Answer
$\frac{64 \pi \sqrt{2}}{3}=v$
Work Step by Step
the volume is determined by
$V=2 \int_{0}^{2 \pi} \int_{1}^{3} \sqrt{9-r^{2}} r d r d \theta$
integrate with respect to $r$
$-\int_{0}^{2 \pi}\left[\frac{\left(9-r^{2}\right)^{3 / 2}}{3 / 2}\right]_{1}^{3} d \theta=v$
$-\left(9-1^{3 / 2}\right)-\frac{2}{3} \int_{0}^{2 \pi}\left[\left(9-3^{2}\right)^{3 / 2}\right] d \theta=v$
$v=-\frac{2}{3} \int_{0}^{2 \pi}-[8]^{3 / 2} d \theta$
$v=\frac{2(8)^{3 / 2}}{3} \int_{0}^{2 \pi} d \theta$
$(2 \pi-0)\frac{2(8)^{3 / 2}}{3}=v$
$v=\frac{4 \pi\left(2^{3 / 2}\right)}{3}=\frac{4 \pi\left(2^{4}\right)(\sqrt{2})}{3}$
$\frac{64 \pi \sqrt{2}}{3}=v$
