Largest area of a revolved quarter-circle arc

Fix $0<r<1$ and take the curve $y=\sqrt{r^2-x^2}$ for $r^2\le x\le r$. Rotating it about the $x$-axis gives a surface of area $S_r$. If $S=\sup_{0<r<1}S_r=q\pi$ with $q$ rational, find $q$.

$q=\tfrac{8}{27}$. On a circle of radius $r$ the arc-length factor gives $y\sqrt{1+y'^2}=r$, so $S_r=\int_{r^2}^{r}2\pi r\,dx=2\pi(r^2-r^3)$ (a spherical zone). This is maximised at $r=\tfrac23$, giving $2\pi\cdot\tfrac{4}{27}=\tfrac{8}{27}\pi$.