Consider a sphere of radius R and a cylinder, whose axis passes through the
sphere's centre, of radius r < R; in describing them, I'll take the axis
direction as vertical
. The cylinder cuts the sphere into two parts, the
central core
(think of an apple) and an outer ring. The core is 2.R
tall, of course; while, with h = √(R.R −r.r), the ring's height is
2.h. Now let's consider the volume of the outer ring, first by brute-force
integration:
in which x = s.cos(a), y = s.sin(a); and the integrand is left with no a-dependence, so we get
For each z, compute the integral over s's variation:
That inner integral is just the change in s.s/2 between its bounds
in which R.R −r.r = h.h is a constant; and the integral of z.z is just the change in z.z.z/3
with the surprising result that the volume doesn't depend on R and r other than via h.
This leads to a question, sometimes asked: What's the volume of the outer
part, given only that its height is 2.h ?
for some specified h. This
implicitly claims that the answer doesn't depend on R and r, from which we can
infer that the answer must be the one you'd get if r were 0, requiring R = h, so
4.π.h.h.h/3 has to be the answer. The tricky part is deciding that the
answer does, indeed, only depend on R and r via h.
So let's go back to what that brute-force integral is saying: at each z value, the s-integral is π.(h.h −z.z), which is exactly the area of the h-radius sphere's cross-section in the horizontal plane at the given z; i.e. our ring's cross-sectional area at each height is the same as that of the h-sphere. So picture the ring's cross-section at a given height; it's a difference between two circles, the inner one of radius r from the cylinder and an outer one, from the sphere, whose radius depends on the z value; the square of that outer radius is R.R −z.z and π times this is the area of the outer circle; subtracting the inner circle's π.r.r, we're left with π.(h.h −z.z), exactly as in the cross-section of the h-sphere.
Written by Eddy.