$$\int_0^1 \int_0^{2\sqrt{1-x}} \! \sqrt{x^2+y^2} \, \mathrm{d}y\,\mathrm{d}x$$
Here's where I'm at:
$J(x,y)=4u^2+4v^2$
Substituting $x$ and $y$ into the integral: $\sqrt{(u^2-v^2)^2+4u^2v^2} \rightarrow u^2+v^2$
So now our new integral is $\int \int (u^2+v^2)(4u^2+4v^2)$
How do I find the order of integration and bounds for the new integral?
Notice that the given double integral is of the form: $$ \iint\limits_R f(x,y) \, dA $$ where the boundary of $R$ consists of the following $3$ curves: \begin{align*} R_1&:~~ y = 0 &\text{where }0 \leq x \leq 1\\ R_2&:~~ y = 2\sqrt{1 - x} &\text{where }0 \leq x \leq 1\\ R_3&:~~ x = 0 &\text{where }0 \leq y \leq 2\\ \end{align*} Now since $x,y \geq 0$ on $R$, observe that: $$\begin{cases} u^2 + v^2 = \sqrt{x^2 + y^2} \\ u^2 - v^2 = x \end{cases} \implies \begin{cases} u = \sqrt{\dfrac{\sqrt{x^2 + y^2} + x}{2}} \\ v = \sqrt{\dfrac{\sqrt{x^2 + y^2} - x}{2}} \end{cases}$$ Hence, the boundary of the region $S$ that maps to $R$ (via the given transformation) is defined by the following $3$ curves: \begin{align*} S_1&:~~ v = 0 &\text{where }0 \leq u \leq 1\\ S_2&:~~ u = 1 &\text{where }0 \leq v \leq 1\\ S_3&:~~ v = u &\text{where }0 \leq u \leq 1\\ \end{align*} I'm sure you can take it from here.