Posted by Jeff on Tuesday, February 12, 2008 at 12:19am.
First you find the volume of the whole ball, then the volume of the round hole inside, then, you subtract the hole's volume from the whole ball's volume, leaving you with the volume of the resulting solid.
Perform a volume integration with the center of the sphere at the origin and the circular hole aligned with the z-axis. The integration will end at +/-a, where
a^2 = 10^2 - 5^2 = 75
a = 5 sqrt 3 = 8.6603
The volume will be twice the volume above the x-y plane.
V = 2*INTEGRAL pi (r^2 - 5^2) dz
(z = 0 to 8.6603)
r^2(z) = 100 - z^2
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