==> geometry/hole.in.sphere.s <==
The volume of the leftover material is equal to the volume of a 6" sphere.
First, lets look at the 2 dimensional equivalent of this problem. Two
concentric circles where the chord of the outer circle that is tangent
to the inner circle has length D. What is the annular area between the
circles?
It is pi * (D/2)^2. The same area as a circle with that diameter.
Proof:
big circle radius is R
little circle radius is r
2 2
area of donut = pi * R - pi * r
2 2
= pi * (R - r )
Draw a right triangle and apply the Pythagorean Theorem to see that
2 2 2
R - r = (D/2)
so the area is
2
= pi * (D/2)
Start with a sphere of radius R (where R > 6"), drill out the 6"
high hole. We will now place this large "ring" on a plane. Next to it
place a 6" high sphere. By Archemedes' theorem, it suffices
to show that for any plane parallel to the base plane, the cross-
sectional area of these two solids is the same.
Take a general plane at height h above (or below) the center
of the solids. The radius of the circle of intersection on the sphere is
radius = srqt(3^2 - h^2)
so the area is
pi * ( 3^2 - h^2 )
For the ring, once again we are looking at the area between two concentric
circles. The outer circle has radius sqrt(R^2 - h^2),
The area of the outer circle is therefore
pi (R^2 - h^2)
The inner circle has
radius sqrt(R^2 - 3^2). So the area of the inner circle is
pi * ( R^2 - 3^2 )
the area of the doughnut is therefore
pi(R^2 - h^2) - pi( R^2 - 3^2 )
= pi (R^2 - h^2 - R^2 + 3^2)
= pi (3^2 - h^2)
Therefore the areas are the same for every plane intersecting the solids.
Therefore their volumes are the same.
QED
There also is a meta-theoretic answer to this puzzle. Assume the puzzle
can be solved. Then it must be solvable with a hole of any diameter, even
zero. But if you drill a hole of zero diameter that is six inches long,
you leave behind the volume of a six inch diameter sphere.