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I can't solve this. How should I proceed?

$$\iint_De^{\large\frac{y-x}{y+x}}\mathrm dx\mathrm dy$$

$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-\frac{1}{2}$ but I have problem finding the range of $u$ and $v$ to calculate the integral.

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    $\begingroup$ Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558 $\endgroup$
    – Jean Marie
    Commented Mar 9, 2019 at 21:37
  • $\begingroup$ i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks ! $\endgroup$
    – khoshrang
    Commented Mar 9, 2019 at 22:31

3 Answers 3

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Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

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  • $\begingroup$ thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x? $\endgroup$
    – khoshrang
    Commented Mar 9, 2019 at 21:34
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    $\begingroup$ You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant. $\endgroup$
    – Andrei
    Commented Mar 9, 2019 at 21:46
  • $\begingroup$ "So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0? $\endgroup$
    – khoshrang
    Commented Mar 9, 2019 at 22:10
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    $\begingroup$ If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$. $\endgroup$
    – Andrei
    Commented Mar 9, 2019 at 22:12
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    $\begingroup$ thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question $\endgroup$
    – khoshrang
    Commented Mar 9, 2019 at 22:26
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Your domain is $$D=\{(x,y):0<x<2\,,\,0<y<2\,,\,x+y<2\}$$

Using change of variables $(x,y)\to(u,v)$ with $$u=\frac{x-y}{x+y}\,,\,v=x+y$$

The region is now $$R=\{(u,v):-1<u<1\,,0<v<2\}$$

Therefore,

\begin{align} \iint_D \exp\left({-\frac{x-y}{x+y}}\right)\,\mathrm{d}x\,\mathrm{d}y&=\frac{1}{2}\iint_R ve^{-u}\,\mathrm{d}u\,\mathrm{d}v \\\\&=\frac{1}{2}\int_0^2v\,\mathrm{d}v\int_{-1}^1 e^{-u}\,\mathrm{d}u \end{align}

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    $\begingroup$ Finally the $v$ appears, but with no explanation at all? $\endgroup$
    – NickD
    Commented Mar 9, 2019 at 20:55
  • $\begingroup$ @NickD I have left the calculation of the jacobian to the OP. $\endgroup$
    – StubbornAtom
    Commented Mar 9, 2019 at 20:58
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    $\begingroup$ That's fine, but you should at least mention it, if only for the sake of future readers. $\endgroup$
    – NickD
    Commented Mar 9, 2019 at 21:02
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    $\begingroup$ @khoshrang To see the bounds of $u$ and $v$, replace the string of inequalities in $D$ with $u$ and $v$. (So first of all, solve for $x,y$ in terms of $u,v$ ). And $v/2$ is the jacobian of course, no surprise that it appears at the end. $\endgroup$
    – StubbornAtom
    Commented Mar 9, 2019 at 21:44
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    $\begingroup$ thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question $\endgroup$
    – khoshrang
    Commented Mar 9, 2019 at 22:28
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Computer algebra gives (for the general case):

$$e \left(x^2 \text{Ei}\left(-\frac{2 x}{x+y}\right)-\frac{y^2 \text{Ei}\left(\frac{2 y}{x+y}\right)}{e^2}\right)+\frac{1}{2} e^{1-\frac{2 x}{x+y}} (x+y)^2$$

over your specified region:

$$e-\frac{1}{e}$$

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    $\begingroup$ yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper? $\endgroup$
    – khoshrang
    Commented Mar 9, 2019 at 20:06
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    $\begingroup$ (-1) because this is not an answer to the question. See the other excellent responses! $\endgroup$
    – James
    Commented Mar 9, 2019 at 21:18
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    $\begingroup$ @JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote. $\endgroup$
    – David G. Stork
    Commented Mar 9, 2019 at 22:30
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    $\begingroup$ The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion $\endgroup$
    – Mikey Spivak
    Commented Mar 9, 2019 at 23:00

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