1.
Here is the complete text of the problem, and Marilyn's answer, as it
appeared in my newspaper.
You have a solid sphere and bore a perfect hole
through the center of it. This leaves a cylindrical hole
that is 6 inches long. What is the volume of the
remainder of the sphere?
--Stephen Chase,
Takoma Park, Md.
What an amazing problem! It appears to be missing
enough data for a solution, but it turns out that the
volume of the remainder of the sphere is always the
same, regardless of the size of the sphere!
Whether you bore a slim hole or a fat one in your
6-inch sphere, you always wind up with 113.09724
cubic inches of sphere remaining.
2. I recall saying the solution is interesting, and that is that the result
is independent of the size of the sphere.
I hadn't worked out the problem when I posted it - I just wanted to be able
to throw away the paper I was saving with a good conscience.
A colleague asked me about the problem, so I too attacked it. I found it
a bit challenging, but also eventually solved it with the disk method of
calculus.
To describe my ideas to my colleague, and liking any opportunity to play
with web pages, I started posting my diagrams as .gif images. Then I saw a
chance to play with the solid modeling of my art program and expanded my
work into a site, albeit not polished, on the problem. It's at
http://www.chelmsford.com/home/mccc/MVS-Problem/
Warning: the mccc may change to mahler soon if my ISP adheres to a request I
made a while ago.
3. The list posting which stated that only a calculus solution had been found
caused me to look up mensuration formulas in my ancient CRC Standard
Mathematical Tables, where I found the formula for "the volume of a
spherical segment" given as
(1/3)*pi*h^2(3r - h)
where h is the height of the segment and r is the radius of the sphere.
This permits a non-calculus solution (posted on the same site).
4. Marilyn makes two mistakes in her solution. 36*pi is 113.09734, rounded,
not the value she gave. And she refers to "your 6-inch sphere" which doesn't
fit the statement or solution of the problem.
5. It is indeed interesting that the solution is independent of the size of
the sphere. The way I came to be comfortable with it is to realize that for
a sphere the size of the earth, you would have to "drill out" practically
all of the earth to be left with a cylinder 6 inches high, and it's
thickness would be minuscule. Still, however, not an intuitive result.
6. It is such a good problem I would think it was in some calculus texts.
And I'm sure that none of the problems sent in to Marilyn are original - but
I'm still appreciative that they appear in the popular press.
Phil Mahler
Middlesex CC
Bedford, MA
****************************************************************************
* To post to the list: email mathedcc@archives.math.utk.edu *
* To unsubscribe, send mail to: majordomo@archives.math.utk.edu *
* In the mail message, enter ONLY the words: unsubscribe mathedcc *
* Words in the Subject: line are NOT processed! *
* Archives at http://archives.math.utk.edu/hypermail/mathedcc/ *
****************************************************************************