I have this double integral:
$$\int_{x=0}^{x=4}\int_{y=\sqrt{x}}^{y=2}\frac{1}{y^{3}+1}dydx$$
I change its order. The first I did was the graph of $R_{I}$:
$$R_{I}=\left\{\begin{matrix} 0\leqslant x\leqslant 4\\ \sqrt{x}\leqslant y \leqslant 2 \end{matrix}\right.$$
So, the new order for the double integral is:
$$\int_{y=0}^{y=4}\int_{x=0}^{x=\sqrt{y}}\frac{1}{y^{3}+1}dydx$$
Is this correct?