Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 0$
Work Step by Step
Referring to Figure 3, the lower circle has polar equation $r=1$.
Whereas, the upper circle has equation in rectangular coordinates: ${x^2} + {\left( {y - 1} \right)^2} = 1$.
Step 1. We find the polar equation of the upper circle using $x = r\cos \theta $ and $y = r\sin \theta $. Substituting these in the equation above gives
${r^2}{\cos ^2}\theta + {\left( {r\sin \theta - 1} \right)^2} = 1$
${r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta - 2r\sin \theta + 1 = 1$
${r^2} - 2r\sin \theta = 0$
$r\left( {r - 2\sin \theta } \right) = 0$
So, the equation of the upper circle in polar coordinates is $r = 2\sin \theta $.
Step 2. We find the rays where the two circles intersect
$r = 1 = 2\sin \theta $
$\sin \theta = \frac{1}{2}$
So, the two circles intersect at $\theta = \frac{\pi }{6}$ and $\theta = \frac{{5\pi }}{6}$.
Since ${\cal D}$ is bounded below by $r=1$ and bounded above by $r = 2\sin \theta $, the description is given by
${\cal D} = \left\{ {\left( {r,\theta } \right)|1 \le r \le 2\sin \theta ,\frac{\pi }{6} \le \theta \le \frac{{5\pi }}{6}} \right\}$
Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A$ in polar coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \mathop \smallint \limits_{r = 1}^{2\sin \theta } {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \left( {{r^3}|_1^{2\sin \theta }} \right)\cos \theta {\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \left( {8{{\sin }^3}\theta \cos \theta - \cos \theta } \right){\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} {\sin ^3}\theta \cos \theta {\rm{d}}\theta - \frac{1}{3}\mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \cos \theta {\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} {\sin ^3}\theta {\rm{d}}\left( {\sin \theta } \right) - \frac{1}{3}\mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \cos \theta {\rm{d}}\theta $
$ = \frac{8}{{12}}\left( {{{\sin }^4}\theta |_{\pi /6}^{5\pi /6}} \right) - \frac{1}{3}\left( {\sin \theta |_{\pi /6}^{5\pi /6}} \right)$
$ = \frac{8}{9}\left( {\frac{1}{{16}} - \frac{1}{{16}}} \right) - \frac{1}{3}\left( {\frac{1}{2} - \frac{1}{2}} \right) = 0$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = 0$.
