Answer
$$V=\frac{9}{4} \pi$$
Work Step by Step
The solid is bounded by the paraboloid $z=1 +2x^{2}+2y^{2}$.
If we put $z = 7$ in the equation of the paraboloid, we get $2x^{2}+2y^{2} = 6 \text { or } x^{2}+y^{2} = 3 $.
This means that the $xy-$plane intersects the paraboloid in the circle $x^{2}+y^{2} = 3 $ in the first octant, so the volume is
$$
\begin{aligned}
V& =\iint_{x^{2}+y^{2} \leq 3 ,\, x\geq 0 ,\, y\geq 0}\left[7-\left(1+2 x^{2}+2 y^{2}\right)\right] d A\\
\end{aligned}
$$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $; thus the disk D is given by:
$$ D=\left\{(r, \theta) | 0 \leq r \leq \sqrt {3}, \quad 0 \leq \theta \leq \frac{\pi }{2}\right\} $$
and, the required volume is given by:
$$
V =\iint_{x^{2}+y^{2} \leq 3 ,\, x\geq 0 ,\, y\geq 0}\left[7-\left(1+2 x^{2}+2 y^{2}\right)\right] d A \\
=\int_{0}^{\pi / 2} \int_{0}^{\sqrt{3}}\left[7-\left(1+2 r^{2}\right)\right] r d r d \theta\\
=\int_{0}^{\pi / 2} d \theta \int_{0}^{\sqrt{3}}\left(6 r-2 r^{3}\right) d r \\
=[\theta]_{0}^{\pi / 2}\left[3 r^{2}-\frac{1}{2} r^{4}\right]_{0}^{\sqrt{3} }\\
=\frac{\pi}{2} \cdot \frac{9}{2}\\
=\frac{9}{4} \pi.
$$
