Answer
$$\dfrac{3\pi-4}{6}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume =\int^{\pi/2}_0 \int^{\cos\theta}_0 \int^{3\sqrt{1-r^2}}_0 \space dz \space r \space dr \space d\theta \\=\int^{\pi/2}_0 \int^{\cos\theta}_0 3 \times \space r \space \sqrt{1-r^2} \space dr \space d\theta \\=\int^{\pi/2}_0 [-(1-r^2)^{3/2}]^{cos\theta}_0 \space d\theta\\=\int^{\pi/2}_0[-(1-cos^2\theta)^{3/2}+1] \space d\theta \\=\dfrac{\pi}{2}+\dfrac{2}{3}[\cos\theta]^{\pi/2}_0 \\=\dfrac{3\pi-4}{6}$$