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Hole in a Sphere


Date: 12/25/96 at 13:16:17
From: Vern Klein
Subject: A mathematical oddity...

Is anyone there acquainted with the mathematical oddity implied by 
this exchange that appeared in the "Ask Marilyn" column in _Parade_ 
magazine on 12-15-96?  We hope someone may be able to state the 
"proposition" in unambiguous terms.  Considering the mail volume, who 
knows when an answer might be had from Marilyn?
      
Quoting from the column of 12-15-96:

Recently you published a question about boring through a sphere and 
leaving a 6-inch cylindrical hole through the center.  What is the
volume of the remainder of the sphere?  You replied, Amazing!  The 
problem appears to be missing enough data - like the size of the 
sphere - for a solution, but it turns out that the volume of the 
remainder of the sphere is always the same, no matter what!  Whether 
you bore a slim hole or a fat one IN YOUR 6-INCH SPHERE, you'll always 
wind up with 113.09724 cubic inches of sphere remaining.  This is 
incorrect.  You inadvertently referred to the sphere as a 6-inch
sphere.  But as you said yourself, the size of the sphere is unknown.  
It is the cylindrical hole that is 6 inches long.  This factor 
accounts for the curious constancy of the net volume of the remainder 
of the sphere.

Marilyn responds:

The sentence should have read, "Whether you bore a slim hole or a fat 
one IN THIS CASE [meaning a 6-inch hole, not a 6-inch sphere], you'll 
always wind up with 113.09724 cubic inches of sphere remaining.

I (Vern) can make no sense of this whatever and feel I must not 
understand the problem as originally presented.  Any comment or 
reference to a source where we might find a statement of the problem 
will be much appreciated.

Thank you for your consideration and may the new year be good to all 
of you there!
           
Vern and Klein


Date: 12/26/96 at 13:46:30
From: Doctor Rob
Subject: Re: A mathematical oddity...

Vern,

This is one of my favorite oddities.  I have known about this one for
about 20 years, so it isn't very new.

The idea is that the six inches is measuring the height of the 
cylindrical surface of the hole in the piece that is left, not the 
height of the part removed.  A typical diagram might be something like 
the following (pardon the attempt at ASCII Art!):

      |<----2r--->|
                  A                      A
      ,- - - - - -.     -----            /|
     /|           |\      ^             / |
    / |           | \     |            /  |
   ,  |           |  .    |           /   |
   |  |     .O    |  |    6"         /    |
   |  |           |  |    |       2R/     |6
   `  |           |  '    |        /      |
    \ |           | /     |       /       |
     \|           |/      v      /        | (angle C = 90 degrees)
      `- - - - - -'     -----   ----------
      B           C             B    2r   C
   |<------2R------->|

Here I am using R for the radius of the sphere and r for the radius of 
the cylindrical hole.  There is a relationship between r and R given 
by the Pythagorean theorem in triangle ABC:  

(2r)^2 + 6^2 = (2R)^2, or r^2 + 9 = R^2

The volume of the remaining piece can be computed by integration since 
it is a solid of rotation obtained by revolving a segment of a circle 
(say the one bounded by the chord AC and the nearby arc of the circle 
in the above diagram) about the axis of the cylinder.  The result is 
V = 6*Pi*(R^2 - r^2 - 3), and when you use R^2 - r^2 = 3^2 = 9, you 
get V = 36*Pi = 113.09724 cubic inches.  This is completely 
independent of both R and r, somewhat surprisingly.

Another way of computing the volume of the remaining piece is to start
with the volume of the sphere, then subtract the volumes of the two 
caps cut off the top and bottom of the sphere, then subtract the 
volume of the right circular cylinder.  The cylinder has volume 
6*Pi*r^2, and each cap has volume Pi*((2/3)R^3 - 3*R^2 + 9), and the 
sphere has volume (4/3)*Pi*r^3.  You get the same result:

V = 6*Pi*(R^2 - r^2 - 3) = 36*Pi

If you are trying to find the volume of a 6-inch diameter sphere with 
no hole, this corresponds to a sphere with a hole with diameter zero.  
In fact, when this problem was posed to a famous mathematician, he 
gave the correct answer of 36*Pi almost immediately.  His reasoning 
was that if the problem made sense, the answer must be independent of 
the hole diameter, which he then set to zero.  The remaining volume 
was then the volume of a sphere with diameter 6 inches, which is 
(1/6)*Pi*6^3 = 36*Pi.

If you need more help, write back again.  Happy New Year to you too!

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

    
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