Answer
Please see the figure attached.
$\mathop \smallint \limits_{ - 1}^2 \mathop \smallint \limits_0^{\sqrt {4 - {x^2}} } \left( {{x^2} + {y^2}} \right){\rm{d}}y{\rm{d}}x \simeq 9.244$
Work Step by Step
We have $\mathop \smallint \limits_{ - 1}^2 \mathop \smallint \limits_0^{\sqrt {4 - {x^2}} } \left( {{x^2} + {y^2}} \right){\rm{d}}y{\rm{d}}x$.
From the order of the integration, we obtain the region of integration:
$ - 1 \le x \le 2$, ${\ \ \ \ }$ $0 \le y \le \sqrt {4 - {x^2}} $
Write $f\left( {x,y} \right) = {x^2} + {y^2}$. Using $x = r\cos \theta $ and $y = r\sin \theta $, in polar coordinates $f\left( {r\cos \theta ,r\sin \theta } \right) = {r^2}$.
We sketch the region and see that the region is part of a disk of radius $2$, and bounded left by $x=-1$. We divide the region ${\cal D}$ into two subregion ${{\cal D}_1}$ and ${{\cal D}_2}$. So, first we need to find the angle of the ray that separate ${{\cal D}_1}$ and ${{\cal D}_2}$.
Substituting $x=-1$ in ${x^2} + {y^2} = 4$ gives
$1 + {y^2} = 4$
$y = \pm \sqrt 3 $
Since $y>0$, so the ray intersects the circle at $\left( { - 1,\sqrt 3 } \right)$. We find the angle of this ray by evaluating:
$\tan \theta = - \sqrt 3 $
$\theta = \frac{2}{3}\pi $
Notice that as $\theta$ varies from $0$ to $\frac{2}{3}\pi $, $r$ ranges from $0$ to $2$. We describe this part of ${\cal D}$ as
${{\cal D}_1} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2,0 \le \theta \le \frac{2}{3}\pi } \right\}$
However, from $\theta = \frac{2}{3}\pi $ to $\theta = \pi $, the rays intersect the line $x=-1$.
Using $x = r\cos \theta $, we find the polar equation of the line $x=-1$:
$r\cos \theta = - 1$
$r = - \frac{1}{{\cos \theta }}$
Thus, the polar description of this part of ${\cal D}$ is
${{\cal D}_2} = \left\{ {\left( {r,\theta } \right)|0 \le r \le - \frac{1}{{\cos \theta }},\frac{2}{3}\pi \le \theta \le \pi } \right\}$
Using the linearity property of the double integral, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta$
$ + \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_{\theta = 0}^{2\pi /3} \mathop \smallint \limits_{r = 0}^2 f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 2\pi /3}^\pi \mathop \smallint \limits_{r = 0}^{ - 1/\cos \theta } f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi /3} \mathop \smallint \limits_{r = 0}^2 {r^3}{\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{\theta = 2\pi /3}^\pi \mathop \smallint \limits_{r = 0}^{ - 1/\cos \theta } {r^3}{\rm{d}}r{\rm{d}}\theta $
$ = \left( {\theta |_0^{2\pi /3}} \right)\left( {\frac{1}{4}{r^4}|_0^2} \right) + \mathop \smallint \limits_{\theta = 2\pi /3}^\pi \left( {\frac{1}{4}{r^4}|_0^{ - 1/\cos \theta }} \right){\rm{d}}\theta $
$ = \frac{8}{3}\pi + \frac{1}{4}\mathop \smallint \limits_{\theta = 2\pi /3}^\pi {\sec ^4}\theta {\rm{d}}\theta $
Recall from Eq. (14) of the Table of Trigonometric Integrals (Section 8.2):
$\mathop \smallint \limits_{\theta = 2\pi /3}^\pi {\sec ^4}\theta {\rm{d}}\theta = \frac{{\tan \theta {{\sec }^2}\theta }}{3}|_{2\pi /3}^\pi + \frac{2}{3}\mathop \smallint \limits_{\theta = 2\pi /3}^\pi {\sec ^2}\theta {\rm{d}}\theta $
$ = \frac{4}{{\sqrt 3 }} + \frac{2}{3}\mathop \smallint \limits_{\theta = 2\pi /3}^\pi {\sec ^2}\theta {\rm{d}}\theta $
Again using Eq. (14) of the Table of Trigonometric Integrals, we get
$\mathop \smallint \limits_{\theta = 2\pi /3}^\pi {\sec ^4}\theta {\rm{d}}\theta = \frac{4}{{\sqrt 3 }} + \frac{2}{3}\left( {\tan \theta |_{2\pi /3}^\pi } \right)$
$ = \frac{4}{{\sqrt 3 }} + \frac{2}{3}\left( {\sqrt 3 } \right) = \frac{6}{{\sqrt 3 }} = 2\sqrt 3 $
Substituting the result $\mathop \smallint \limits_{\theta = 2\pi /3}^\pi {\sec ^4}\theta {\rm{d}}\theta = 2\sqrt 3 $ in the integration above gives
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta = \frac{8}{3}\pi + \frac{1}{2}\sqrt 3 \simeq 9.244$
Thus,
$\mathop \smallint \limits_{ - 1}^2 \mathop \smallint \limits_0^{\sqrt {4 - {x^2}} } \left( {{x^2} + {y^2}} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^3}{\rm{d}}r{\rm{d}}\theta \simeq 9.244$
