# Find The Volume of a Frustum Using Calculus

Use the method of the disk around the x axis to find the volume of a frustum.

## 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 $y = m x$ 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
$\qquad V = \displaystyle \int_a^b \pi y^2 dx \quad (I)$

The slope $m$ is given by
$\qquad m = \dfrac{R - r}{h}$
where $h$ is the height of the frustum given by
$\qquad h = b - a$

Substitute $y$ by $mx$ in (I) and write
$\qquad V = \displaystyle m^2 \pi \int_a^b x^2 dx$

Evaluate the integral
$\qquad V = m^2 \pi \left[\dfrac{1}{3} x^3 \right]_a^b$

$\qquad = \dfrac{1}{3} m^2 \pi (b^3 - a^3) \quad (II)$

Note that
$\qquad r = m \; a$ and $R = m \; b$

Hence
$\qquad a = \dfrac{r}{m}$ 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
$\qquad V = \dfrac{1}{3 \; m} \pi \left(R^3 - r^3\right)$

Substitute $m$ by $\dfrac{R - r}{h}$ in the above and rewrite as

$\qquad V = \dfrac{ \pi h}{3} \dfrac{ \left(R^3 - r^3\right)}{R-r} \quad (III)$

Note that using division of polynomials in two variables, $\dfrac{\left(R^3 - r^3\right)}{R-r}$ may be simplified as
$\qquad \dfrac{\left(R^3 - r^3\right)}{R-r} = R^2 + r R + r^2$

We now substitute the above in (III) to obtain the final formula for the volume of the frustum
$\boxed {V = \dfrac{\pi h}{3} \left( R^2 + r R + r^2 \right) }$