From: Lee Corbin (lcorbin@tsoft.com)
Date: Tue Oct 29 2002 - 00:14:45 MST
Spike writes
> >>Cooooool! Cool question John! I get 4*pi cubic
> >>inches, regardless of the diameter of the sphere.
> >>I did it with only one integral using cylindrical
> >>shells. {8-] spike
> >
> > Actually the answer is that the remaining material is always equal to the
> > volume of a 6 inch sphere [ (4/3)*pi*r^3 where r =3]...
>
> Ja, I found the mistake in my integral. I get 36*pi,
> which is in fact the volume of a 6 inch diameter sphere.
> Way cool! spike
Yeah, the joke is that *that* is how one is supposed to
solve the problem.
There is a class of problems where you *assume* that
the asker knows what he is talking about. In this
case, since he didn't provide any other information,
you are supposed to be entitled to assume an infinitesimal
hole in the sphere, of 6 inches!
Then the answer is immediate: find the volume of a
6 inch sphere.
An extraordinarily similar one is the following:
A ten unit long line just fits inside an annulus
by being tangent to the inner circle and a chord
of the outer one. What is the area of the annulus?
I wish I knew more problems like that (hint, hint).
Also, I wonder if there already exists a name for
this class of math problems. If not, let's invent
one!
Lee
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