Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}A = \frac{{16}}{3}$
Work Step by Step
We have the domain ${\cal D} = \left\{ {\left( {x,y} \right)|{x^2} + {y^2} \le 4,y \ge 0} \right\}$. Notice that this is the upper half of a disk of radius $2$.
Since ${x^2} + {y^2} = 4$, so $y = \sqrt {4 - {x^2}} $.
We can consider the domain as a vertically simple region. So, we re-write the description as
${\cal D} = \left\{ {\left( {x,y} \right)| - 2 \le x \le 2,0 \le y \le \sqrt {4 - {x^2}} } \right\}$
Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - 2}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } \left( {x + y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 \left( {\mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } x{\rm{d}}y + \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } y{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 x\left( {\mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } {\rm{d}}y} \right){\rm{d}}x + \mathop \smallint \limits_{x = - 2}^2 \left( {\mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } y{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 x\sqrt {4 - {x^2}} {\rm{d}}x + \mathop \smallint \limits_{x = - 2}^2 \left( {\frac{1}{2}{y^2}|_0^{\sqrt {4 - {x^2}} }} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 x\sqrt {4 - {x^2}} {\rm{d}}x + \frac{1}{2}\mathop \smallint \limits_{x = - 2}^2 \left( {4 - {x^2}} \right){\rm{d}}x$
Notice that the antiderivative of $x\sqrt {4 - {x^2}} $ is $ - \frac{1}{3}{\left( {4 - {x^2}} \right)^{3/2}}$ because
$\frac{d}{{dx}}\left( { - \frac{1}{3}{{\left( {4 - {x^2}} \right)}^{3/2}}} \right) = x\sqrt {4 - {x^2}} $
So, the double integral above becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}A = - \frac{1}{3}{\left( {4 - {x^2}} \right)^{3/2}}|_{ - 2}^2 + \frac{1}{2}\left( {4x - \frac{1}{3}{x^3}} \right)|_{ - 2}^2$
$ = 0 + \frac{1}{2}\left( {8 - \frac{8}{3} + 8 - \frac{8}{3}} \right)$
$ = \frac{{16}}{3}$
