## Thomas' Calculus 13th Edition

$32\pi$
We know that the paraboloid will intersect the xy-plane when: $9-x^2-y^2=0$ Solving, we get: $x^2+y^2=9$ Now, calculate the volume: $V=4\int^{\pi/2}_0 \int^3_1 \int^{9-r^2}_0$ dz r dr $d\theta$ =$4\int^{\pi/2}_0 \int^3_1 (9r-r^3)dr$ $d\theta$ =$4\int^{\pi/2}_0 [\frac{9r^2}{2}-\frac{r^4}{4}]^{3}_1 d\theta$ =$4\int^{\pi/2}_0 (\frac{81}{4}-\frac{17}{4})d\theta$ =$64 \int^{\pi/2}_0 d\theta$ =$32\pi$