Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sin \left( {{x^2} + {y^2}} \right){\rm{d}}A = \pi $
Work Step by Step
The region ${\cal D}$ is between the circles of radii $\sqrt {\frac{\pi }{2}} $ and $\sqrt \pi $, so it can be described in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|\sqrt {\frac{\pi }{2}} \le r \le \sqrt \pi ,0 \le \theta \le 2\pi } \right\}$
Evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sin \left( {{x^2} + {y^2}} \right){\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = \sqrt {\pi /2} }^{\sqrt \pi } \sin \left( {{r^2}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \left( {{r^2}} \right)|_{\sqrt {\pi /2} }^{\sqrt \pi }} \right){\rm{d}}\theta $
$ = - \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \left( \pi \right) - \cos \left( {\frac{\pi }{2}} \right)} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \pi $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sin \left( {{x^2} + {y^2}} \right){\rm{d}}A = \pi $.
