Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\left( {{x^2} + {y^2}} \right)^{ - 2}}{\rm{d}}y{\rm{d}}x = \frac{1}{4}$
Work Step by Step
We have $f\left( {x,y} \right) = {\left( {{x^2} + {y^2}} \right)^{ - 2}}$ and the given region: ${x^2} + {y^2} \le 2$, $x \ge 1$.
Notice that the region ${\cal D}$ is part of a disk of radius $\sqrt 2 $ bounded by the line $x=1$. This is illustrated in the figure attached.
To determine the range of $\theta$, we find the angles of the rays that intersect the circle. Substituting $x=1$ in ${x^2} + {y^2} = 2$, we get $y = \pm 1$. So,
$\tan \theta = \pm 1$
${\theta _1} = - \frac{\pi }{4}$, ${\ \ \ \ }$ ${\theta _2} = \frac{\pi }{4}$
Thus, $\theta$ varies from ${\theta _1} = - \frac{\pi }{4}$ to ${\theta _2} = \frac{\pi }{4}$.
From the figure attached, we see that $r$ varies from the line $x=1$ to $r = \sqrt 2 $.
Using $x = r\cos \theta $, the equation of the line $x=1$ in polar coordinates is
$1 = r\cos \theta $
$r = \frac{1}{{\cos \theta }}$
Thus, the polar description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)| - \frac{\pi }{4} \le \theta \le \frac{\pi }{4},\sec \theta \le r \le \sqrt 2 } \right\}$
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}^{} {\left( {{x^2} + {y^2}} \right)^{ - 2}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{r = \sec \theta }^{\sqrt 2 } \left( {{r^{ - 4}}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{r = \sec \theta }^{\sqrt 2 } {r^{ - 3}}{\rm{d}}r{\rm{d}}\theta $
$ = - \frac{1}{2}\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \left( {{r^{ - 2}}|_{\sec \theta }^{\sqrt 2 }} \right){\rm{d}}\theta $
$ = - \frac{1}{2}\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \left( {\frac{1}{2} - {{\cos }^2}\theta } \right){\rm{d}}\theta $
Using the identity ${\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2}$, write
$ - \frac{1}{2}\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \left( {\frac{1}{2} - {{\cos }^2}\theta } \right){\rm{d}}\theta = - \frac{1}{2}\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \left( {\frac{1}{2} - \frac{{1 + \cos 2\theta }}{2}} \right){\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \cos 2\theta {\rm{d}}\theta $
$ = \frac{1}{8}\sin 2\theta |_{ - \pi /4}^{\pi /4}$
$ = \frac{1}{8}\left( {1 + 1} \right) = \frac{1}{4}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\left( {{x^2} + {y^2}} \right)^{ - 2}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = - \pi /4}^{\pi /4} \mathop \smallint \limits_{r = \sec \theta }^{\sqrt 2 } \left( {{r^{ - 4}}} \right)r{\rm{d}}r{\rm{d}}\theta = \frac{1}{4}$.
