Old Boniface he took his cheer, Then he bored a hole through a solid sphere, Clear through the center,...

The volume of the leftover material is equal to the volume of a 6" sphere.

First, let's 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 Archimedes' 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.

[ Editor: the two-dimensional version, posted in 1994 with an added
restriction: ]

You have two concentric circles, of radii r1>r2.  You wish to find the
area of the ring-shaped region between r1 and r2.  There is a problem,
though: you don't know r1 or r2, but you only know the length of the
chord s which is tangent to the inner circle.  You know that there is a
formula A(s) which gives the area of this ring solely in terms of s,
but you don't know what the formula is.  And, most importantly, you
*don't* know Pythagoras' Theorem, the Laws of Sines and Cosines, or any
trigonometry at all (and no, you can't derive it).  The only things you
know are the formulae for the areas of a circle, a rectangle, and a
triangle.

Find A(s).

-- The Archon <archon@physics.physics.wm.edu>

If the area depends only on the length of the chord, consider the
degenerate case where r2 = 0, so that the chord becomes the diameter
of the outer circle.  Then A(s) = pi s^2 / 4.

-- Dave Dodson <dodson@convex.com>