This incredible problem was first made famous by Martin Gardner with his ‘Mathematical Games’ column published in Scientific American throughout the 60s and 70s.
A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere?
No additional data is needed to solve this. (really!)
Without using calculus, one of the key equations to solving this problem is the volume of a spherical “cap”. That is the volume above a plane that has bisected a sphere. It is (1/6)πh(3r2 + h2) (where h is the altitude of the cap, and r is its radius). (See the mathworld page for more info)
The volume remaining in the sphere will be the volume of the sphere – (the volume of the “hole” + 2*(volume of the cap)). Refer to the drawing below for clarification.
The volume of the sphere is simply 4/3πR3. The volume of the hole is 6π(R2 – 9). The volume of both of the caps can be obtained by plugging in h = R – 3, and r = sqrt(R2 – 9)) .
After a long, boring simplification, we find that all the terms cancel to leave 36pi!
Try it for yourself.