Answer
Please see the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\left( {{x^2} + {y^2}} \right)^{ - 1}}{\rm{d}}A \simeq 0.685$
Work Step by Step
We have $f\left( {x,y} \right) = y{\left( {{x^2} + {y^2}} \right)^{ - 1}}$ and the region ${\cal D}$ is given by
${\cal D} = \left\{ {\left( {x,y} \right)|y \ge \frac{1}{2},{x^2} + {y^2} \le 1} \right\}$
Using $x = r\cos \theta $ and $y = r\sin \theta $, we obtain in polar coordinates:
$f\left( {r\cos \theta ,r\sin \theta } \right) = \frac{{r\sin \theta }}{{{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta }} = \frac{{\sin \theta }}{r}$
From the figure attached, we see that the line $y = \frac{1}{2}$ intersects the upper half of the circle at two points. Substituting $y = \frac{1}{2}$ in ${x^2} + {y^2} = 1$ we find the corresponding angles at these points:
${x^2} + \frac{1}{4} = 1$
$x = \pm \frac{1}{2}\sqrt 3 $
$\tan {\theta _1} = \frac{{1/2}}{{\frac{1}{2}\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$, ${\ \ \ }$ ${\theta _1} = \frac{\pi }{6}$
$\tan {\theta _2} = \frac{{1/2}}{{ - \frac{1}{2}\sqrt 3 }} = - \frac{1}{{\sqrt 3 }}$, ${\ \ \ }$ ${\theta _2} = \frac{{5\pi }}{6}$
Now, the line $y = \frac{1}{2}$ in polar coordinates is $r\sin \theta = \frac{1}{2}$. So,
$r = \frac{1}{{2\sin \theta }}$
From the figure attached we see that the region is a part of a unit disk, so $r$ varies from $\frac{1}{{2\sin \theta }}$ to $1$.
Thus, the description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|\frac{1}{{2\sin \theta }} \le r \le 1,\frac{\pi }{6} \le \theta \le \frac{{5\pi }}{6}} \right\}$
Using the Change of Variables Formula for ${\cal D}$, we evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\left( {{x^2} + {y^2}} \right)^{ - 1}}{\rm{d}}A = \mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \mathop \smallint \limits_{r = 1/\left( {2\sin \theta } \right)}^1 f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \mathop \smallint \limits_{r = 1/\left( {2\sin \theta } \right)}^1 \sin \theta {\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \sin \theta \left( {r|_{1/\left( {2\sin \theta } \right)}^1} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \sin \theta \left( {1 - \frac{1}{{2\sin \theta }}} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = \pi /6}^{5\pi /6} \left( {\sin\theta - \frac{1}{2}} \right){\rm{d}}\theta $
$ = \left( { - \cos\theta - \frac{1}{2}\theta } \right)|_{\pi /6}^{5\pi /6}$
$ = \frac{1}{2}\sqrt 3 - \frac{{5\pi }}{{12}} + \frac{1}{2}\sqrt 3 + \frac{\pi }{{12}}$
$ = \sqrt 3 - \frac{\pi }{3} \simeq 0.685$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\left( {{x^2} + {y^2}} \right)^{ - 1}}{\rm{d}}A \simeq 0.685$.
