Answer
$$\dfrac{9\pi(8\sqrt{2}-7)}{2}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^3_0 \int^{\sqrt{18-r^2}}_{r^2/3} dz \space r \space dr \space d \theta = \int^{2\pi}_0 \int^{3}_0[r \times (18-r^2)^{1/2}-\dfrac{1}{3} r^3] \space dr \space d\theta \\=\int^{2\pi}_0 [\dfrac{-1}{3} \times (18-r^2)+\dfrac{1}{12} r^4]^{3}_0 d\theta \\=\dfrac{9\pi(8\sqrt{2}-7)}{2}$$
