## Thomas' Calculus 13th Edition

$\frac{37\pi}{15}$
$\int^{\pi}_0 \int^{\theta/\pi}_0 \int^{3\sqrt{4-r^2}}_{-\sqrt{4-r^2}}$ z dz dr $d\theta$ =$\int^\pi_0 \int^{\theta/\pi} _0 \frac{1}{2}(9-r^2)-(4-r^2)]r$ dr d $\theta$ =$\int^{\pi}_0 \int^{\theta/\pi} _0[4r-r^3]^drd\theta$ =$4\int^\pi_0 [2r^2-\frac{r^4}{4}]^{\theta/\pi}_0$ =$4\int^{\pi}_0 (\frac{2\theta^2}{\pi}-\frac{\theta^4}{4\pi^4})d\theta$ =$\frac{37\pi}{15}$