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I want to calculate this integral with substitution $x=u+v , \ y=u-v$: $$\iint_R \sin\left(\frac{y-x}{y+x}\right)dydx$$ $$R:= \{(x,y):x+y≤\pi, y≥0,x≥0\}$$ but I don't know how to set new bounds for $u$ and $v$.

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    $\begingroup$ What are the bounds of $R$ in the first place? $\endgroup$ Jun 12, 2020 at 16:31
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    $\begingroup$ You probably should include into the problem what is $R$. You could also check similar problems already posted on this site. E.g., $\iint_D \sin(\frac{y-x}{y+x})dydx$ over a trapezoid $D$ is found both by SearchOnMath and Approach0. See also: How to search on this site? $\endgroup$ Jun 12, 2020 at 16:32
  • $\begingroup$ yes I forgot it!!! $\endgroup$ Jun 12, 2020 at 16:33
  • $\begingroup$ @AlekosRobotis yes I'v seen showing regions in both forms. in this form I mean the region is bounded by lines $x=0$ and $y=0$ and $x+y=\pi$ $\endgroup$ Jun 12, 2020 at 16:39
  • $\begingroup$ I suspect you want $x+y\leq\pi$; otherwise the integral is one-dimensional, not two... $\endgroup$ Jun 12, 2020 at 16:43

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We have a transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ where the coordinates of the first $\mathbb{R}^2$ are $(u,v)$ and those of the second $\mathbb{R}^2$ are $(x,y)$. We know the transformation is given by $T(u,v)=(u+v,u-v)$. This is a linear transformation with matrix given by $$ M= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}. $$ It has inverse matrix given by $$ M^{-1}=\frac{-1}{2} \begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}.$$ You want to integrate over the convex triangular region $R$ with vertices $(0,0), (\pi,0),(0,\pi)$. We know that $(0,0)$ has unique preimage $(0,0)$, and we can compute the preimages of $(\pi,0)$ and $(0,\pi)$ using $M^{-1}$. $M^{-1}(\pi,0)=(\pi/2,\pi/2)$, and $M^{-1}(0,\pi)=(\pi/2,-\pi/2).$ So, $T^{-1}(R)$ is the convex region spanned by $(0,0), (\pi/2,\pi/2),(\pi/2,-\pi/2)$. Put another way, this is is the region $$ T^{-1}(R)=\{(u,v):u\le \lvert v\rvert, v\le \pi/2\}.$$

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The region $R$ can be written as the set

$$\{(x,y)\mid0\le x\le\pi\land0\le y\le\pi-x\}$$

With the given change of variables, we have

$$\begin{cases}x=u+v\\y=u-v\end{cases}\implies\begin{cases}u=\frac{x+y}2\\v=\frac{x-y}2\end{cases}$$

The boundary of $R$ in the $(u,v)$ plane consists of the lines,

$$x=u+v=0\implies v=-u$$

$$y=u-v=0\implies v=u$$

$$x+y=2u=\pi\implies u=\frac\pi2$$

and together with $x\ge0$ and $y\ge0$, it follows that $u\ge0$.

Then in the new coordinates, the region $R$ is the set

$$R=\left\{(u,v)\mid0\le u\le\frac\pi2\land-u\le v\le u\right\}$$

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firstly,if we have: $$x=u+v,y=u-v$$ then we notice that: $$x+y=2u$$ and so: $$u=\frac{x+y}{2}\le\frac{\pi}{2}$$ now we will try at the boundaries of $R$: $$x=0\Rightarrow u+v=0\therefore u=-v$$ $$y=0\Rightarrow u-v=0\therefore u=v$$ and so: $$-v\le u\le v$$ also notice that if: $$x,y\ge0\Rightarrow u\ge 0$$ which gives our new region as: $$R=\left\{(u,v):0\le u\le \frac \pi 2,-u\le v\le u\right\}$$ I have switch around the first inequality calculated because we want one for $u$ and one for $v$.


Now we know that: $$\frac{\partial x}{\partial u}=1$$ $$\frac{\partial x}{\partial v}=1$$ $$\frac{\partial y}{\partial u}=1$$ $$\frac{\partial y}{\partial v}=-1$$ and so the determinant of the Jacobian is: $$\det(J)=\det\begin{pmatrix}1&1\\1&-1\end{pmatrix}=-2$$ Which gives our integral as: $$\iint_{R}\sin\left(\frac{-2v}{2u}\right)(-2)dudv=2\iint_{R}\sin\left(\frac vu\right)dudv$$

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