Answer
$$32\pi $$
Work Step by Step
The paraboloid will intersect the $ xy $-plane when $9-x^2-y^2=0 \implies x^2+y^2=9$
Our aim is to integrate the integral as follows:
$$ volume=(4) \times \int^{\pi/2}_0 \int^3_1 \int^{9-r^2}_0 dz \space r dr \space d\theta \\=(4) \times \int^{\pi/2}_0 \int^3_1 (9r-r^3)dr \space d\theta \\=(4) \times \int^{\pi/2}_0 [\dfrac{9}{2} \times r^2-\dfrac{r^4}{4}]^{3}_1 d\theta \\=(4) \int^{\pi/2}_0 (\dfrac{81}{4}-\dfrac{17}{4})d\theta\\=32\pi $$
