Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \frac{{11}}{{60}}$
Work Step by Step
We have the region ${\cal D}$ bounded above by $y = x\left( {2 - x} \right)$ and below by $x = y\left( {2 - y} \right)$.
From the figure attached we see that ${\cal D}$ is a vertically simple region.
First, we apply the quadratic formula to the lower boundary curve to solve for $y$ as a function of $x$:
$x = y\left( {2 - y} \right)$
$ - {y^2} + 2y - x = 0$
$y = \frac{{ - 2 \pm \sqrt {{2^2} - 4\cdot\left( { - 1} \right)\cdot\left( { - x} \right)} }}{{2\cdot\left( { - 1} \right)}} = 1 \mp \sqrt {1 - x} $
We notice that the lower boundary curve has two branches corresponding to the two signs. From the figure attached we see that the correct choice of $y$ as a lower boundary of ${\cal D}$ is $y = 1 - \sqrt {1 - x} $ because $y \le 1$.
Next, we find the intersection points between the two curves: $y = x\left( {2 - x} \right)$ and $y = 1 - \sqrt {1 - x} $ by solving the following equation (actually the graph suggests that these points occur at $x=0$ and $x=2$):
$x\left( {2 - x} \right) = 1 - \sqrt {1 - x} $
$2x - {x^2} = 1 - \sqrt {1 - x} $
$\sqrt {1 - x} = {x^2} - 2x + 1$
Squaring both sides we get
$1 - x = {x^4} - 4{x^3} + 6{x^2} - 4x + 1$
${x^4} - 4{x^3} + 6{x^2} - 3x = 0$
$x\left( {{x^3} - 4{x^2} + 6x - 3} \right) = 0$
$x\left( {x - 1} \right)\left( {{x^2} - 3x + 3} \right) = 0$
But ${x^2} - 3x + 3 = 0$ does not give real solutions because
$x = \frac{{3 \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\cdot1\cdot3} }}{{2\cdot1}} = \frac{{3 \pm \sqrt { - 3} }}{2}$
are complex numbers.
So, the intersection occurs at $x=0$ and $x=1$.
Thus, the domain description is
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,1 - \sqrt {1 - x} \le y \le x\left( {2 - x} \right)} \right\}$
We evaluate the integral of $f\left( {x,y} \right) = x$ over ${\cal D}$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 1 - \sqrt {1 - x} }^{x\left( {2 - x} \right)} x{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 x\left( {y|_{1 - \sqrt {1 - x} }^{x\left( {2 - x} \right)}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 x\left( {x\left( {2 - x} \right) - 1 + \sqrt {1 - x} } \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {2{x^2} - {x^3} - x + x\sqrt {1 - x} } \right){\rm{d}}x$
Notice that the antiderivative of $x\sqrt {1 - x} $ is $ - \frac{2}{{15}}{\left( {1 - x} \right)^{3/2}}\left( {2 + 3x} \right)$ because
$\frac{d}{{dx}}\left( { - \frac{2}{{15}}{{\left( {1 - x} \right)}^{3/2}}\left( {2 + 3x} \right)} \right) = \frac{1}{5}\sqrt {1 - x} \left( {2 + 3x} \right) - \frac{2}{5}{\left( {1 - x} \right)^{3/2}}$
$ = \sqrt {1 - x} \left( {\frac{2}{5} + \frac{3}{5}x - \frac{2}{5} + \frac{2}{5}x} \right)$
$ = x\sqrt {1 - x} $
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \left( {\frac{2}{3}{x^3} - \frac{1}{4}{x^4} - \frac{1}{2}{x^2} - \frac{2}{{15}}{{\left( {1 - x} \right)}^{3/2}}\left( {2 + 3x} \right)} \right)|_0^1$
$ = \frac{2}{3} - \frac{1}{4} - \frac{1}{2} + \frac{4}{{15}} = \frac{{11}}{{60}}$
