This works just like calculus, except you don't actually use an integral.

The strategy is to find the ratio of the volumes of a cone and it's respective cylinder, since once you find that, that remains constant no matter how you scale them.

Let the radius of the base = r, and without loss of generality (you'll see why), assume that r is an integer. Let the height = r also (to spare us of complexity)

We can approximate the volume of the cone by adding the volumes of r discs, where each disk has an integral radius from 1 to r.

(Ex, the first disk has radius 1, the next has 2, the next has 3 etc... the last has radius r)

Let the height of each disk be 1.

The volume is therefore:

summation of k as k ranges from 1 to r of

times the sum of the first r perfect squares

, by the formula for the sum of the first r perfect squares.

The volume of a cylinder with the same base and height is

The ratio of the area of the cone to the cylinder is therefore

Note that if we had used a height other than r, the height would've canceled out anyway once we divided by the volume of the cylinder.

Obviously, the more disks we have (the bigger r is), the more accurate the ratio will be.

As r gets very very big,

get's very very small, and

get's very very small.

At this point, note that since r is getting very very big, the assumption that it is an integer get's less and less important, because we are taking an infinte number of "samples".

So effectively, the ratio is

, and the volume of the cone is

the volume of the cylinder.

Volume of a cylinder is Area of base x Height, so the volume of a cone is 1/3 x Area of Base x Height

Again, these kinds of approximations work just like calculus and are the foundation of calculus, but you should be able to understand it without calculus.

PS - what is the latex tag on this forum? I'm sick of writing without latex...

PPS - Thanks John