# 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?

-
Use a change of coordinates to make the symmetry more apparent. – Daniel Fischer Mar 22 '14 at 23:42
can you show me a little bit how to do it? – user108297 Mar 22 '14 at 23:44
Take $u = y-x;\; v = y+x$. Find out which region in the $uv$-plane $S$ becomes. – Daniel Fischer Mar 22 '14 at 23:45
thanks, one more question, what will be the boundary then. I know v=1 and v=2, how about u? – user108297 Mar 23 '14 at 0:06

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ If you exchange $x$ with $y$ the integral changes its sign. The result is $\color{#00f}{\Large\mbox{zero}}$.

-

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, $$\int\int_S \sin (\frac{y-x}{x+y})dA=\int_{v=1}^{v=2}\int_{u=-v}^{u=v}\sin \left(\frac{u}{v} \right)\cdot \left|-\frac{1}{2}\right|dudv=$$ $$=\frac{1}{2}\int_{v=1}^{v=2}\left( v\int_{u=-v}^{u=v}\sin\left(\frac{u}{v}\right)\frac{1}{v}du \right)dv =... =0.$$

-

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.

-