Answer
$\dfrac{32\pi}{3}-4\pi\sqrt 3$
Work Step by Step
Consider $V=\int_0^{2\pi} \int_{0}^{1}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}[rz]_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}\times r dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}2r(\sqrt{4-r^2}) \times dr d\theta$
Suppose $a=4-r^2$ and $da=-2rdr$
$=\int_0^{2\pi} \int_{3}^{4} a^{(1/2)} da d\theta$
$=\int_0^{2\pi} [(2/3) a^{3/2}]_{3}^{4} \times a^{1/2} d\theta$
$= [\dfrac{16}{3} \theta-2 \times \theta \sqrt 3]_0^{2\pi}$
$=\dfrac{32}{3}\pi-4\sqrt 3 \pi$
