Answer
The iterated integral after changing the order of integration:
$\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y{\rm{d}}x$
Evaluate:
$\mathop \smallint \limits_{x = 0}^2 \left( {\mathop \smallint \limits_{y = 0}^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y} \right){\rm{d}}x = \frac{{152}}{{15}}$
Work Step by Step
We have $\mathop \smallint \limits_0^4 \mathop \smallint \limits_{\sqrt y }^2 \sqrt {4{x^2} + 5y} {\rm{d}}x{\rm{d}}y$.
From the limits of the integrals we obtain the domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 4,\sqrt y \le x \le 2} \right\}$
Notice that this is a horizontally simple region.
Using this description we sketch the the domain of integration. Please see the figure attached.
We change the order of integration such that the domain becomes a vertically simple region. In this case, the left and right boundaries are $x=0$ and $x=2$, respectively. Whereas, the lower boundary is $y=0$. To find the upper boundary, we use $x = \sqrt y $ and get $y = {x^2}$. So, the new domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le {x^2}} \right\}$
So, the iterated integral after changing the order of integration:
$\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y{\rm{d}}x$
Next, we evaluate this integral:
$\mathop \smallint \limits_{x = 0}^2 \left( {\mathop \smallint \limits_{y = 0}^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^2 \left( {\frac{2}{{15}}{{\left( {4{x^2} + 5y} \right)}^{3/2}}|_0^{{x^2}}} \right){\rm{d}}x$
$ = \frac{2}{{15}}\mathop \smallint \limits_{x = 0}^2 \left( {{{\left( {9{x^2}} \right)}^{3/2}} - {{\left( {4{x^2}} \right)}^{3/2}}} \right){\rm{d}}x$
$ = \frac{2}{{15}}\mathop \smallint \limits_{x = 0}^2 \left( {27{x^3} - 8{x^3}} \right){\rm{d}}x$
$ = \frac{{38}}{{15}}\cdot\frac{1}{4}{x^4}|_0^2$
$ = \frac{{152}}{{15}}$
