Volume of a Hemisphere Using Cavalieri's Theorem

Date: 09/09/99 at 09:00:29
From: Theron Pappas
Subject: Derive v = (2/3)pi R^3

The Volume of a Hemisphere (a classical application of Cavalieri's 
theorem):

Derive the formula v = (2/3)pi R^3 for the volume of a hemisphere of 
radius R by comparing its cross sections with the cross sections of a 
solid right circular cylinder of radius R and Height R from which a 
solid right circular cone of base radius R and height R has been 
removed.

Date: 09/09/99 at 13:08:02
From: Doctor Peterson
Subject: Re: Derive v=(2/3)pi R^3

Hi, Theron.

People are often told these days that you need calculus to find the 
volume of a sphere; but there are several ways to approach it. Here's 
what you have:

             ---------------------------
      -------                           -------
    +-               *****+*****               -+
    | -------   *****     |     *****   ------- |
    |   \   *---------------------------*   /   |
    |     **              |              **     |
    |    *  \             |     a       /  *    |
    |  **      \          +----------+      **  |
    | *          \        |        /          * |
    |*             \      |h     /             *|
    |*               \    |    /               *|
    *                  \  |  /                  *
    *                     X---------------------*
    *    solid         /     \        R         *
    |*               /         \               *|
    |*             /             \             *|
    | *          /                 \          * |
    |  **      /     hollow          \      **  |
    |    *  /                           \  *    |
    |     **                             **     |
    |   /   ****---------------------****   \   |
    | -------   *****           *****   ------- |
    +-               ***********               -+
      -------                           -------
             ---------------------------

(I've drawn a whole sphere, though your problem deals only with a 
hemisphere.)

If we cut through this figure and look at the cross section, it will 
look like this:

                     -----------
                -----           -----
            ----     ***********     ----
          --     ****           ****     --
         -    ***    .........../   ***    -
       --    *    ...     |    /...    *    --
      -    **   ..        |   /    ..   **    -
     -    *    .         a|  / b     .    *    -
     -    *   .           | /         .   *    -
    -    *   .            |/           .   *    -
    -    *   .            *----------------*-----
    -    *   .                        R.   *    -
     -    *   .                       .   *    -
     -    *    .                     .    *    -
      -    **   ..                 ..   **    -
       --    *    ...           ...    *    --
         -    ***    ...........    ***    -
          --     ****           ****     --
            ----     ***********     ----
                -----           -----
                     -----------

Here R is the radius of the sphere and of the circumscribing cylinder; 
a is the radius of a cross-section of the cone at the height where 
we've cut it; and b is the radius of a cross-section of the sphere.

You want to show that the area of the cross-section of the sphere,

     pi b^2

is the same as the area of the "washer shape' (annulus) that is the 
cross-section of the cylinder-with-a-cone cut out,

     pi R^2 - pi a^2

See if you can write equations for a and b in terms of the height h of 
the cross-section, and then prove what I just said. Cavalieri's 
theorem will finish the job for you.

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