Consider the region D defined by $1 \leq x^2-y^2 \leq 4$ and $ 0 \leq y \leq \frac{3x}{5}$. In the problem, set up an integral to compute $\int\int_{D} e^{x^2-y^2} dA$. Consider the change of coordinates $G(u,v)=(\frac{v}{2}+\frac{u}{2v},\frac{v}{2}-\frac{u}{2v})$. Find a region R of the uv-plane so that $G:R \rightarrow D$ is a change of coordinates map( so G is onto and one-to-one on the interior of R).
After I find the 4 curves in the uv-plane that map to the 4 curves forming the boundary of D, what am I supposed to do then? I am a little lost.
Added: A bounty is awarded to a solution to this problem.
We want to find some bounds on $(u,v)$ so that $G$ maps that domain to $D$. In other words:
$$1 \leq (\frac{v}{2}+\frac{u}{2v})^2- (\frac{v}{2}-\frac{u}{2v})^2 \leq 4$$ This is equivalent to: $$1 \leq 4\frac{vu}{4v} \leq 4$$ $$1 \leq u \leq 4$$
So the domain $R$ is the set $(u,v)$ where $u \in [1,4]$ and $v$ is any number in $\mathbb{R}$.
Edit: I missed the second condition. Including this gives $$0 \leq \frac{v}{2}-\frac{u}{2v} \leq \frac{3}{5}(\frac{v}{2}+\frac{u}{2v})$$ Simplifying this gives: $$u \leq v^2$$ if $v \geq 0$ and some other condition. Thus $R$ is the set of $(u,v)$ that satisfy all the inequalities.
To answer your question, after you find the $4$ curves in the $UV$ plane , $R$ is the region inside, bounded by the $4$ curves.