Answer
- in the order $dxdy$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = {y^2}}^y xy{\rm{d}}x{\rm{d}}y = \frac{1}{{24}}$
- in the order $dydx$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^{\sqrt x } xy{\rm{d}}y{\rm{d}}x = \frac{1}{{24}}$.
The two results agree.
Work Step by Step
We have the domain ${\cal D}$ between $y=x$ and $y = \sqrt x $.
From $y = \sqrt x $, we see that $x,y \ge 0$. So, ${\cal D}$ is in the first quadrant.
We find the intersection of $y=x$ and $y = \sqrt x $:
$y = x = \sqrt x $
Squaring both sides and after arranging gives
${x^2} - x = 0$
$x\left( {x - 1} \right) = 0$
So, the intersection is at $x=0$, $x=1$.
1. as an iterated integral in the order $dxdy$
In this order, ${\cal D}$ is described as a horizontally simple region:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 1,{y^2} \le x \le y} \right\}$
Evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = {y^2}}^y xy{\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \left( {{x^2}|_{{y^2}}^y} \right)y{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \left( {{y^3} - {y^5}} \right){\rm{d}}y$
$ = \frac{1}{2}\left( {\left( {\frac{1}{4}{y^4} - \frac{1}{6}{y^6}} \right)|_0^1} \right)$
$ = \frac{1}{2}\left( {\frac{1}{4} - \frac{1}{6}} \right) = \frac{1}{{24}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = {y^2}}^y xy{\rm{d}}x{\rm{d}}y = \frac{1}{{24}}$.
2. as an iterated integral in the order $dydx$
In this order, ${\cal D}$ is described as a vertically simple region:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,x \le y \le \sqrt x } \right\}$
Evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^{\sqrt x } xy{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 x\left( {{y^2}|_x^{\sqrt x }} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{x^2} - {x^3}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\left( {\frac{1}{3}{x^3} - \frac{1}{4}{x^4}} \right)|_0^1} \right)$
$ = \frac{1}{2}\left( {\frac{1}{3} - \frac{1}{4}} \right) = \frac{1}{{24}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^{\sqrt x } xy{\rm{d}}y{\rm{d}}x = \frac{1}{{24}}$.
The two results agree.
