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.

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