0  1 10 1−  1 1 3 y 3 y (c) A y  bh  1    23 1    2 2  V 3 4 3 y 2 1     3 1 4 1  0 3 dy  4 y 3 y  1   3 y 1  3 y 2 (d) A y  ab  21    1     2 2 2 2 V 3 y 2  3 3 1  10 40  2  1 0 3 y 1  dy  2 a 1− 3 1− y 3 y b 1− 65. V     3 R 2 r 2 R 2 r 2 R 2 r 2  2 1− y R 2  x 2  2  r 2  dx R 2  r 2  x 2 dx 0   2 R 2  r 2 x   x3 3  2 R 2  r 23 2  4   R 2  r 2 3 2 3 R r R2 − r 2 R  R 2 r 2 0 R 2  r 23 2 3  3  y 1 67. V   0 y 2 dy   y3 3  1 0   3  1   2 10 20 328 Chapter 7  Applications of Integration 1 69. V    1  y dy 1 x 2  x 4 dx 71. V   0 0  x3  x5   y  y2 2   3  5  1  1 2  5 1 0 1 1 2 15   1 0   2   When a  2: x2   y2  1 represents a circle.     r V  2 a=2   2  3 3 1 2 0 1 1  6 x  R ± r2  y2 R  r2  y2  R  r2  y2 0 2 2  dy r a=1 1  y2  y3  77. (a) x  R2  y 2  r 2 y −1   75. (a) When a  1: x  y  1 represents a square. 1  y  y 2 dy 0  3  1 73. V    2 4Rr2  y2 dy 0 x r  8R r2  y2 dy 0 −1 y   (b) y  1  x a1a  1 A2 1 1  xa1a dx  4  1 1  x a1a dx x 0 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. R  r (b) r2  y2 dy is one-quarter of the area of a circle 0 of radius r, 14 r2. V  8R14  r 2  2 2 r 2R Section 7.3 Volume: The Shell Method 3. px  x, hx  x 1. px  x, hx  x  V  2 xx dx V  2  2 x 3 3  2  0 4 16 3  2 5. px  x, hx  x2  V  2 3  2 x  4 0 4 52 0  128 5   y 2 4 x 3 dx 0   45 x  7. px  x, hx  4x  x 2  x 2  4x  2x 2 y 2 2 x 32 dx 0  V  2 xx dx 0 0    4 2 x4x  2x 2 dx 4 0 2  8 2  4 0 1 −1 3 2x 2  x 3 dx x 1 2 3  2  1 2  4 x 3  x 4 3 4 2 0 16  3 1 −1 x 1 2 3 Section 7.3 9. px  x  4  4x  4 x2 V  2  x  4x  4x dx  2 1   ex 2 ex 2 x ex 2 dx 22 dx y 1 3 4 0 2 x4 4 3  2  x  2x 2 4 3 1 329 2 2 2 1 2 0  x 0 2 3   1 V  2 3 2 1 11. px  x, hx  y hx  4  4x  x 2 Volume: The Shell Method  −1 1 2 3 0 2 8 3  ...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture