Answer
$\frac{4\pi{r^{3}}}{3\sqrt 3}$
Work Step by Step
The cylinder has volume
$V$ = $\pi{y^{2}}(2x)$
$x^{2}+y^{2}$ = $r^{2}$
$y^{2}$ = $r^{2}-x^{2}$
$V(x)$ = $\pi(r^{2}-x^{2})(2x)$ = $2\pi(r^{2}x-x^{3})$ where $0$ $\leq$ $x$ $\leq$ $r$
$V'(x)$ = $2\pi(r^{2}-3x^{2})$
$V'(x)$ = $0$
$2\pi(r^{2}-3x^{2})$ = $0$
$x$ = $\frac{r}{\sqrt 3}$
$V(0)$ = $V(r)$ = $0$
so there is a maximum when $x$ = $\frac{r}{\sqrt 3}$ and
$V\left(\frac{r}{\sqrt 3}\right)$ = $\pi\left(r^{2}-\frac{r^{2}}{3}\right)\left(\frac{2r}{\sqrt 3}\right)$ = $\frac{4\pi{r^{3}}}{3\sqrt 3}$
