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 $$v$$.

• What are the bounds of $R$ in the first place? Jun 12, 2020 at 16:31
• 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? Jun 12, 2020 at 16:32
• yes I forgot it!!! Jun 12, 2020 at 16:33
• @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 $x+y=\pi$ Jun 12, 2020 at 16:39
• I suspect you want $x+y\leq\pi$; otherwise the integral is one-dimensional, not two... Jun 12, 2020 at 16:43

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 $$T^{-1}(R)=\{(u,v):u\le \lvert v\rvert, v\le \pi/2\}.$$

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

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

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: $$\iint_{R}\sin\left(\frac{-2v}{2u}\right)(-2)dudv=2\iint_{R}\sin\left(\frac vu\right)dudv$$