Volume and Surface Area of a Cone Frustum

Date: 08/30/99 at 14:48:34
From: Chris Reddy
Subject: Cone surface area

I have looked at your examples of the different types of cones, but I 
am unable to figure out how you derived the formula for the volume and 
total surface area for the frustum of a right circular cone.

Date: 08/30/99 at 17:00:33
From: Doctor Peterson
Subject: Re: Cone surface area

Hi, Chris.

You haven't said where you saw a derivation of these formulas; I found 
these pages for the volume, and none for the area:

  http://mathforum.org/dr.math/problems/taylor5.6.98.html   
  http://mathforum.org/dr.math/problems/rizza08.09.99.html   

I'm going to assume that you are happy with the formulas for a cone, 
and only want to see how we can get from there to the formulas for a 
frustum. If you want more than I give you, feel free to write back.

First let's do the volume. Rather than repeat what the other pages 
explain, I'll try a slightly different approach, still using similar 
triangles. Here's my picture:

     --------------------- +P --------------------
       |                  /|\                  |
       |                 / | \                 |
       |                /  |  \ S-s            |H-h
       |               /   |   \               |
       |              /    |    \              |
       |             /*****|**r**\             |
       |            **    C+-----*+D--------------
      H|         S /  ***********  \           |
       |          /        |        \          |
       |         /         |         \         |
       |        /          |          \ s      |
       |       /           |           \       |h
       |      /            |            \      |
       |     /             |             \     |
       |    /    **********|**********    \    |
       |   /*****          |       R  *****\   |
     -----*               A+----------------+B----
           ******                     ******
                 *********************

We know R, r, and h, but not H, the total height of the cone from 
which the frustum was cut. If we can find it, then the volume of the 
frustum will be the volume of the whole cone, pi R^2 H/3, minus the 
volume of the cone we cut off the top, pi r^2 (H-h)/3.

The triangles PAB and PCD are similar, so we can write the equation

     AB   CD        R    r
     -- = --   or   - = ---
     PA   PC        H   H-h

Cross-multiplying [that is, multiplying both sides by H(H-h)], we get

      R(H-h) = rH

We can distribute the left side and collect H terms, then divide:

     RH - Rh = rH

     RH - rH = Rh

      (R-r)H = Rh

               Rh
           H = ---
               R-r

Now let's write the volume formula and substitute this formula for H:

         pi         pi
     V = -- R^2 H - -- r^2 (H-h)
          3          3

         pi
       = -- (R^2 H - r^2 H + r^2 h)
          3

         pi
       = -- [(R^2 - r^2) H + r^2 h]
          3

         pi              Rh
       = -- [(R^2 - r^2) --- + r^2 h]
          3              R-r

         pi               R
       = -- [(R^2 - r^2) --- + r^2] h
          3              R-r

We can write R^2 - r^2 as (R - r)(R + r) and cancel:

         pi
       = --- [(R + r) R + r^2] h
          3

         pi
       = --- [R^2 + Rr + r^2] h
          3

That's the formula.

Now let's work on the lateral surface area. The formula for a complete 
cone is:

     A = pi R S

where R is the radius and S is the slant height of the whole cone. For 
the frustum, we will subtract the area of the cut-off cone (whose 
slant height is S-s) from the whole:

     A = pi R S - pi r (S-s)

       = pi (RS - rS + rs)

       = pi ((R-r)S + rs)

By the same similar triangles as before, we can write

     AB   CD        R    r
     -- = --   or   - = ---
     PB   PD        S   S-s

Again solving for S,

      R(S-s) = rS

     RS - Rs = rS

     RS - rS = Rs

      (R-r)S = Rs

               Rs
           S = ---
               R-r

Now the area is

                  Rs
     A = pi ((R-r)--- + rs) = pi (Rs + rs) = pi(R+r)s
                  R-r

and we're done.

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