Answer
$$\dfrac{37\pi}{15}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{\pi}_0 \int^{\theta/\pi}_0 \int^{3\sqrt{4-r^2}}_{-\sqrt{4-r^2}} \space z \space dz \space dr \space d\theta =\int^\pi_0 \int^{\theta/\pi} _0 \dfrac{1}{2}(9-r^2)-(4-r^2)]r \space dr \space d \theta \\=\int^{\pi}_0 \int^{\theta/\pi} _0[4r-r^3] \space dr \space d\theta \\= 4 \times \int^\pi_0 [2r^2-\dfrac{r^4}{4}]^{\theta/\pi}_0 \\=4 \times \int^{\pi}_0 (\dfrac{2\theta^2}{\pi}-\dfrac{\theta^4}{4\pi^4})d\theta \\=\dfrac{37\pi}{15}$$
