## Thomas' Calculus 13th Edition

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