How do you change the order of integration in this double integral? I need to reverse the order of integration for the following problem:

$\int_0^{324}{\int_{\sqrt{y}/2}^9}\sqrt{x^3+5}dxdy$

I plotted the function to find new limits of integration:
$0\leq y\leq 2x^2, 0\leq x \leq 9$

\begin{align}
\int_0^{9}{\int_0^{2x^2}\sqrt{x^3+5}dydx} &amp; = \int_0^{9}{\sqrt{x^3+5}y\Big|_0^{2x^2}dx}\\
&amp; = \int_0^{9}{\sqrt{x^3+5}\cdot2x^2dx}\\
&amp; = \frac{4}{9}(x^3+5)^{2/3}\Big|_0^{9}\\
&amp; = \frac{2936\cdot\sqrt{734}-20\sqrt{5}}{9}
\end{align}

However, this is not the correct answer. I think my limits of integration are wrong.

Edit: There is a mistake in the initial problem. The limits are supposed to be $\sqrt{y}/2 \leq x \leq 9$.
In your first integral, your region is bound by 

$x= \sqrt {\frac  y2}\\
x = 9\\
y = 0\\
y = 324$

Graphing of these curves



We get two regions.

If this is indeed what you are trying to integrate.

flipping the order of integration

$\int_0^9 \int_0^{2x^2} f(x,y) \ dy \ dx + \int_9^{9\sqrt{2}}\int_{2x^2}^{324} f(x,y) \ dy \ dx$

Or, as Daniel Fisher points out, your curve is supposed to be $x= \frac {\sqrt y}{2}.$

$\int_0^9 \int_0^{4x^2} f(x,y) \ dy \ dx $
In your first integral, your region is bound by 

$x= \sqrt {\frac  y2}\\
x = 9\\
y = 0\\
y = 324$

Graphing of these curves



We get two regions.

If this is indeed what you are trying to integrate.

flipping the order of integration

$\int_0^9 \int_0^{2x^2} f(x,y) \ dy \ dx + \int_9^{9\sqrt{2}}\int_{2x^2}^{324} f(x,y) \ dy \ dx$

Or, as Daniel Fisher points out, your curve is supposed to be $x= \frac {\sqrt y}{2}.$

$\int_0^9 \int_0^{4x^2} f(x,y) \ dy \ dx $