Given a sphere of radius R which has been drilled down its center.
Center the sphere at the origin. The drilled-out ring is height h.
What is the volume of the ringlike remainder?
Since this reeks of an undergrad calc problem, we take that approach.
Answer: Vsphere - Vcylinder - 2 * Vendcap (1)
Vsphere = 4/3 pi R^3
Vcylinder = 2 h pi r^2
where r^2 = R^2 - h^2
For Vendcap, we do basic integration. We sum the disks making up the endcap.
Vendcap = Integral from h (at the bottom of the endcap) to R (at the top) of
pi r^2 with respect to y
where r^2 = R^2 - y^2
ie R
pi * Integ R^2 - y^2 dy
h
Now, this yields pi * ( y R^2 - y^3/3 )
and we evaluate this at y=R and y=h and take their difference.
This yields pi * ( R^3 - R^3/3 - h R^2 + h^3/3 ) .
Now, remembering to subtract twice this value when plugging into (1),
we find that all terms except 4/3 pi h^3 fall out. Ie, the volume
depends only on the height; and we note that the dependancy is exactly
that of a sphere's volume on its radius. (Cymbal flourish)
(Thus the surprise typical of a good pedantic calc problem if you get the
right answer.)
