Consider the ordinate set of the function
on the interval
. Revolving this ordinate set about the
-axis generates a right circular cone. Using integration, compute the volume of this right circular cone. Show that the result agrees with our usual formula for the volume of a right circular cone, namely,
, where
is the area of the base.
Proof. First, the area of a cross-section of this solid of revolution is
. So, using the formula for the volume of a Cavalieri solid (Theorem 2.7 in Apostol), we have

Since the area of the base is
(since
is the height of
at
, and this is then rotated around the
-axis), we have,
![Rendered by QuickLaTeX.com \[ V = \frac{1}{3} a(B) h = \pi c^2 \frac{b^3}{3}. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-28008b4494d06a212e74f139b6e4e4f7_l3.png)