Volume of a cone by revolution

Text

The region under the line $y=\tfrac12 x$, with $0\le x\le6$, is revolved about the $x$-axis. Compute the volume of the cone obtained and check it with $V=\tfrac13\pi r^{2}h$.

Solution

$V=\pi\displaystyle\int_0^6\Big(\tfrac{x}{2}\Big)^{2}dx=\dfrac{\pi}{4}\Big[\tfrac{x^{3}}{3}\Big]_0^6=18\pi\approx56.549$. Check: $r=3$, $h=6$, $V=\tfrac13\pi\cdot9\cdot6=18\pi$. ✓ $\boxed{V=18\pi\approx56.549}$