## Thomas' Calculus 13th Edition

$16\pi$
$V=\int^{2\pi}_0 \int^2_0 \int^{4-r\cos\theta-r\sin\theta}$ dz r dr $d\theta$ =$\int^{2\pi}_0 \int^{2}_0 [4r-r^2(cos\theta+\sin\theta)]dr d\theta$ =$\frac{8}{3} \int^{2\pi}_0(3-\cos\theta)-\sin\theta d\theta$ =$16\pi$