Answer
The volume is $\frac{9\pi}{4}$.
Work Step by Step
Find the domain $R$ of the paraboloid $z=1+2x^2+2y^2$ under the plane $z=7$ in the first octant:
$z\leq 7$
$1+2x^2+2y^2\leq 7$
$2x^2+2y^2\leq 6$
$x^2+y^2\leq 3$
The last inequality describes a region inside circle center at $(0,0)$ with radius $\sqrt{3}$.
Then, $R=\{(r,\theta)|0\leq r\leq \sqrt{3},0\leq \theta\leq \pi/2$.
The volume of the solid bounded the paraboloid and the plane $z=7$ in the first octant is given by $V=\int_0^\sqrt{3}\int_0^{\pi/2}(7-(1+2r^2))rd\theta dr$.
Evaluate the volume:
$V=\int_0^\sqrt{3}\int_0^{\pi/2}(6r-2r3)d\theta dr$
$=\int_0^\sqrt{3}(6r-2r^3)\theta]_0^{\pi/2}dr$
$=\int_0^\sqrt{3}(3r-r^3)\pi dr$
$=(\frac{3r^2}{2}-\frac{r^4}{4})\pi ]_0^{\sqrt{3}}$
$=(\frac{9}{2}-\frac{9}{4})\pi-0$
$=\frac{9\pi}{4}$
