Solution
= Solution
In <cylindrical coordinates>, the two surfaces meet when $9-r^2=r^2$, giving $r=3/\sqrt2$ and $z=9/2$. The solid has $0\leq r\leq3/\sqrt2$, $0\leq\theta\leq2\pi$, and $r^2\leq z\leq9-r^2$. With <volume element> $r\,dz\,dr\,d\theta$,
$$
\boxed{V=2\pi\int_0^{3/\sqrt2}(9-2r^2)r\,dr
=2\pi\left[\frac92r^2-\frac12r^4\right]_0^{3/\sqrt2}
=\frac{81\pi}{4}}.
$$
The upper-minus-lower height is nonnegative throughout that radial interval.