Let $R$ be the region bounded by $x+y=1, x=0, y=0.$ Show
$$\iint_R \cos\frac{x-y}{x+y}\, dx\,dy=\frac{\sin1}{2}.$$
So I've let $u=x-y$, $x+y=v$.
Graphically, the domain is a triangle with the slope as the $y=1-x$ line with vertices $(0,1)$ and $(1,0)$.
It seems pretty obvious that the domain for $v$ is $0$ to $1$.
For $u$, I've imagined a bunch of $y=x-u$ lines whereby $u$ is the variable. So in my mind as $u$ varies we get a stack of lines with gradient $1.$ The domain for $u$ will be when these lines intersect $(0,1)$ and $(1,0)$. Graphically, it seems like $u$ will vary from $1$ to $-1$.
Calculating the inverse Jacobian:
$$J=
\begin{vmatrix}
1 & -1\\
1 & 1
\end{vmatrix}=2
$$
$$\frac{1}{J}=\frac{1}{2}.$$
But when I start plugging in the substitution the problem starts:
$$\iint_R \cos\frac{x-y}{x+y}\, dx\,dy$$
$$\frac{1}{2}\int^1_0\int^1_{-1}\cos\frac{u}{v}\,du\,dv$$
$$\frac{1}{2}\int^1_0 2v\sin\frac{1}{v}\,dv.$$
After this I'm stuck. Am I on the right track?
Best Answer
You are not using the correct bounds for the integral as correctly mentioned in the other answer.
The region of interest is $$R=\{(x,y): x\ge 0, y\ge 0, x+y\le 1\}$$
A different change of variables, namely $$(x,y)\to (u,v)\text{ such that } u=\frac{x-y}{x+y}, v=x+y$$
makes the new ranges independent of each other. Since the region $R$ is transformed to some $$D=\{(u,v): -1\le u\le 1, 0\le v\le 1\}$$
The absolute value of the Jacobian is $v/2$, so that the integral is
\begin{align} I=&\iint_D \cos u \left(\frac{v}{2}\right)\, du\, dv \\&=\frac{1}{2}\int_{-1}^{1} \cos u\,du\int_0^1 v\,dv \end{align}