Very Tricky Double Integral Problem Evaluate the Integral:
\begin{align}
\int_{0}^{4}\int_{\sqrt{x}}^{2}{\mathrm{d} y\,\mathrm{d}x \over 1 + y^{3}}
\end{align}
I can't understand how this would be possible. There IS a formula for evaluating $1/\left(1+y^{3}\right)$, but it becomes an extremely complicated mess that I can't imagine could be re-integrated. Is there a simplification technique that I'm missing here $?$.
 A: Reverse the order of integration.  How?  Plot the region of integration.  It is all $y \in [\sqrt{x},2]$, and then $x \in [0,4]$.  So, imagine enclosing the graph of $y=\sqrt{x}$ in a box $[0,4] \times [0,2]$.  Then we are integrating the top part of the box, above the curve, $y$ first, then $x$.
Reversing means integrate over the same region, $x$ first, then $y$.  The resulting integral is very straightforward, and the answer is $(2/3) \log{3}$.
A: change the order of integral to get 
$$\int_{0}^{2}\int_{0}^{y^2}\frac{1}{1+y^3}dxdy$$
A: There are many double integral questions like this. The idea is that by changing the order of integration, an integral that might be difficult or even impossible to compute becomes much easier.
We can see from the integration bounds that we are integrating over the region bounded by $y=\sqrt{x}$, $y=2$, $x=0$ and $x=4$.
Swapping integration means integrating by $x$ first, so that $x$ goes from $0$ to $y^2$ and $y$ goes from $0$ to $2$.
$$\int_{0}^2 \int_{0}^{y^2} \frac{1}{1+y^3} dx dy = \int_{0}^2 \frac{y^2}{1+y^3} dy$$
This can be completed by using the substitution method.
