Answer
$$16\pi $$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume=\int^{2\pi}_0 \int^2_0 \int^{4-r\cos\theta-r\sin\theta} \space dz \space r \space dr \space d\theta \\=\int^{2\pi}_0 \int^{2}_0 [4r-r^2(cos\theta+\sin\theta)]dr \space d\theta \\=\dfrac{8}{3} \times \int^{2\pi}_0(3-\cos\theta)-\sin\theta d\theta \\=16\pi $$
