Volume of the cone by integration

Text

A cone has base radius $3$ and height $6$. Obtain its volume by revolving the line $y=\tfrac{x}{2}$, $0\le x\le 6$, about the $x$-axis.

Solution

$V=\pi\int_{0}^{6}\left(\frac{x}{2}\right)^{2}dx=\frac{\pi}{4}\left[\frac{x^{3}}{3}\right]_{0}^{6}=\frac{\pi}{4}\cdot 72=18\pi\approx56.549.$ This matches $V=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\pi\cdot9\cdot6=18\pi$. $\boxed{V=18\pi\approx56.549}$