Evaluate $$\int_{x=0}^{1/2}\int_{y=x}^{y=1-x}\frac{y-x}{(x+y)^2\sqrt{1-(x+y)^2}}dydx$$ using change of variables $r=x+y$, $s=y-x$.
I found the Jacobian of transformation to be $J^{(x, y)}_{(r, s)} =\frac{1}{2}$, and the integrand in new variables is obviously $$\frac{s}{r^2\sqrt{1-r^2}}dsdr .$$
The problem I have is that I don't really know how to get from the limits $$0\le x\le 1/2,\quad x\le y\le 1-x$$ to limits for $r$ and $s$. From the first inequality I obtained $s\le r\le 1+s$, and from the second inequality I obtained $0\le s$ and $r\le 1$. Would it be okay to say that then $0\le s\le r$ and $0\le s\le r\le 1$, so that new limits are $$ 0\le s\le r,\quad 0\le r\le 1$$?
If you draw the original integration region, you will see that you can "cover" it with straight lines of the form $y - x = s$, with $s$ starting in zero and going up to 1. So clearly you can have any value $s \in [0,1]$. Now, taking one of these straight lines, intersect it with lines of the type $x+y = r$ and observe what are the values of $r$ so that the intersection lies within the integration region. The conclusion is that you must have $s \leq r \leq 1$.