Use the method of the disk around the x-axis to find the volume of a frustum.
In what follows, \( C \) represents the constant of integration.
Volume of Frustum Formula
Problem: Find the volume of a frustum with height \( h \) and radii \( r \) and \( R \) as shown below.
Solution to the problem:
A frustum may be obtained by revolving \( y = mx \) between \( x = a \) and \( x = b \) around the x-axis as shown below. The height \( h = b - a \).
Rotating a disk (red) of radius \( y \) (hence of area \( \pi y^2 \)) and thickness \( dx \), the volume \( V \) of the frustum may be written as:
\[ V = \int_a^b \pi y^2 \, dx \quad (I) \]The slope \( m \) is given by:
\[ m = \dfrac{R - r}{h} \]where \( h \) is the height of the frustum given by:
\[ h = b - a \]Substitute \( y \) by \( mx \) in equation (I) and write:
\[ V = m^2 \pi \int_a^b x^2 \, dx \]Evaluate the integral:
\[ V = m^2 \pi \left[\dfrac{1}{3} x^3 \right]_a^b \] \[ \qquad = \dfrac{1}{3} m^2 \pi (b^3 - a^3) \quad (II) \]Note that:
\[ r = m a \quad \text{and} \quad R = m b \]Hence:
\[ a = \dfrac{r}{m} \quad \text{and} \quad b = \dfrac{R}{m} \]Substitute these into equation (II):
\[ V = \dfrac{1}{3} m^2 \pi \left(\left(\dfrac{R}{m}\right)^3 - \left(\dfrac{r}{m}\right)^3\right) \]Simplify:
\[ V = \dfrac{1}{3 m} \pi \left(R^3 - r^3\right) \]Substitute \( m \) by \( \dfrac{R - r}{h} \) in the above and rewrite as:
\[ V = \dfrac{\pi h}{3} \dfrac{\left(R^3 - r^3\right)}{R - r} \quad (III) \]Note that using polynomial division in two variables, \( \dfrac{R^3 - r^3}{R - r} \) may be simplified as:
\[ \dfrac{R^3 - r^3}{R - r} = R^2 + rR + r^2 \]Substitute the above into equation (III) to obtain the final formula for the volume of the frustum:
\[ \Large \displaystyle \boxed{V = \dfrac{\pi h}{3} \left(R^2 + rR + r^2\right)} \]