1. ## Cylinder/Sphere

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 ring-shaped solid that would remain.

Then, show that the volume will depend on only the height of the ring when its tallest.

2. 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 ring-shaped 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 .= . . . y .= .√R² - x²

The region cut off by y = d is revolved about the x-axis.
The intersections are: .(±h, d)
. . Note that: . .= .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!)