You are not logged in.

- Topics: Active | Unanswered

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

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:

I wrote:

Using Calculus:

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 physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

good.

The sum usually is proven by induction, but here's a direct way :

http://www.mathsisfun.com/forum/viewtopic.php?id=3146

IPBLE: Increasing Performance By Lowering Expectations.

Offline