You have a 3x3x3 cube. You bore a circular hole, of diameter 1, from the center of the top face to the center of the bottom face. You bore a similar hole from the left face center to the right face center, and another hole from front center to back center. How much material was removed?

(In reply to Overcomplicated? by TomM)

Remember, they want the volume that's removed, not the volume that's left, so the volume of the cube (9) is irrelevant.

You have to add the three cylinders, but then those volumes that have been covered by 2 cylinders twice.  So subtract out three times the volume of a 2-cylinder Steinmetz solid. If the cylinders are called A, B andC, that takes care of (A and B), (A and C) and (B and C); but now the 3-cylinder Steinmetz solid has been added in three times and subtracted out three times, so it has to be added in again.

The purpose of the pictures was to get a better feel for the shapes of the various Steinmetz solids, with the idea that someone could integrate them without looking up the answers.

Taking the easy way out, however, the Steinmetz solid, according to the Wolfram site referenced previously, has

For two cylinders: V=16*r^3 / 3
For three cylinders: V=8*(2-sqrt(2))*r^3

Since r=1/2, the total volume sought is

3*3*pi*/4    (for the three cylinders)

- 3*16/(8*3)       (for the three 2-cylinder solids)

+ 8*(2-sqrt(2)) / 8     (for the 3-cylinder solid)

This comes out to approximately 5.65436990820392, confirming the result of 5.654, given in my Monte Carlo comment.

The above total simplifies to 9*pi/4 - sqrt(2).

 

Posted by Charlie on 2005-03-08 14:26:12