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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.

Working dollar
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2 Answers2

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Alternatively we have $$1\leqslant y\leqslant 2,\quad y\leqslant x\leqslant y^{3},\quad x,y>0$$ Then $$1\leqslant y^{2}\leqslant 4,\quad 1\leqslant \frac{x}{y}\leqslant y^{2}$$ Setting $$u=\frac{x}{y},\quad v=y^{2}, \quad u,v>0$$ Therefore by the change of variables for double integrals, we have $$\iint_{S}e^{\frac{x}{y}}\, {\rm d}x\, {\rm d}y=\int_{1}^{4}\int_{1}^{v}e^{u}\cdot \color{red}{\frac{1}{2}}\, {\rm d}u\, {\rm d}v=\frac{1}{2}\left(e^{4}-4e\right).$$

Just a small remark:

  • In your direct approach you have $$S=\{(x,y): a\leqslant y\leqslant b,\quad f(y)\leqslant x \leqslant g(y)\}$$ Then you can calculate directly the double integral as $$\iint_{S}f(x,y)\, {\rm d}\, {\rm d}y=\int_{a}^{b}\int_{f(y)}^{g(y)}f(x,y)\, {\rm d}x\, {\rm d}y.$$
A. P.
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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 $[1,2]$.

duality
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