Answer
$\frac{4\pi(8-3\sqrt{3})}{3}$
Work Step by Step
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 $
=$\frac{4\pi(8-3\sqrt{3})}{3}$
