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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 $R_{I}$:

$$R_{I}=\left\{\begin{matrix} 0\leqslant x\leqslant 4\\ \sqrt{x}\leqslant y \leqslant 2 \end{matrix}\right.$$

Graph of R_{I}

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?

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I think you've made a mistake. If $y = \sqrt x$ then $x = y^2$. Here we have $0\le x \le 4 $, $\sqrt x \le y \le 2$. This is equivalent to $0 \le y \le 2$ and $0 \le x \le y^2$. Thus \begin{align*} \int_0^4 \int_{\sqrt x}^2 \frac{1}{y^3+1} dy \,dx &= \int^2_0 \int_0^{y^2} \frac{1}{y^3 +1}dx \,dy \\&= \int^2_0 \frac{y^2}{y^3+1}dy\\ &=\left. \frac{1}{3}\log(y^3+1)\right|^2_0 = \frac 1 3 \log(9) = \frac 2 3 \log (3). \end{align*}

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$$y=\sqrt x\implies x=y^2$$ and in your region $y$ ranges from $0$ to $2$. So the integral you're looking for is $$ \int_0^2\int_0^{y^2}{1\over y^3+1}\color{red}{dx}\color{blue}{dy} $$

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