Answer
$\dfrac{32\pi}{3}-4\pi\sqrt 3$
Work Step by Step
Here, we have $V=\int_0^{2\pi} \int_{0}^{1}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dz dr d\theta$
and $x=r \cos \theta; y=r \sin \theta, z=z$
Then, we get
$=\int_0^{2\pi} \int_{0}^{1}[rz]_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dz dr d\theta=\int_0^{2\pi} \int_{0}^{1}2r(\sqrt{4-r^2}) dr d\theta$
Plug in $p=4-r^2$ and $dp=-2rdr$
Then, we have $V=\int_0^{2\pi} \int_{3}^{4} \sqrt p dp d\theta$
or, $=\int_0^{2\pi} [\dfrac{2}{3}(p^{3/2})]_{3}^{4} \sqrt p d\theta$
or, $= [\dfrac{16}{3} \times \theta-2\theta \sqrt 3]_0^{2\pi}$
Hence, $V=\dfrac{32\pi}{3}-4\pi\sqrt 3$
