Compute this double integral Compute the integral
$$
\iint_{S}\sin\left(\,y - x \over x + y\,\right)\,{\rm d}A
$$
 where S is the trapezoidal region of the plane bounded by the lines
$x + y = 1$, $x + y = 2$ and the coordiante axis.
So when I just computed as normal way and I found out by software that the integral of $\sin\left(\,y - x \over x + y\right)$ is some crazy expression.
So I think there should be some other way to solve this question, is it something related to change of coordinate ? If yes, how?
 A: $\newcommand{\+}{^{\dagger}}
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If you exchange $x$ with $y$ the integral changes its sign.
The result is $\color{#00f}{\Large\mbox{zero}}$.
A: Is for change of coordinates. Write $u=y-x$ and $v=x+y$. Then $y=\frac{1}{2}u+\frac{1}{2}v$ and $x=\frac{1}{2}v-\frac{1}{2}u$. Now, we must transform the region in the $xy$ plane in a region of the plane $uv$ as follow: Let be $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ the transform
$$T(x,y)=(y-x, x+y):=(u,v). $$
This transform send  the trapezoidal region $S$ on $xy$ plan in the region in $uv$ plan with coordinates $(-1,1); (1,1); (2,2)$ and $(-2,2)$ (make a drawing).
The determinant of Jacobian transformation is
$$\det\frac{J(x,y)}{(u,v)}= -\frac{1}{2},$$
Then,
$$\int\int_S \sin (\frac{y-x}{x+y})dA=\int_{v=1}^{v=2}\int_{u=-v}^{u=v}\sin \left(\frac{u}{v} \right)\cdot \left|-\frac{1}{2}\right|dudv=$$
$$=\frac{1}{2}\int_{v=1}^{v=2}\left( v\int_{u=-v}^{u=v}\sin\left(\frac{u}{v}\right)\frac{1}{v}du \right)dv =... =0.$$
A: Let $u = -x + y$, and $v = x + y$. Then $y = \frac{u + v}2$, and $x = \frac{v - u}2$. Now we need to find the region in the $uv$-plane. 
$$x + y = 1 \implies v = 1.$$
$$x + y = 2 \implies v = 2.$$
$$y = 0\ \text{(}x\text{-axis in }xy\text{-plane)} \implies u + v = 0 \implies u = -v.$$
$$x = 0\ \text{(}y\text{-axis in }xy\text{-plane)} \implies v - u = 0 \implies u = v.$$
Thus the region $S'$ in the $uv$-plane is also a trapezoid and is bounded by the lines:
$$u = v, u = -v, v = 1,\text{ and }v = 2.$$
So the set up for the new integral is: 
$$\int_1^2\int_{-v}^v -\frac12\sin\left(\!\frac uv\!\right)\,du\,dv.$$
You can continue.
