Answer
The solid is bounded by the paraboloids $z= 6 -x^{2}-y^{2} $ and $z= 2x^{2}+2y^{2} $. The required volume is given by: $$ \begin{aligned} V &=\iint_{x^{2}+y^{2} \leq 2} \left( 6 -x^{2}-y^{2}-( 2x^{2}+2y^{2} ) \right) d A \\ &=6 \pi. \end{aligned} $$
Work Step by Step
The solid is bounded by the paraboloids $z= 6 -x^{2}-y^{2} $ and $z= 2x^{2}+2y^{2} $.
The two paraboloids intersect when $ 2x^{2}+2y^{2}= 6 -x^{2}-y^{2} $ or $ x^{2}+y^{2}= 2 . $
So the volume is $$ \begin{aligned} V& =\iint_{D}(Z_{1}-Z_{2}) d A\\ &=\iint_{x^{2}+y^{2} \leq 2} \left( 6 -x^{2}-y^{2}-( 2x^{2}+2y^{2} ) \right) d A\\ \end{aligned} $$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $, thus the disk D is given by: $$ D=\left\{(r, \theta) | 0 \leq r \leq \sqrt {2}, \quad 0 \leq \theta \leq 2\pi \right\} $$ and, the required volume is given by: $$ \begin{aligned} V &=\iint_{x^{2}+y^{2} \leq 2} \left( 6 -x^{2}-y^{2}-( 2x^{2}+2y^{2} ) \right) d A \\
&=\int_{0}^{2 \pi} \int_{0}^{\sqrt {2}} 3\left(2-r^{2}\right) r d r d \theta \\ &=\int_{0}^{2 \pi} d \theta \int_{0}^{\sqrt {2}}\left(6 r-3 r^{3}\right) d r \\
&=[\theta]_{0}^{2 \pi}\left[3 r^{2}-\frac{3}{4}r^{4}\right]_{0}^{\sqrt {2}} \\
&=2 \pi (6-3)\\
& =6 \pi.
\end{aligned} $$
