Solutions To Selected Problems

Chapter 6

Exercises 6.1

  1. 1.

    T

  2. 3.

    Answers will vary.

  3. 5.

    4π+π222.436

  4. 7.

    π

  5. 9.

    1/2

  6. 11.

    4.5

  7. 13.

    2-π/2

  8. 15.

    1/6

  9. 17.

    98

  10. 19.

    643

  11. 21.

    27/2

  12. 23.

    9/2

  13. 25.

    All enclosed regions have the same area, with regions being the reflection of adjacent regions. One region is formed on [π/4,5π/4], with area 22.

  14. 27.

    1

  15. 29.

    9/2

  16. 31.

    1/12(9-22)0.514

  17. 33.

    1

  18. 35.

    4

Exercises 6.2

  1. 1.

    T

  2. 3.

    Recall that “dx” does not just “sit there;” it is multiplied by A(x) and represents the thickness of a small slice of the solid. Therefore dx has units of in, giving A(x)dx the units of in3.

  3. 5.

    48π3/5 units3

  4. 7.

    π/6 units3

  5. 9.

    9π/2 units3

  6. 11.

    2π/15 units3

  7. 13.
    (a) π/2 (b) 5π/6 (c) 4π/5 (d) 8π/15
  8. 15.
    (a) 4π/3 (b) 2π/3 (c) 4π/3 (d) π/3
  9. 17.
    (a) 8π (b) 8π (c) 16π/3 (d) 8π/3
  10. 19.

    Placing the tip of the cone at the origin such that the x-axis runs through the center of the circular base, we have A(x)=πx2/4. Thus the volume is 250π/3 units3.

  11. 21.

    Orient the cone such that the tip is at the origin and the x-axis is perpendicular to the base. The cross-sections of this cone are right, isosceles triangles with side length 2x/5; thus the cross-sectional areas are A(x)=2x2/25, giving a volume of 80/3 units3.

Exercises 6.3

  1. 1.

    T

  2. 3.

    F

  3. 5.

    9π/2 units3

  4. 7.

    96π5

  5. 9.

    48π3/5 units3

  6. 11.

    768π7

  7. 13.
    (a) 4π/5 (b) 8π/15 (c) π/2 (d) 5π/6
  8. 15.
    (a) 4π/3 (b) π/3 (c) 4π/3 (d) 2π/3
  9. 17.
    (a) 16π/3 (b) 8π/3 (c) 8π (d) 8π
  10. 19.
    (a) Disk: π01[12-(y4)2]dy=π3 Shell: 2π01xx4dx=π3 (b) Disk: π01(x4)2dx=π9 Shell: 2π01y(1-y4)dy=π9.
  11. 21.
    (a) Disk: π-21[(-4x+8)2-(4x2)2]dx=1152π5 Shell: 2π04yydy+2π416y[(2-y4)+y2]dy=128π5+1024π5 (b) Disk: π04[1+y2]2-[1-y2]2dy+π416[1+y2]2-[1-(2-y4)]2dy=32π3+130π3 Shell: 2π-21(1-x)[(-4x+8)-4x2]dx=54π (c) Disk: π-21[(16-4x2)2-(16-(-4x+8))2]dx=1728π5 Shell: 2π04(16-y)[y]dy+2π416(16-y)[(2-y4)+y2]dy=2176π15+3008π15.

Exercises 6.4

  1. 1.

    In SI units, it is one joule, i.e., one newton-meter, or kgm/s2m. In Imperial Units, it is ft-lb.

  2. 3.

    Smaller.

  3. 5.
    (a) 500 ft-lb (b) 100-50229.29 ft
  4. 7.
    (a) 12dl2 ft-lb (b) 75 % (c) (1-2/2)0.2929
  5. 9.
    (a) 756 ft-lb (b) 60,000 ft-lb (c) Yes, for the cable accounts for about 1% of the total work.
  6. 11.

    575 ft-lb

  7. 13.

    0.05 J

  8. 15.

    5/3 ft-lb

  9. 17.

    fd/2 J

  10. 19.

    5 ft-lb

  11. 21.
    (a) 52,929.6 ft-lb (b) 18,525.3 ft-lb (c) When 3.83 ft of water have been pumped from the tank, leaving about 2.17 ft in the tank.
  12. 23.

    212,135 ft-lb

  13. 25.

    187,214 ft-lb

  14. 27.

    4,917,150 J

Exercises 6.5

  1. 1.

    Answers will vary.

  2. 3.

    499.2 lb

  3. 5.

    6739.2 lb

  4. 7.

    3920.7 lb

  5. 9.

    2496 lb

  6. 11.

    602.59 lb

  7. 13.
    (a) 2340 lb (b) 5625 lb
  8. 15.
    (a) 1597.44 lb (b) 3840 lb
  9. 17.
    (a) 56.42 lb (b) 135.62 lb
  10. 19.

    5.1 ft

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