Engineering Board Exam Reviewer · Applications: Volume, Arc Length, Surface Area · hard

The base of a solid is the region between y=x2 and y=8-x2. Cross sections perpendicular to the x-axis are semicircles whose diameters lie in the base. Find the volume of the solid.

  1. 256 π/15 53.62
  2. 512 π/15 107.23
  3. 128 π/15 26.81
  4. 256 π/5 160.85

Answer

A. 256 π/15 53.62

Solution

Diameter =8-2x2, so radius =4-x2 and area = π2(4-x2)2. V= π2 -22(4-x2)2dx=π 02≤ft(16-8x2+x4 )dx=π≤ft(32-64/3+32/5 )=256π/15.

More Applications: Volume, Arc Length, Surface Area questions

  1. The base of a solid is the circle x2+y2=16. Cross sections perpendicular to the x-axis are squares. Find the volume of the solid.
  2. The region bounded by y=sqrt(x), the x-axis, and x=4 is revolved about the line y=2. Find the volume generated.
  3. The region bounded by y=x and y=x2 is revolved about the line x=2. Find the volume generated.
  4. The ellipse x2/9+y2/4=1 is revolved about the x-axis. Find the volume of the ellipsoid.
  5. The region bounded by y=x3, the y-axis, and y=8 is revolved about the y-axis. Find the volume generated.
  6. The region under y=e-x for x≥0 is revolved about the x-axis. Find the volume of the resulting solid.
  7. The region bounded by y=x and y=x2 is revolved about the x-axis. Find the volume of the solid generated.
  8. The region bounded by y=x2, x=0, x=2, and the x-axis is revolved about the x-axis. Find the volume generated.
  9. The region bounded by y=sqrt(x), x=0, and x=4 is revolved about the x-axis. Find the volume generated.
  10. The region bounded by y=x, x=0, and y=3 is revolved about the y-axis. Find the volume generated.

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