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 \( y = m x \) 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 thikness \( \Delta x \), 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 (I) and write
\[ V = \displaystyle m^2 \pi \int_a^b x^2 dx \]
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
\[ r = m \; a \] and \[ R = m \; b \]
Hence
\[ a = \dfrac{r}{m} \] and \[ b = \dfrac{R}{m} \]
Substitute in (II)
\[ \qquad V = \dfrac{1}{3} m^2 \pi \left(\left(\dfrac{R}{m}\right)^3 - \left(\dfrac{r}{m}\right)^3\right) \]
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 division of polynomials in two variables, \( \dfrac{\left(R^3 - r^3\right)}{R-r} \) may be simplified as
\[ \dfrac{\left(R^3 - r^3\right)}{R-r} = R^2 + r R + r^2 \]
We now substitute the above in (III) to obtain the final formula for the volume of the frustum
\[ \boxed {V = \dfrac{\pi h}{3} \left( R^2 + r R + r^2 \right) } \]