Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = 0$
Work Step by Step
We have $f\left( {x,y} \right) = y$ and the region: ${x^2} + {y^2} \le 1$, ${\left( {x - 1} \right)^2} + {y^2} \le 1$.
In polar coordinates: $f\left( {r\cos \theta ,r\sin \theta } \right) = r\sin \theta $.
Notice that the description implies that the region is bounded by an unit circle centered at the origin and by another unit circle centered at $\left( {1,0} \right)$. It is illustrated in the figure attached.
1. Find the range of $\theta$
First we find the the angle $\theta$ where the two circles intersect:
${\left( {x - 1} \right)^2} + {y^2} = 1$
${x^2} - 2x + 1 + {y^2} = 1$
${x^2} + {y^2} - 2x = 0$
Substituting ${x^2} + {y^2} = 1$ in the equation above gives
$1 - 2x = 0$, ${\ \ \ \ }$ $x = \frac{1}{2}$
Substituting $x = \frac{1}{2}$ in ${x^2} + {y^2} = 1$ we obtain $y = \pm \frac{1}{2}\sqrt 3 $.
Using $\left( {x,y} \right) = \left( {\frac{1}{2},\frac{1}{2}\sqrt 3 } \right)$ and $\left( {x,y} \right) = \left( {\frac{1}{2}, - \frac{1}{2}\sqrt 3 } \right)$ we find the angles the rays make with the $x$-axis:
$\tan {\theta _1} = - \sqrt 3 $, ${\ \ \ \ }$ $\tan {\theta _2} = \sqrt 3 $
So, ${\theta _1} = - \frac{\pi }{3}$ and ${\theta _2} = \frac{\pi }{3}$.
2. Next, we find the range of $r$.
Recall from Example 2, the unit circle centered at $\left( {1,0} \right)$ has polar equation $r = 2\cos \theta $. Whereas, the unit circle centered at $\left( {0,0} \right)$ is $r=1$. From the figure attached, we see that $r$ varies from $r = 2\cos \theta $ to $r=1$.
Referring to the figure attached, we notice that the ray intersects the two circles depending on the angle of the ray. So, we divide ${\cal D}$ into three subregions and list the angular sectors in the following table:
$\begin{array}{*{20}{c}}
\theta &r&{{\rm{Region}}}\\
{ - \frac{\pi }{2} \le 0 \le - \frac{\pi }{3}}&{0 \le r \le 2\cos \theta }&{{{\cal D}_1}}\\
{ - \frac{\pi }{3} \le 0 \le \frac{\pi }{3}}&{0 \le r \le 1}&{{{\cal D}_2}}\\
{\frac{\pi }{3} \le 0 \le \frac{\pi }{2}}&{0 \le r \le 2\cos \theta }&{{{\cal D}_3}}
\end{array}$
Thus, the polar description of the region is ${\cal D} = {{\cal D}_1}\bigcup {{\cal D}_2}\bigcup {{\cal D}_3}$, where
${{\cal D}_1} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2\cos \theta , - \frac{\pi }{2} \le \theta \le - \frac{\pi }{3}} \right\}$
${{\cal D}_2} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1, - \frac{\pi }{3} \le \theta \le \frac{\pi }{3}} \right\}$
${{\cal D}_3} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2\cos \theta ,\frac{\pi }{3} \le \theta \le \frac{\pi }{2}} \right\}$
Using Eq. (4) of Theorem 1, and the linearity property of the double integral 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}_1}}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_3}}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = - \pi /2}^{ - \pi /3} \mathop \smallint \limits_{r = 0}^{2\cos \theta } \left( {r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{\theta = - \pi /3}^{\pi /3} \mathop \smallint \limits_{r = 0}^1 \left( {r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = \pi /3}^{\pi /2} \mathop \smallint \limits_{r = 0}^{2\cos \theta } \left( {r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = - \pi /2}^{ - \pi /3} \mathop \smallint \limits_{r = 0}^{2\cos \theta } {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{\theta = - \pi /3}^{\pi /3} \mathop \smallint \limits_{r = 0}^1 {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = \pi /3}^{\pi /2} \mathop \smallint \limits_{r = 0}^{2\cos \theta } {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{ - \pi /3} \left( {{r^3}|_0^{2\cos \theta }} \right)\sin \theta {\rm{d}}\theta + \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /3}^{\pi /3} \left( {{r^3}|_0^1} \right)\sin \theta {\rm{d}}\theta $
$ + \frac{1}{3}\mathop \smallint \limits_{\theta = \pi /3}^{\pi /2} \left( {{r^3}|_0^{2\cos \theta }} \right)\sin \theta {\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{ - \pi /3} {\cos ^3}\theta \sin \theta {\rm{d}}\theta + \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /3}^{\pi /3} \sin \theta {\rm{d}}\theta $
$ + \frac{8}{3}\mathop \smallint \limits_{\theta = \pi /3}^{\pi /2} {\cos ^3}\theta \sin \theta {\rm{d}}\theta $
Consider the integral: $\smallint {\cos ^3}\theta \sin \theta {\rm{d}}\theta $
By changing of variable, we write $u = \cos \theta $. So, $du = - \sin \theta d\theta $. The integral becomes
$\smallint {\cos ^3}\theta \sin \theta {\rm{d}}\theta = - \smallint {u^3}{\rm{d}}u$
Thus, the integral above becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - \frac{8}{3}\mathop \smallint \limits_{u = 0}^{1/2} {u^3}{\rm{d}}u + \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /3}^{\pi /3} \sin \theta {\rm{d}}\theta - \frac{8}{3}\mathop \smallint \limits_{u = 1/2}^0 {u^3}{\rm{d}}u$
$ = - \frac{8}{3}\mathop \smallint \limits_{u = 0}^{1/2} {u^3}{\rm{d}}u - \frac{1}{3}\left( {\cos \theta |_{ - \pi /3}^{\pi /3}} \right) - \frac{8}{3}\mathop \smallint \limits_{u = 1/2}^0 {u^3}{\rm{d}}u$
The second integral is zero. Reversing the limits of integration in the third integral gives
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - \frac{8}{3}\mathop \smallint \limits_{u = 0}^{1/2} {u^3}{\rm{d}}u + \frac{8}{3}\mathop \smallint \limits_{u = 0}^{1/2} {u^3}{\rm{d}}u = 0$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta = 0$.
