Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 5\pi $
Work Step by Step
We have in polar coordinates: $r = 2\left( {1 + \cos \theta } \right)$.
From Figure 2, we see that $\theta $ varies from $0$ to $\pi $. So, the domain description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2\left( {1 + \cos \theta } \right),0 \le \theta \le \pi } \right\}$
Using $x = r\cos \theta $, we evaluate the integral in polar coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^\pi \mathop \smallint \limits_{r = 0}^{2\left( {1 + \cos \theta } \right)} {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^\pi \left( {{r^3}|_0^{2\left( {1 + \cos \theta } \right)}} \right)\cos \theta {\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^\pi {\left( {1 + \cos \theta } \right)^3}\cos \theta {\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^\pi \left( {\cos \theta + 3{{\cos }^2}\theta + 3{{\cos }^3}\theta + {{\cos }^4}\theta } \right){\rm{d}}\theta $
Since ${\cos ^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$, so
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^\pi \cos \theta {\rm{d}}\theta + 4\mathop \smallint \limits_{\theta = 0}^\pi \left( {1 + \cos 2\theta } \right){\rm{d}}\theta $
${\ \ \ }$ $ + 8\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^3}\theta {\rm{d}}\theta + \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^4}\theta {\rm{d}}\theta $
$ = \frac{8}{3}\left( {\sin \theta |_0^\pi } \right) + 4\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_0^\pi $
${\ \ \ }$ $ + 8\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^3}\theta {\rm{d}}\theta + \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^4}\theta {\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 4\pi + 8\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^3}\theta {\rm{d}}\theta + \frac{8}{3}\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^4}\theta {\rm{d}}\theta $
Using Eq. (6) of the Table of Trigonometric Integrals in Section 8.2:
$\smallint {\cos ^n}x{\rm{d}}x = \frac{{{{\cos }^{n - 1}}x\sin x}}{n} + \frac{{n - 1}}{n}\smallint {\cos ^{n - 2}}x{\rm{d}}x$
we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 4\pi + 8\left( {\frac{{{{\cos }^2}\theta \sin \theta }}{3}|_0^\pi + \frac{2}{3}\mathop \smallint \limits_{\theta = 0}^\pi \cos \theta {\rm{d}}\theta } \right)$
${\ \ \ \ \ \ \ }$ $ + \frac{8}{3}\left( {\frac{{{{\cos }^3}\theta \sin \theta }}{4}|_0^\pi + \frac{3}{4}\mathop \smallint \limits_{\theta = 0}^\pi {{\cos }^2}\theta {\rm{d}}\theta } \right)$
Since $\mathop \smallint \limits_{\theta = 0}^\pi \cos \theta {\rm{d}}\theta = \sin \theta |_0^\pi = 0$
and
$\mathop \smallint \limits_{\theta = 0}^\pi {\cos ^2}\theta {\rm{d}}\theta = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^\pi \left( {1 + \cos 2\theta } \right){\rm{d}}\theta = \frac{1}{2}\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_0^\pi = \frac{\pi }{2}$
so,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 4\pi + \frac{8}{3}\left( {\frac{3}{8}\pi } \right) = 5\pi $
