Anyway, the solution to Flip's problem is as follows:
Let r be the sphere radius. In order to have a hole of length 6, the radius of the hole must be sqrt(r
2-9). After drilling the hole through the sphere, the remainder is a
solid of revolution formed by taking the area below y=sqrt(r
2-x
2) and above y=sqrt(r
2-9) from -3 to 3, and rotating around the x-axis.
Disk integration gives:
volume = π*int(-3..3) {sqrt(r
2-x
2)
2-sqrt(r
2-9)
2}dx = π*int(-3..3) {(r
2-x
2)-(r
2-9)}dx = π*int(-3..3) (9-x
2)dx = π*int(-3..3) sqrt(9-x
2)
2 dx
which is the volume of a sphere of radius 3cm (disk integration on y=sqrt(9-x
2) rotated around the x-axis), which is (4/3)π*3
3=36π cm
3.
What's interesting about the solution? It doesn't depend on r (the sphere radius). So, no matter what r is, the answer is the same as a volume of a sphere of radius 3cm (which is also the shape of the remaining sphere as r approaches 3cm).