Solid of revolution of an arc

Text

The curve $y=\dfrac{x}{4}\sqrt{x(8-x)}$ (with $0\le x\le 8$) revolves about the $x$-axis. Compute the volume of the resulting solid.

Solution

$y^{2}=\dfrac{x^{2}}{16}\,x(8-x)$, hence $V=\pi\int_{0}^{8}y^{2}\,dx=\frac{\pi}{16}\int_{0}^{8}\left(8x^{3}-x^{4}\right)dx=\frac{\pi}{16}\left[2x^{4}-\frac{x^{5}}{5}\right]_{0}^{8}.$ $=\frac{\pi}{16}\left(8192-\frac{32768}{5}\right)=\frac{512\pi}{5}\approx321.699.$ $\boxed{V=\dfrac{512\pi}{5}\approx321.70}$