Changing variables double integral I want to know if there's a way to know the best variables change for an integral calculation. For example, if we consider the integral
$$ \iint_{R}\left(x+y\right)dx\,dy $$
Where $ R $ is bounded by the line $ y=x,\thinspace\thinspace\thinspace y=2x,\thinspace\thinspace\thinspace\thinspace y+x=2 $.
This integral is very easy to calculate directly, without variables change. But I wonder how simple it can get if one knows how to change the variables in an efficient way.
I'll give another example;
$$ \iint_{R}\left(x-y\right)e^{x^{2}-y^{2}}dx\,dy $$
Where $ R $ is bounded by the curves $ x^{2}-y^{2}=1,\thinspace\thinspace\thinspace\thinspace x^{2}-y^{2}=-1,\thinspace\thinspace\thinspace\thinspace x+y=1,\thinspace\thinspace\thinspace\thinspace x+y=3 $.
In this example it is pretty obvious that we should change the variables in the following way:
$$ \left(x-y\right)=u,\thinspace\thinspace\thinspace\thinspace\left(x+y\right)=v $$
But how can I find a convenient variables change when its not very clear and implied, like in the first example I gave
Thanks in advance.
 A: Sometimes the region in an integral is such that you cannot compute without splitting the region into $2$ or more sub-regions, and you have to found individual bounds. The idea of finding a change of variable is such that the region becomes simpler and you do not have to break the integral into multiple. But when you simplify the region, the integrand may transform such that the integral becomes difficult. So the change of variables needs to be chosen keeping both in mind.
The same is true when the integral is difficult and we are trying to simplify that by using change of variable - we need to keep in mind what it does to the region.
In your first case,
$\displaystyle \iint_{R}\left(x+y\right)dx\,dy$
Where $R: \ y = x, y = 2x, x+y=2$
For the given region, the integral needs to be split into two. So if the idea is to transform this into a region such that we can compute it with one integral,
Writing the bounds as $1 \leq \frac{y}{x} \leq 2, 0 \leq x + y \leq 2$ makes it obvious that $u = \frac{y}{x}, v = x + y$ will transform the region into a rectanle with $1 \leq u \leq 2, 0 \leq v \leq 2$.
$|J| = \frac{v}{(1+u)^2}$
The integral becomes $\displaystyle \int_0^2 \int_1^2 \frac{v^2}{(1+u)^2} \ du \ dv \ $ which is still not difficult to compute.
Your second case -
$\displaystyle \iint_{R}\left(x-y\right)e^{x^{2}-y^{2}}dx\,dy$
Where $ \ R: \ -1 \leq x^{2}-y^{2} \leq 1, 1 \leq x+y \leq 3$
The integral in fact needs to be split into $3$. So to simplify the region, the natural choice that I see is
$1 \leq  u (= x+y) \leq 3, -1 \leq v ( = x^2-y^2) \leq 1$, which transforms the region into a rectangle. $|J| = \frac{1}{2u}$. So integral becomes,
$\displaystyle \int_{-1}^{1} \int_{1}^{3}  \frac{v e^v} {2u^2} \ du \ dv = \frac{2}{3e}$
A: Note that the integration boundaries are  $\frac yx \in (1,2)$ and $x+y\in(0,2)$, which naturally leads to the variable changes
$$u=x+y,\>\>\>\>\>v = \frac yx$$ and the corresponding Jacobian $\frac u{(1+v)^2}$. Thus, the integral becomes
$$\int_S(x+y)dxdy =\int_1^2 \int_0^2 \frac{u^2}{(1+v)^2}dudv =\frac49
$$
