A 'Boring' Volume Problem or "If You Find Yourself in a Hole, Stop Digging!"

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)

I always assigned one or both of these to my BC Calculus classes, most often as Extra Credit problems. Because of the extra points they could earn, most students tried these and submitted solutions. My feeling is that if a student could do both of these by both methods, they really understood disk and shell!

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?

'Rigor Mathis' - A Calculus Paradox or...

Update: I've added another 'paradox' in the comments. With the AP Calculus (BC) Exam looming, AP teachers may want to share this with their students for review.

[The following AP level question is designed for upper level students.]

Why do mathematicians have to be so rigid, um, I mean, rigorous?

Here's an AP Calculus problem brought to my attention this morning by one of our outstanding Calculus teachers, Mr. D. He found this in an AP Review book.

[Rather than play with the symbols, I'll 'write it out']:

The definite integral of sec²(x) from x = 0 to x = 3pi/4 is?
It was multiple choice and the answer given was -1.

I shared this problem with my AP group later on in the morning and I asked them why that answer makes no sense. R.J. immediately replied, "The answer can't be negative since sec²(x) is never negative."
Of course, but let's work it out!

[I intentionally did it incorrectly at first]:
By the Fundamental Theorem, the integral equals tan(3pi/4) - tan(0) = -1!

What's going on here! I was gratified that one of my students recognized that the function sec²(x) has an infinite discontinuity at x = pi/2, so the original integral is improper. When we integrate from 0 to pi/2, then from pi/2 to 3pi/4, and apply the rigorous limit definition of an improper integral, we see that the integral diverges! If anything, the 'area' is infinite or unbounded.

Using these kinds of 'paradoxical' examples and asking students to 'FIND THE ERROR' is a wonderful device many educators use to deepen student understanding of mathematics and demonstrate the need to be rigorous!

Now why isn't the definite integral of 1/x from -1 to 1 equal to zero, since the region in the first quadrant 'clearly cancels' the part in the 3rd quadrant?? Hmmm... I'll bet some of you could explain this and find many other such 'paradoxes'!