Answer
1. Domain ${\cal D}$ as a vertically simple region:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} xy{\rm{d}}y{\rm{d}}x = \frac{1}{{12}}$
2. Domain ${\cal D}$ as a horizontally simple region
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^{\sqrt {1 - y} } xy{\rm{d}}x{\rm{d}}y = \frac{1}{{12}}$.
The two answers agree.
Work Step by Step
Method 1. ${\cal D}$ as a vertically simple region
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1 - {x^2}} \right\}$
So, the integral of $f\left( {x,y} \right) = xy$ over ${\cal D}$ is
$\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} xy{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 0}^{1 - {x^2}} xy{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 x\left( {\frac{1}{2}{y^2}|_0^{1 - {x^2}}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 x{\left( {1 - {x^2}} \right)^2}{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {x - 2{x^3} + {x^5}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{1}{2}{x^2} - \frac{1}{2}{x^4} + \frac{1}{6}{x^6}} \right)|_0^1$
$ = \frac{1}{2}\left( {\frac{1}{2} - \frac{1}{2} + \frac{1}{6}} \right) = \frac{1}{{12}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} xy{\rm{d}}y{\rm{d}}x = \frac{1}{{12}}$.
Method 2. ${\cal D}$ as a horizontally simple region
We have $y = 1 - {x^2}$, so $x = \pm \sqrt {1 - y} $. Since the region is in the first quadrant, we choose $x = \sqrt {1 - y} $. So,
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 1,0 \le x \le \sqrt {1 - y} } \right\}$
So, the integral of $f\left( {x,y} \right) = xy$ over ${\cal D}$ is
$\mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^{\sqrt {1 - y} } xy{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 0}^{\sqrt {1 - y} } xy{\rm{d}}x} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 y\left( {\frac{1}{2}{x^2}|_0^{\sqrt {1 - y} }} \right){\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 y\left( {1 - y} \right){\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \left( {y - {y^2}} \right){\rm{d}}y$
$ = \frac{1}{2}\left( {\frac{1}{2}{y^2} - \frac{1}{3}{y^3}} \right)|_0^1$
$ = \frac{1}{2}\left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{1}{{12}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^{\sqrt {1 - y} } xy{\rm{d}}x{\rm{d}}y = \frac{1}{{12}}$.
The two answers agree.
