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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/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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