Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V = 0$
Work Step by Step
We have $f\left( {x,y,z} \right) = x$ and the region ${\cal W}:{x^2} + {y^2} \le z \le 4$.
Referring to the figure attached, we see that ${\cal W}$ is a region between the surface $z = {x^2} + {y^2}$ and the plane $z=4$. So, it is a $z$-simple region which lies over the disk of radius $2$, namely the domain ${\cal D}$ in the $xy$-plane. We notice that ${\cal D}$ is both vertically and horizontally simple region. However, for convenience, we choose to describe ${\cal D}$ as a horizontally simple region. Thus,
${\cal D} = \left\{ {\left( {x,y} \right)| - 2 \le y \le 2, - \sqrt {4 - {y^2}} \le x \le \sqrt {4 - {y^2}} } \right\}$
Thus, the triple integral is equal to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = {x^2} + {y^2}}^4 x{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } \left( {\mathop \smallint \limits_{z = {x^2} + {y^2}}^4 x{\rm{d}}z} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } x\left( {z|_{{x^2} + {y^2}}^4} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } x\left( {4 - {x^2} - {y^2}} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } \left( {4x - {x^3} - {y^2}x} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 2}^2 \left( {\left( {2{x^2} - \frac{1}{4}{x^4} - \frac{1}{2}{y^2}{x^2}} \right)|_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} }} \right){\rm{d}}y$
Notice that the power functions in $x$ have exponents that are even numbers, so the integrand in the integral above yields $0$, that is,
$\left( {2{x^2} - \frac{1}{4}{x^4} - \frac{1}{2}{y^2}{x^2}} \right)|_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } = 0$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V = 0$.
