Calculate $\int \limits_{A}^{ }\frac{y}{x^3}\, dy\, dx$ with $A=\{(x,y)\in \mathbb{R}^2|1\leq xy\leq 2,\, 1\leq \frac xy\leq 2\}$, coordinate-transfo.

Question

Calculate $\int \limits_{A}^{ }\frac{y}{x^3}\, dy\, dx$ with $A=\{(x,y)\in (\mathbb{R^+})^2|1\leq xy\leq 2,\, 1\leq \frac xy\leq 2\}$.

Define new coordinates $u=xy\,v=\frac yx$.

Now define $B:=[1,2]^2$ and $h: A\to B, (x,y)\mapsto (u,v)=(xy,\frac yx)$

It is known that $h$ is bijective and continuously differentiable on $B$.

Now The following theorem $(*)$ gets used:

"Let $U\subset \mathbb{R}^n$ be open, $f:U\to \mathbb{R}^n$ continuously differentiable (a $C^1$-function) and $a \in U$. If $J_f(a)$ is invertable then $f$ is a local $C^1$-diffeomorphism. "

Now $J_h((x,y))=\begin{pmatrix} y & x \\ \frac{-y}{x^2} & \frac{1}{x} \end{pmatrix}\Rightarrow \det \begin{pmatrix} y & x \\ \frac{-y}{x^2} & \frac{1}{x} \end{pmatrix}=2\frac{y}{x} >0$ for every $(x,y)\in A$
So $J_h$ is ivertable.

First Question In the solution to the problem it is stated that $h$ satifys the conditions of $(*)$ and is a $C^1$-diffeomorphism but $A$ is not an open set. Why is $(*)$ applicable?

Next step:

Define $g=h^{-1}$. Now one can apply the transformation formula:

$\int \limits_{g(B)}^{}f(x,y)d(x,y)=\int \limits_{B}^{ }f(g(u,v))\cdot|\det J_g(u,v)|d(u,v)$

Now the second question:

It is stated that the following is true:

$f(g(u,v))\cdot|\det J_g(u,v)|= f(x,y)\cdot \frac{1}{|\det J_h(u,v)|}$

Do you have an idea, what the justification could be?

Important Edit

I made a mistake. I wrote $A=\{(x,y)\in \mathbb{R}^2|1\leq xy\leq 2,\, 1\leq \frac xy\leq 2\}$ instead of $A=\{(x,y)\in (\mathbb{R^+})^2|1\leq xy\leq 2,\, 1\leq \frac xy\leq 2\}$ !