I didn't really understand it until this post.. but now I do.

Going back to the calculus answer, in order to find the volume, you'd have the sphere, the cylinder, and two end caps. Adding the cylinder and the end caps gives you the total volume removed from the sphere.

The condition is that the cylinder must be 6 inches long. The diameter is not specified or necessary; it simply must be 6 inches long. Or high. Whichever variable makes you happy. Also, note that this is the cylinder

*not counting the end caps.*Consider a large sphere and a really, really large person wanting a 6 inch bracelet. You have a drill and various drill bits all infinitely long (or, for all practical purposes, long enough to drill through the sphere with room to spare) but with varying diameter (length isn't important; diameter is). In order to make that 6 inch bracelet, you would have to choose a drill bit with a diameter large enough such that when you drill a hole completely through the sphere (going through the center, of course), the cylinder that you took out (again, not counting the end caps) would be 6 inches long. Thus, when you lay what's left over (the bracelet) flat on a table, it will be 6 inches high.

That is, assuming, that the sphere is large enough to accomplish this.

Consider a sphere infinitesimally

*larger* than 6 inches in diameter (take the limit as diameter approaches 6). If you wanted to drill a hole through the sphere such that the cylinder which is removed (or, alternatively, the bracelet which is left behind) is 6 inches high, well, the sphere is already nearly 6 inches high itself, so you would need an infinitesimally thin drill. At 6 inches exactly, there's no way you can drill a hole which, when calculating the volume removed, would be composed of a 6 inch cylinder and two end caps.

Just my two cents. Hope it helps