Answer
The volume of the region is $\frac{{512}}{3}$.
Work Step by Step
Recall that the double integral defines the signed volume between the graph of $f\left( {x,y} \right)$ and the $xy$-plane. That is, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A$.
Referring to the figure attached, we see that the solid region lies over the domain ${\cal D}$, which is a vertically simple region. The left and right boundaries of ${\cal D}$ are $x=-2$ and $x=2$, respectively. Whereas, the lower and upper boundaries are $y = {x^2}$ and $y = 8 - {x^2}$, respectively. Thus, the domain description of ${\cal D}$ is
${\cal D} = \left\{ {\left( {x,y} \right)| - 2 \le x \le 2,{x^2} \le y \le 8 - {x^2}} \right\}$
From the figure attached, we see that the solid is bounded below by $z=y$ and bounded above by $z=16-y$. Thus, by Theorem 3, $f\left( {x,y} \right) = 16 - y - y = 16 - 2y$.
We evaluate the volume of the solid as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - 2}^2 \mathop \smallint \limits_{y = {x^2}}^{8 - {x^2}} \left( {16 - 2y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 \left( {\left( {16y - {y^2}} \right)|_{y = {x^2}}^{8 - {x^2}}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 \left( {16\left( {8 - {x^2}} \right) - {{\left( {8 - {x^2}} \right)}^2} - 16{x^2} + {x^4}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 \left( {64 - 16{x^2}} \right){\rm{d}}x$
$ = \left( {\left( {64x - \frac{{16}}{3}{x^3}} \right)|_{ - 2}^2} \right)$
$ = 128 - \frac{{128}}{3} + 128 - \frac{{128}}{3}$
$ = \frac{{512}}{3}$
So, the volume of the region is $\frac{{512}}{3}$.
