## Thomas' Calculus 13th Edition

$\pi(6\sqrt{2}-8)$
$\int^{2\pi}_0 \int^1_0 \int^{(2-r^2)^{-1/2}}_r$3 dz r dr $d\theta$ =$3\int^{2\pi}_0 \int^1_0 [r(2-r^2)^{-1/2}-r^2]$ dr $d\theta$ =$3\int^{2\pi}_0[-(2-r^2)^{1/2}-\frac{r^3}{3}]^1_0$ d $d \theta$ =$3\int^{2\pi}_0 (\sqrt{2}-\frac{4}{3})d\theta$ =$\pi(6\sqrt{2}-8)$