Answer
$\dfrac{32\pi}{3}-4\pi\sqrt 3$
Work Step by Step
Conversion of rectangular to cylindrical coordinate system gives: $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
Here, $x=r \cos \theta; y=r \sin \theta, z=z$
Let $V=\int_0^{2\pi} \int_{0}^{1}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dz dr d\theta$
or, $=\int_0^{2\pi} \int_{0}^{1}[rz]_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dz dr d\theta=\int_0^{2\pi} \int_{0}^{1}2r(\sqrt{4-r^2}) dr d\theta$
Plug $k=4-r^2$ and $dk=-2rdr$
Thus, $V=\int_0^{2\pi} \int_{3}^{4} k^{1/2} dk d\theta=\int_0^{2\pi} [\dfrac{2}{3}k^{3/2}]_{3}^{4} k^{1/2} d\theta$
or, $V= [\dfrac{16}{3} \theta-2\theta \sqrt 3]_0^{2\pi}=\dfrac{32\pi}{3}-4\pi\sqrt 3$
