Hello, Ideasman!
This is a classic problem . . .
A cylindrical drill, which has radius d, is used to drill a hole
through the center of a sphere which has radius R.
Determine the volume of the ringshaped solid that would remain.
Then, show that the volume depends on only the height of the ring. Code:

* * *
*::::::::::*
*    +    *
*  h * *
d * R
*  * *
  *     +     *  
*  *

*  *
*  *
*  *
* * *

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .______
We have the circle: .x² + y² .= .R² . → . y .= .√R²  x²
The region cut off by y = d is revolved about the xaxis.
The intersections are: .(±h, d)
. . Note that: .h² .= .R²  d²
. . . . . . . . . . . . . . . . . . . . . . . . . . .______
The volume of the ring is: .2 × π ∫ [(√R²  x²)²  d²] dx ... from 0 to h
We have: .V .= .2π ∫(R²  x²  d²) dx .= .2π ∫(R²  d²  x²) dx
Since R²  d² .= .h², we have: .V .= .2π ∫ (h²  x²) dx .= .2π(h²x  x³/3)
Evaluate from 0 to h: .V .= .2π(h³  h³/3) .= .4πh³/3
Since h is half the height of the hole (H): .h = ½H
Therefore: .V .= .4π(H/2)³/3 .= .πH³/6
The volume of the ring is a function of H only.
(The radius of the sphere and the size of the drill are irrelevant!)