Answer
$81\pi$
Work Step by Step
Find the region $R$ such that the paraboloid $z=18-2x^2-2y^2$ (in polar coordinates $z=18-2r^2$) is above the $xy-$plane:
$z\geq 0$
$18-2x^2-2y^2\geq 0$
$2x^2+2y^2\leq 18$
$x^2+y^2\leq 9$
Then, $R$ is a disk center at $(0,0)$ with the radius $3$ or in polar coordinates $R=\{(r,\theta)|0\leq r\leq 3,0\leq\theta\leq 2\pi\}$.
The volume of the solid below the paraboloid and above the $xy-$plane is given by $V=\int_0^3\int_0^{2\pi}(18-2r^2)\cdot rd\theta dr$
$=\int_0^3\int_0^{2\pi}(18r-2r^3)d\theta dr$
$=\int_0^3(18r-2r^3)\pi\theta]_0^{2\pi}dr$
$=\int_0^3(36r-4r^3)\pi dr$
$=(18r^2-r^4)\pi]_0^3$
$=(18\cdot 9-81)\pi-0$
$=81\pi$
