The following problem is part of a larger problem asking to use a Jacobian to calculate a double integral:
Show that $r(u,v) = (u^2-v^2,2uv)$ maps the triangle $R$ to the domain $D$, where
$$R = \{(u,v) : 0 \leq v \leq u \leq 2)\}$$
and $D$ is bounded by
$$ \begin{cases} x=0,\\ y=0\\ y^2 = 64 - 16x \end{cases} $$
So far, I have established that:
$$ \begin{cases} x = u^2 - v^2\\ y = 2uv \end{cases} $$
and that my triangle $R$ is bounded by
$$ \begin{cases} v=0\\ u=2\\ u=v \end{cases} $$
I'm unsure as to where to go from here. I've tried substituting equations into others but they give me unclear results. I don't think it's meant to be a difficult problem, I'm just failing to understand the core concept.
Each of the equations bounding $R$ can be rewritten in parametric form, then mapped under the transformation and converted into coordinates for $D$.
For $v=0$ we parametrise it as $(t,0)$ for $t\in[0,2]$. Applying the transformation we get $(t^2-0^2,2t0)=(t^2,0)$, which corresponds to a line $y=0$.
For $u=2$ we get $(2,t)$ for $t\in[0,2]$, mapping to $(x,y)=(4-t^2,4t)$. We see that $4-y^2/16=4-t^2=x$, so $u=2$ corresponds to $y^2=64-16x$ as desired.
Finally, $u=v$ parametrises as $(t,t)$ for $t\in[0,2]$, maps to $(0,2t^2)$ and thus to $x=0$.