Which of the following integrals correctly computes the volume formed when the region bounded by the curves x2 + y2 = 100, x = 6 and y = 0 is rotated around the y-axis?

Question

Which of the following integrals correctly computes the volume formed when the region bounded by the curves x² + y² = 100, x = 6 and y = 0 is rotated around the y-axis?

Answer 1

The question is worded somewhat ambiguously, given that the region in question is asymmetrical over the y-axis, ie the axis of rotation. The region bounded by x² + y² = 100 and y = 0 is the top half of a circle or radius 10 centered on the origin. The vertical line x = 6 cuts into that region that would otherwise extend all the way to x = 10. I will show the integral to compute the volume assuming that the region is also meant to stop at x = -6.

We can do this by means of cylindrical shells, integrating an expression representing 2πrh, which is the lateral surface area of a cylindrical shell:

∫₀⁶ 2πx√(100 - x²)dx

If x = 6 was not meant to be included, we could compute the volume by integrating with respect to y instead, using disks (in which case the integrand represents the area of a circle, πr²):

∫₀¹⁰ π · (100 - y²) dy

The latter integral should yield 2,000π / 3 , as that is the volume of a hemisphere with r = 10.