I was recently asked this question by email "How would you prove that a Cone has one third the Volume of a Cylinder", and I responded like this:
Firstly, the cylinder is easy, just have lots of little discs, each disc has a radius of "r" and an area of πr², and we need to integrate over height:
V = ∫πr² dh = πr²h
That was fairly easy!
Now for the cone, the little disc's radius get smaller as you get higher. The rate they get smaller is a constant, the slope of the sides, which we can call s.
For simplicity I will turn the cone upside-down, so the disc's radius get bigger with height. Each disc will have a radius of "sh" and an area of π(sh)², and we need to integrate over height:
V = ∫π(sh)² dh = ∫πs²h² dh = πs² (1/3)h³
Now we know that the slope s = r/h, so: V = π(r/h)² (1/3)h³ = πr² (1/3)h
Which is one third of the cylinder's volume!
Now, I recall seeing proofs without calculus that did something similar, they turn the cone into "N" small discs, and add them up.
Disc number "i" will have a height of h/N, and a radius of r(i/N), with a volume of πr²(i/N)²(h/N) = πr²i²h/N³
Summing over "i" : V = ∑πr²i²h/N³ = πr²h/N³ ∑i²
The only thing in the way now is "∑i²", which I know can be simplified to a few terms, but am not sure how. If that could be done, hopefully we would see the formula come out to be πr² (1/3)h
So, I haven't totally solved this for you, but I hope I have helped.
I would like to do a page about this, can someone correct my work and give an easy way to complete the ∑i² term?
The sum usually is proven by induction, but here's a direct way :
IPBLE: Increasing Performance By Lowering Expectations.