## Trigonometry (11th Edition) Clone

$V=\frac{1}{2}\theta (r_{1}^{2}-r_{2}^{2})$
The base area represents the difference between two sectors of circles of radius $r_{2}$ and $r_{1}$ where $r_{1}\gt r_{2}$. The area of a sector of a circle is $A=\frac{1}{2}r^{2}\theta$ where $r$ is the radius of the sector of the circle and $\theta$ is the central angle in radians. Since the base area represents the difference between two sectors of circles of radius $r_{1}$ and $r_{2}$: $A=\frac{1}{2}r_{1}^{2}\theta-\frac{1}{2}r_{2}^{2}\theta$ $A=\frac{1}{2}\theta(r_{1}^{2}-r_{2}^{2})$ Multiplying this area by the height to find the volume $V$, $V=A\times h$ $V=\frac{1}{2}\theta(r_{1}^{2}-r_{2}^{2})\times h$ $V=\frac{1}{2}\theta (r_{1}^{2}-r_{2}^{2})$