Area in $uv$ plane corresponding to two areas in $xy$? The inequalities $1 \leq x^2 - y^2 \leq 4$ and $3 \leq xy \leq 5$ describe the area $D$. See picture below. 
The excerice is to calculate the integral: $\iint_D 2(x^4 - y^4) \,dx\,dy$
I would do a vaiable substitution where $u = x^2 - y^2$ and $v = xy$ where $1 \leq u \leq 4, 3 \leq v \leq 5$, with the corresponding Jacobian $\frac{\partial(u, v)}{\partial(x, y)} = 2(x^2 + y^2) = 2u \Leftrightarrow \frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{2(x^2 + y^2)}$.
Doing the integral, we get
$\int_{1}^{4} \int_{3}^{5} 2 (x^4 - y^4) \, |\frac{\partial(x, y)}{\partial(u, v)}| \, du \, dv = \int_{1}^{4} \int_{3}^{5} 2 (x^2 + y^2) (x^2 - y^2) \frac{1}{2(x^2 + y^2)} \,du\,dv =  \int_{1}^{4} \int_{3}^{5} u \,du \,dv = 15$. 

As one clearly sees, the inequalities corresponds to two areas in $xy$; one in the first quadrant and one in the third quadrant. 
How do I know if which area I've integrated over when I do the variable substitution? Is it the one in the first or third quadrant or both? Or is it indifferent? 
To me it looks like the $uv$ inequalities correspond to both areas in $xy$. Is that correct? 

To integrate we need also to express the intagrand function $2(x^4 - y^4)$ in terms of $u$ and $v$
We can proceed as follow


*

*$x^4 - y^4=(x^2-y^2)(x^2+y^2)=u\sqrt{u^2+4v^2}$

