Volume of a Cone

Date: 01/29/2001 at 21:55:37
From: cari
Subject: Volume of cones

Dr. Math,

I know HOW to find the volume of a cone(1/3area of base times height 
divided by three) but my teacher wants to know WHY. 

We know that filling the cone with water can prove it, but how does 
it work with the actual shapes? If you put a cone inside a cylinder, 
then you obviously have extra room, but how do two other cones fit in 
there? I understand that they won't keep their regular shape, but 
please explain. 

Thank you,
Cari

Date: 01/29/2001 at 23:29:46
From: Doctor Peterson
Subject: Re: Volume of cones

Hi, Cari.

We have explanations of many such formulas in our archives; but the 
cone is probably one of the hardest to explain without using calculus. 
Here's one answer I've given about pyramids, which are closely 
related:

  Volume of a Pyramid
  http://mathforum.org/dr.math/problems/terence9.1.98.html   

You'll see there how you can fit three pyramids with the same volume 
into a prism; from there, geometrical knowledge lets us build up to 
any pyramid, and then to the cone.

The best proof I'm familiar with comes very close to the spirit of 
calculus, without requiring you to know any of it. Let's try doing 
this directly, rather than starting with a pyramid.

You can't actually fit three cones together into a cylinder. Instead, 
we can dismantle a cone into lots of little near-cylinders. Picture 
one of those baby toys that look like a cone made up of several rings 
stacked up; or imagine a cone sliced like a pineapple, and the slices 
trimmed to make flat cylinders. You can imagine that if you make the 
slices thin enough, the scrap from the trimming will be as little as 
you like; so the sum of the volumes of the cylinders will be very 
close to the volume of the cone itself.

How can we find the volume of those slices? Here's a cross-section of 
the cone, showing the slices:

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

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 volume of cylinder k will be

    pi r[k]^2 h  = pi (kR/N)^2 (H/N) = pi R^2 H * k^2/N^3

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 = pi R^2 H / N^3 Sum(k^2)
      = pi R^2 H / N^3 (0 + 1 + 4 + ... + (N-1)^2)

At this point I have to either do some magic and tell you the formula 
for the sum of squares, and hope you trust me, or try to convince you. 
The formula is:

    0 + 1 + 4 + ... + N^2 = N(N+1)(2N+1)/6

I show a proof by induction in the page I referred to above; another 
proof can be found here:

  Formula For the Sum Of the First N Squares
  http://mathforum.org/dr.math/problems/sandin2.20.98.html   

If we replace N with N-1, we get

    0 + 1 + 4 + ... + (N-1)^2 = (N-1)(N)(2N-1)/6

Put this into our formula and you get

    V = pi R^2 H (N-1)(N)(2N-1)/(6N^3)

      = pi R^2 H/6 (N-1)/N N/N (2N-1)/N

      = pi R^2 H/6 (1-1/N)(1)(2-1/N)

Now, if N is very large, 1/N is very small, in fact, as close to zero 
as you want if N is large enough. So to find the volume of the cone 
itself, we can just replace it with 0. (Proving this thoroughly is 
where calculus begins.) We get

    V = pi R^2 H/6 (1)(1)(2)
      = 1/3 pi R^2 H

Whew! There's the formula.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/