Answer
Please see the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt x y{\rm{d}}A = \frac{{256\sqrt 2 }}{{693}}$
Work Step by Step
We have $f\left( {x,y} \right) = \sqrt x y$ and the domain ${\cal D} = \left\{ {0 \le x \le 2,0 \le y \le 2x - {x^2}} \right\}$.
Considering ${\cal D}$ as a vertically simple region, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt x y{\rm{d}}A = \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^{2x - {x^2}} \sqrt x y{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^2 \sqrt x \left( {{y^2}|_0^{2x - {x^2}}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^2 \sqrt x \left( {4{x^2} - 4{x^3} + {x^4}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^2 \left( {4{x^{5/2}} - 4{x^{7/2}} + {x^{9/2}}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{8}{7}{x^{7/2}} - \frac{8}{9}{x^{9/2}} + \frac{2}{{11}}{x^{11/2}}} \right)|_0^2$
$ = \frac{1}{2}\left( {\frac{8}{7}\cdot8\sqrt 2 - \frac{8}{9}\cdot16\sqrt 2 + \frac{2}{{11}}\cdot32\sqrt 2 } \right) = \frac{{256\sqrt 2 }}{{693}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt x y{\rm{d}}A = \frac{{256\sqrt 2 }}{{693}}$.
