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