<s>[INST] Calculate double integral $\int_Se^{\frac{x}{y}}dxdy$ Calculate double integral $\int_Se^{\frac{x}{y}}dxdy$ for the region $1 \le y \le2$ and $y \le x \le y^3$
What I have tried:
$y = 1, y=2 \\ x = y, x=y^3$
$1 \le x \le 2, \text{when }x=y \\
1 \le x \le 2^{\frac{1}{3}}, \text{when }x=y^3 \\
2 \le x \le 8, \text{ when} y=2, \text{when y=1, we have x=1}$
However, we want to integrate with respect to $x$ first and then $y$. How can I correctly derive the calculations with respect to $x$?
From looking at a graph of the bounds, I have got the following:
$$\int_1^2\int_y^2e^{\frac{x}{y}}dxdy+\int_{2^{\frac{1}{3}}}^2 \int_{2}^{y^3}e^{\frac{x}{y}}dxdy$$
I cannot seem to figure out the calculation to get these bounds without the need for visualisation.
 [/INST] Integrating wrt $x$ first (the inner integral in equation below),
\begin{equation}
 \int_1^2 dy \int_y^{y^3} dx \,  e^{x/y} =  \int_1^2 dy \, y  \,  \left[ e^{x/y} \right]_{y}^{y^3} = \int_{1}^2 dy \, y \, [e^{y^2} - e] = \frac{e^4 - e }{2}  -\frac{3}{2}e = \frac{1}{2} e^4 - 2 e \,.
\end{equation}
We can see that $x\le y^3$ is the same as $y \ge x^{1/3}$, so we are interested in the region enclosed by $y=x^{1/3}, y = x, y=1,y=2$. Three of these curves meet at $(x=1,y=1)$, so the other boundaries are $y=x^{1/3}, y = x, x=2$. We have integrated wrt to $x$ first so that $y$ varies freely over the interval $[1,2]$.
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