I see I should restate the problem; there was an omission as presented.
"You have a sphere. You drill a hole right through the center and measure the cylinder that the hole forms. The side of that cylinder is six inches long [or three, or two, or whatever]. How much volume is left in the sphere?"
ASSUMING THAT THE PROBLEM HAS A SOLUTION -- not always a valid assumption but we have to make it -- since they've been so adamant about leaving out the other sizes, they mustn't really matter. The volume left behind (volume-of-sphere minus volume-of-hole) must be the same, no matter what the diameter of the sphere is: a larger-diameter hole will require a larger-diameter sphere to make the straight side measure six inches! And if that's so then you can specify whatever numbers makes your life easiest. Conveniently, the smaller the hole, the closer the remaining volume is to the volume of the sphere itself, and with a microscopically-small hole the volume removed is microscopic and negligible, and with a zero-size hole what's left is volume-of-sphere minus volume-of-hole, i.e. volume-of-sphere minus zero, or just plain ol' volume-of-sphere = 4/3 times pi times radius cubed. The diameter of the sphere will be the height of a zero-diameter hole, and the radius half of that, in this case 3 inches.
You have to take it as a leap of faith that the problem is honest and so we can make those assumptions. Or you can invoke a mathematician to prove it. (I'm a cardiologist, Jim, not a mathematician!) But the problem has been around long enough that I suspect it's true as presented.
(It does make it easier to have the problem stated correctly...)
So, eta, you were right the first time. As the problem was originally given, the answer is a formula: 4/3 times pi time the cube of (1/2 the height of the side of the hole).
And if we're reduced to having me as Resident Mathematician, it's a good thing this is a Words board!
