## Thomas' Calculus 13th Edition

$\frac{4\pi(8-3\sqrt{3})}{3}$
We calculate the volume as follows: V=$8\int^{2\pi}_0 \int^1_0 \int^{\sqrt{4-r^2}}_0$ dz r dr $d\theta$ =$8\int^{2\pi}_0 \int^1_0 r(4-r^2)^{1/2}dr$ d $\theta$ =$8\int^{2\pi}_0 [-\frac{1}{3}(4-r^2)^{3/2}]^1_0 d\theta$ =$\frac{8}{3} \int^{2\pi}_0 (3^{3/2}-8)d\theta$ =$\frac{4\pi(8-3\sqrt{3})}{3}$