Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^{\sqrt x } 2xy{\rm{d}}y{\rm{d}}x = \frac{1}{{12}}$
Work Step by Step
We have $f\left( {x,y} \right) = 2xy$ bounded by $x=y$ and $x = {y^2}$. We can consider the domain as a vertically simple region (please see the figure attached).
Substituting $y=x$ in $x = {y^2}$ gives
$x = {x^2}$
${x^2} - x = 0$
$x\left( {x - 1} \right) = 0$
So, the intersection occurs at $x=0$ and $x=1$.
Thus, the lower boundary is $y=x$ and the upper boundary is $y = \sqrt x $. Whereas, the left boundary is $x=0$ and the right boundary is $x=1$. Thus, the domain description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,x \le y \le \sqrt x } \right\}$
We integrate $f\left( {x,y} \right) = 2xy$ over ${\cal D}$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^{\sqrt x } 2xy{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 x\left( {{y^2}|_x^{\sqrt x }} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 x\left( {x - {x^2}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {{x^2} - {x^3}} \right){\rm{d}}x$
$ = \left( {\frac{1}{3}{x^3} - \frac{1}{4}{x^4}} \right)|_0^1$
$ = \frac{1}{3} - \frac{1}{4} = \frac{1}{{12}}$
