Volume of a cone: geometric method

We're on the trail of Archimedes. Last time we found a formula for the sum of the squares of the integers from 1 to n:

S = n*(n+1)*(2n+1)/6

We're going to use that in a derivation of the formula for the volume of a cone. To start with, we slice the cone horizontally. In the limit as the number of slices gets very large and each individual slice gets very thin, the triangular shaped wedge pieces and the little bit at the top won't contribute significantly to the volume.

Here is the diagram from the post:

+ /|\ / |h\ +--+--+ /| |h |\ / | |r1| \ +--+--+--+--+ /| |h |\ / | | r2 | \ +--+-----+-----+--+ /| |h |\ / | | r3 | \ +--+--------+--------+--+ /| |h |\ / | | r4 | \ +--+-----------+-----------+--+ R

As Dr. Peterson explains

If the cone has base radius R and height H, and we've cut it into N slices (including that empty slice at the top, with radius r0 = 0), then each cylinder will have height h = H/N, and radius r[k] = kR/N, where k is the number of the cylinder, starting with 0 at the top and ending with N-1 for the bottom cylinder.

The only thing that's tricky about this is the numbering. The triangle at the top will be ignored. This includes the cylinder numbered k = 0 with radius = kR/N = 0. The next section below it contains a cylinder with h for its height and r1 for its radius. Since we started counting at 0, the last of the N segments has k = N-1.

The volume of each individual cylinder will be:

= π * h * rk2 = π * H/N * (kR/N)2 = π * R2 * H * k2 / N3

The total volume will be the sum of these, for all k from 0 to N-1; since only k is different from one cylinder to the next, we can factor everything else out from the sum and get:

V = [π * R2 * H / N3 ] * Sum(k2) Sum(k2) = 0 + 1 + 4 + ... + (N-1)2

Use our formula from last time:

Sum(k2) = (N-1)*(N)*(2N-1) / 6

Divide by N³:

Sum(k2) = (1-1/N)*(1)*(2-1/N) / 6

As N gets very large this converges to 2/6 = 1/3, so we have finally:

V = π*R2*H / 3

Very nice.