Answer
$\dfrac{162\pi}{5}$
Work Step by Step
$z=9-x^2-y^2$ and $z=9-r^2$
For cylindrical coordinate system: $I=\int_0^{\pi} \int_{0}^{3}\int_{0}^{9-r^2} (r) \times (r)\times dz dr d\theta$
$=\int_0^{\pi} \int_{0}^{3} [r^2z]_{0}^{9-r^2} \times dr d\theta$
$=\int_0^{\pi} \int_{0}^{3} r^2(9-r^2) \times dr d\theta$
$=\int_0^{\pi} \int_{0}^{3} (9r^2-r^4)dr d\theta$
$=\int_0^{\pi} [3r^3-\dfrac{r^5}{5}]_{0}^{3} d\theta$
$=[\dfrac{162\theta}{5}]_0^{\pi}$
$=\dfrac{162\pi}{5}$
