Answer
$$16\pi $$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume=\int^{2\pi}_0 \int^2_0 \int^{4-r\sin\theta}_0 \space dz\space r \space dr \space d\theta \\=\int^{2\pi}_0 \int^2_0 (4r-r^2\sin\theta) dr \space d\theta \\=8 \times \int^{2\pi}_0 (1-\frac{\sin\theta}{3}) \space d\theta \\=16\pi $$
