Answer
$\dfrac{162\pi}{5}$
Work Step by Step
Here, we have $z=9-x^2-y^2 \implies z=9-r^2$
We need to set up the integral.
$I=\int_0^{\pi} \int_{0}^{3}\int_{0}^{9-r^2} (r )(r) dz dr d\theta=\int_0^{\pi} \int_{0}^{3} [zr^2]_{0}^{9-r^2}dr d\theta$
$=\int_0^{\pi} \int_{0}^{3} (9r^2-r^4)dr d\theta$
$=\int_0^{\pi} [3r^3-\dfrac{r^5}{5}]_{0}^{3} d\theta$
$=\int_0^{\pi} [3(3^3-0)-(\dfrac{(3)^5}{5}-0)] d\theta$
$=[\dfrac{162\theta}{5}]_0^{\pi}$
Thus, we have $I=[\dfrac{162(\pi-0)}{5}]=\dfrac{162\pi}{5}$
