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