Tim writes: > I said the height of the _remaining_ portion of the sphere, with the > core drilled out as described, is 10 cm:
> "Appendix: A math puzzle. Imagine a solid sphere. Maybe the sphere is > made of plutonium. A drill bit is lowered onto the sphere, going right > through the center, centered on the center (that is, the drilled-out > core is not off-center in any way. What is left is a sphere with a > cylindrical section (and the two end caps) removed. The height of the > remaining part of the sphere is 10 centimeters. What is its volume?" Assuming Tim is telling the truth, the answer does not depend on the radius of the sphere whose remaining part has height 10 centimeters. It therefore suffices to consider the limiting case, in which the drill bit has vanishingly small diameter, and carves the core out of a sphere infinitesimally close to 10 centimeters high. The volume of what remains is then easily seen to be the same as the volume of a sphere with diameter 10 centimeters, which can be computed with the usual formula of (4/3)*Pi*R^3. -- Eric Michael Cordian 0+ O:.T:.O:. Mathematical Munitions Division "Do What Thou Wilt Shall Be The Whole Of The Law"
