Important Note: It took forever but I finally posted the detailed video explanation of this problem here.
Please don't gag on my feeble attempt at humor in the title (my wife actually had bought a sign with that quote -- it's hanging on the dining room wall).
There are a couple of classic volume problems in calculus which have always been my favorites:
- The Volume of the Torus Problem (using 2 methods: cylindrical shells and by disks)
- The Hole in the Sphere Problem (also by 2 methods)
In this post we will focus on the 2nd problem as it always seems to generate curiosity and interest. I'm guessing that most of you know the puzzle version of this question that was answered by Marilyn vos Savant in her Ask Marilyn column over a decade ago. It's just possible that some calculus student in some second semester class is feeling some anxiety over this problem!
Here's one version of that famous conundrum. There are many approaches here, even the clever mathematical approach of assuming that the problem is well-defined and therefore independent of the radii involved (I expect at least one of our readers to do it that way!).
A hole is drilled (bored) completely through a solid sphere, symmetrically through its center. If the resulting hole is 6 inches in height (or depth), show that the remaining volume must be 36π inches cubed.
That's right, the answer is independent of the radius of the sphere and the diameter of the hole! The total volume of the sphere and the volume removed however do depend on the radii. Note that the volume removed is a cylinder with two spherical caps.
The original problem was worded ambiguously in Marilyn's column and then clarified somewhat. My version is not perfect but hopefully you'll get the 'picture', although a real picture would be far better. I will probably do a video presentation of the solution and a discussion of the problem because the diagram and the math expressions are cumbersome and it's not worth the time to play with Draw programs or LaTeX right now. I plan on presenting in detail the disk-washer and cylindrical shells method using a general depth of h inches for the hole.
For now, have fun playing with this. This is a well-known problem and therefore searchable on the web but try it yourself first. Try to use calculus to set up the integral and if you're brave you'll evaluate those integrals without Mathematica or the TI-89! Can you see why the answer for the volume remaining depends only on the depth of the hole?