Answer to old chestnut: hole in sphere:

The remaining volume is equal to that of a sphere six inches in diameter, or 4/3 pi (3 in.)3 = 36 pi cubic inches.

The answer turns out to not depend on the radius of the sphere, only on the height of the piece remaining. As the radius increases, the radius of the hole also increases, so that whether you have something the looks like a bead, or something that looks like a thin ring, it still has the same volume.

Having been told that you have enough information to solve the problem, you could assume that it must still be true if you drilled a zero-radius hole in a six-inch sphere, and come up with the above answer. However, I provide a proof below.

Proof (using calculus):

The volume equals the integral of the cross-sectional area of the ring (cross-sections taken perpendicular to the axis of the hole).

Each cross-section is an annulus (a circle with a smaller circle cut out of the middle). The radius of the cut out circle is a constant, equal to the radius of the outer circle at the top and bottom of the hole.

The radius of the outer circle varies. To calculate this, construct a right triangle with one leg stretching along the axis of the hole from the center of the sphere to the position of cross-section, the other leg in the plane of the cross-section from the axis to the edge of the sphere, and the hypotenuse is a radius.

First calculate the radius of the hole in terms of the radius of the sphere, R. At the end of the sphere, the vertical leg is 3 inches, the hypotenuse is R, so the radius of the hole is Rh=sqrt(R2-32).

Next, calculate the outer circle radius in terms of the height x above or below the center of the sphere and the radius of the sphere. This is Ro=sqrt(R2-x2).

The cross-sectional area is then pi Rh2 - pi Ro2 or
pi (R2-x2) - pi (R2-32) =
pi (32-x2) =
pi (9-x2).

Then the volume is the integral of this area from x = -3 to 3, or
pi (9x - x3/3) {at x = 3} - pi (9x - x3/3) {at x = -3}, or
pi (27 - 9) - pi (-27 + 9) =
pi (27 - 9 + 27 - 9) =
36 pi.

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