Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V = \frac{{128}}{{15}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = x$ and the region ${\cal W}$ in the first octant ($x \ge 0$, $y \ge 0$, $z \ge 0$) above $z = {y^2}$ and below $z = 8 - 2{x^2} - {y^2}$. So, the triple integral is equal to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = {y^2}}^{8 - 2{x^2} - {y^2}} x{\rm{d}}z} \right){\rm{d}}A$
where ${\cal D}$ is the projection of ${\cal W}$ onto the $xy$-plane.
To evaluate the integral over ${\cal D}$, we must find the boundary of ${\cal D}$.
Step 1. Find the boundary of ${\cal D}$
We find the intersection of the surfaces $z = {y^2}$ and $z = 8 - 2{x^2} - {y^2}$ by solving the equation
${y^2} = 8 - 2{x^2} - {y^2}$
$2{x^2} + 2{y^2} = 8$
${x^2} + {y^2} = 4$
Since ${\cal W}$ is in the first octant ($x \ge 0$, $y \ge 0$, $z \ge 0$), we see that the projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$, a disk of radius $2$.
Step 2. Express ${\cal D}$ as a simple domain
We notice that ${\cal D}$ is both vertically and horizontally simple region. However, for convenience (since the limit of the integral involves the square root and the integrand is $f\left( {x,y,z} \right) = x$), we choose to describe ${\cal D}$ as a horizontally simple region. Thus,
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 2,0 \le x \le \sqrt {4 - {y^2}} } \right\}$
Step 3. Write the triple integral as an iterated integral and evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = {y^2}}^{8 - 2{x^2} - {y^2}} x{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{\sqrt {4 - {y^2}} } \left( {\mathop \smallint \limits_{z = {y^2}}^{8 - 2{x^2} - {y^2}} x{\rm{d}}z} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{\sqrt {4 - {y^2}} } x\left( {z|_{{y^2}}^{8 - 2{x^2} - {y^2}}} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{\sqrt {4 - {y^2}} } \left( {8x - 2{x^3} - 2{y^2}x} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \left( {\left( {4{x^2} - \frac{1}{2}{x^4} - {y^2}{x^2}} \right)|_0^{\sqrt {4 - {y^2}} }} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \left( {4\left( {4 - {y^2}} \right) - \frac{1}{2}{{\left( {4 - {y^2}} \right)}^2} - {y^2}\left( {4 - {y^2}} \right)} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \left( {16 - 4{y^2} - 8 + 4{y^2} - \frac{1}{2}{y^4} - 4{y^2} + {y^4}} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \left( {8 - 4{y^2} + \frac{1}{2}{y^4}} \right){\rm{d}}y$
$ = \left( {8y - \frac{4}{3}{y^3} + \frac{1}{{10}}{y^5}} \right)|_0^2$
$ = 16 - \frac{{32}}{3} + \frac{{16}}{5}$
$ = \frac{{128}}{{15}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V = \frac{{128}}{{15}}$.
