How to evaluate $\iint_R \sin(\frac{y-x}{y+x})dydx$ with Jacobian substitution? 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 $v$.
 A: 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
$$ T^{-1}(R)=\{(u,v):u\le \lvert v\rvert, v\le \pi/2\}.$$
A: 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
$$R=\left\{(u,v)\mid0\le u\le\frac\pi2\land-u\le v\le u\right\}$$
A: 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:
$$\iint_{R}\sin\left(\frac{-2v}{2u}\right)(-2)dudv=2\iint_{R}\sin\left(\frac vu\right)dudv$$
