Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}y{\rm{d}}x = 0$
Work Step by Step
We have $f\left( {x,y} \right) = x$ and the region $2 \le {x^2} + {y^2} \le 4$.
In polar coordinates we have
$f\left( {r\cos \theta ,r\sin \theta } \right) = r\cos \theta $
The description implies that the region is an annulus located between the disk of radius $\sqrt 2 $ and the disk of radius $2$. Thus, the region description in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|\sqrt 2 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
Using the Change of Variables Formula for ${\cal D}$, we evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = \sqrt 2 }^2 \left( {r\cos \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = \sqrt 2 }^2 {r^2}{\rm{d}}r} \right)$
$ = \left( {\sin \theta |_0^{2\pi }} \right)\left( {\frac{1}{3}{r^3}|_{\sqrt 2 }^2} \right) = 0$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = \sqrt 2 }^2 \left( {r\cos \theta } \right)r{\rm{d}}r{\rm{d}}\theta = 0$.