Volume of a Cone

Date: 02/19/99 at 11:35:04
From: Jodi Anderson
Subject: Volume of a Cone

I'm teaching a tenth grade geometry class, and I'm not happy with my 
discussion on finding the volume of a cone. We have actually measured 
the volume of a cone and compared it to the volume of a cylinder with 
the same height and radius, and found that it is about 1/3 the volume 
of the cylinder, but my students still want to know why. I do not 
remember ever seeing a proof of this, except possibly in Calculus.  
Do you have any suggestions for a proof of this formula?

Thanks.

Date: 02/19/99 at 13:16:48
From: Doctor Rob
Subject: Re: Volume of a Cone

Slice the cone into n pieces with n-1 equally spaced planes parallel to 
its base. Each of these pieces is a frustum of a cone, and has volume 
that is greater than the volume of a cylinder whose radius is the 
radius of the smaller base, and smaller than the volume of a cylinder 
whose radius is the radius of the larger base. That is because the 
small cylinder is contained in the frustum of the cone contained in the 
large cylinder. Here is a diagram of a cross-section of the kth piece, 
from the side:

      A   B               C               D   E
       o--o---------------o---------------o--o
       | /|               |               |\ |
       |/ |               |               | \|
       o--o---------------o---------------o--o
      F   G               H               I   J

The cross-section of the frustum is FJDB; of the larger cylinder it is 
FJEA; and of the smaller cylinder it is GIDB. A segment of the axis of 
the cone and the cylinders is CH, whose length is h/n, the height of 
each frustum and each cylinder. For the kth piece from the top, the 
radius of the smaller base BC = CD = r*(k-1)/n, and the radius of the 
larger base FH = HJ = r*k/n. Then, if V(k) is the volume of the kth 
frustum, we have the inequality

   Pi*[r*(k-1)/n]^2*(h/n) < V(k) < Pi*[r*k/n]^2*(h/n),
   (k-1)^2*Pi*r^2*h/n^3   < V(k) < k^2*Pi*r^2*h/n^3,

Now, add up these inequalities for k = 1, 2, 3, ..., n to get the
volume of the cone in the middle,

   V = V(1) + ... + V(n).

Divide by Pi*r^2*h, and multiply by n^3, and you have:

   0^2 + ... + (n-1)^2 < V*n^3/(Pi*r^2*h) < 1^2 + ... + n^2.

Now you have to know the sum of the first n squares is equal to
n*(n+1)*(2*n+1)/6, so the sum of the first n-1 squares is equal to
(n-1)*n*(2*n-1)/6. This can be proved by Mathematical Induction if you 
don't already know this formula. So,

   (n-1)*n*(2*n-1)/6 < V*n^3/(Pi*r^2*h) < n*(n+1)*(2*n+1)/6.

Now, divide by n^3, and rearrange things a little:

   (1-1/n)*(1-1/[2*n])/3 < V/(Pi*r^2*h) < (1+1/n)*(1+1/[2*n])/3.

Now recall that n could be any positive integer, and the above 
inequalities must be true. As n gets large, the quantity on the left 
gets closer and closer to 1/3, and the quantity on the right also gets 
closer and closer to 1/3. The difference between these two quantities 
is just 1/n. As n gets large, this difference gets small, approaching 
zero. In fact, if you want the left side to be larger than 1/3 - a, for 
some real number a > 0, just pick n > 1/(2*a), and if you want the 
right side to be smaller than 1/3 + a, just pick n > 1/(3*a). That 
means that for any positive real number a, we have

   1/3 - a < V/(Pi*r^2*h) < 1/3 + a

The only real number x that satisfies 1/3 - a < x < 1/3 + a, for every 
positive real number a, is 1/3 itself. Thus

   V/(Pi*r^2*h) = 1/3,
   V = (1/3)*Pi*r^2*h.

By the way, a very similar proof, depending on the same formula for 
the sum of the first n squares, works for pyramids, too.

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