## Thursday, January 24, 2008

### 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?

Anonymous said...

Dave is the height of the hole
the equivalent to the diameter
of the sphere or is the height
of the hole the height of the
cutout cylinder in the sphere?

Dave Marain said...

I knew the wording wasn't perfect, but then it never is!
The 6" height refers to the height of the cylindrical hole remaining and therefore is 'less' than the diameter. The quotes are part of the clever approach I mentioned in the post, but I'll be quiet now!

Anonymous said...

Pretty cool, Dave. The disk method seems to work fine. Don't you think the washer method would be somewhat more complicated in this case?

I still fail to see the reasoning behind your "well-defined" method though. Seems like circular reasoning to me, but then, I am usually totally clueless.

I tried also to do it in 3-D spherical coordinates as the difference in volume between a cone with a spherical bottom, and a cone with a flat bottom, but seemed to run into problems (as I am writing this I realize I got what I deserved: the two are certainly not equivalent).

Dave Marain said...

c--
I agree this is a cool problem! I'm guessing by 'disk' method you meant you drew the rectangle all the way to the y-axis, then subtracted the cylindrical hole after completing the calculus. I like the washer set-up for a different reason. The expression for dV simplifies to pi*( (R^2-y^2)-r^2) =
pi*((R^2-r^2) - y^2). The key to this whole problem of course is that R^2-r^2 can be expressed directly in terms of the height of the cylinder (i.e., the 'depth of the hole'). Of course, the same result occurs with the ordinary disk method followed by subtraction of the hole, however, for some reason I prefer the washer setup.

The fact that the expressions R^2 (needed to find the total volume) and r^2 (needed to find the volume of the cylindrical hole) are variable, yet the difference R^2-r^2 is constant should seem no more surprising than the equation x-y=6, i.e., x and y may vary, but it is certainly possible for their difference to be constant! In this problem, that seems geometrically counterintuitive.

The 'well-defined' method is a very famous mathematician's ploy. Assuming there is a unique answer to the problem, then the result should be independent of the values of R and r. So, why not allow r to approach 0, i.e., let the hole vanish, which is equivalent to assuming that h = the diameter. I'll say more later.

I wrote out the complete solution on a whiteboard and may simply take a picture of it, but it's not nearly as effective as a video with narration showing the step-by-step development. I'm working on that...

Anonymous said...

Hi Dave,

I think when I said washers, I actually meant the method of cylindrical shells, wherein you assume that the cylinder is vertical, and the integration is over cylindrical shells that cover the required volume.

TC.

Anonymous said...

I used to love these, but hadn't tried one in ages (high school??? or soon after).

Neat how easy it comes back - and how useful 'that little theorem' is, and how easy it is to get an extra factor of 2... (still trying to get rid of it)

Jonathan

Anonymous said...

I'm so clueless ... I've tried to
simplify the problem and came to a
point where the solution lies in
determining the radius of a circle
when only three points on the circle
are known. Each of them with one unknown coordinate.

To me this looks like I've made the
problem even harder. Or not?

Anonymous said...

Try to relate the radius of the sphere, the radius of the bore, and 3.

Jonathan