How to find limit of integration for marginal densities after transformation? Here is the problem and solution for part b, which I need help with :

How do you get the highlighted limits of integration?
Four equations I have are:
$u = xy$
$v = x/y$
$x = \sqrt{uv}$
$y = \sqrt{u/v}$
I feel like u(v) and v(u) were found, graphed, then used to find the region to integrate. I really have no idea what's going on here, so any help would be greatly appreciated.
Thank you!
 A: Note: I think the answer provided is wrong. For $0<v<1$, the limits of integration should be $\frac{1}{v}, \infty$ instead of $\frac{1}{2}, \infty$.
We begin by noting that $u \in [1,\infty)$ and $v \in [0, \infty)$. This is easily derived from the limits of $x$ and $y$. 
This is meant to be more intuitive than rigorous (although I will do my best to not be hand-wavy), but I hope that the idea is communicated sufficiently:
We first show that the support for $f_{u,v}$ does not include any of the points 
$$\{ (u,v): v>u \}$$ 
(i.e. $v$ must always be smaller than $u$)
We do so by contradiction. Suppose $v>u$. Then $$\frac{v}{u} > 1 \implies \frac{x}{y} \frac{1}{xy} > 1 \implies \frac{1}{y^2} > 1$$
Which is a contradiction since $y \geq 1$
Note that the points $$\{ (u,v): v=u \}$$ are in the support, since $u = v$ whenever $y = 1$. 
Now, we show that the support for $f_{u,v}$ does not include any of the points 
$$\{ (u,v): v < \frac{1}{u} \}$$ 
Again, we show this by contradiction. Suppose $v < \frac{1}{u}$. Then
$$uv < 1 \implies x^2 < 1 $$
Which is a contradiction, since $x \geq 1$
We also note that the points
$$\{ (u,v): v = \frac{1}{u} \}$$ are in the support, since $ v = \frac{1}{u}$ whenever $x = 1$.
So now, we know that the support $S_{u,v}$ must be a subset of the set 
$$S = \{(u,v): v \leq u, v \geq \frac{1}{u}\}$$
It turns out that every point in the set $S$ is a point in the support (If you don't want to see the proof, just skip to the image below). To prove this, suppose $(u,v) \in S$, then we will have:
$$
\begin{align}
&v \leq  u \text{ and } uv \geq 1 \\ 
\implies &y^2 \geq 1 \text{ and } x^2 \geq 1 \\
\implies &y \geq 1 \text{ and } x \geq 1 \\
\implies &(x,y) \in supp(f_{x,y}) \\
\implies &(u,v) \in supp(f_{u,v})
\end{align}
$$
Where the last implication follows from the fact that the mapping $(x,y) \to (xy, \frac{x}{y})$ is bijective. 
Proof of bijection:


*

*Injective: If there are points $(x_1,y_1),(x_2,y_2) \in supp(f_{x,y})$ such that $x_1y_1 = x_2y_2$ and $\frac{x_1}{y_1} = \frac{x_2}{y_2}$, then 
$$\begin{align}
\implies & (x_1 = x_2 = 0 \text{ or } y_1 = y_2) \text{ or } (x_1 = x_2 \text{ or } y_1 = y_2 = 0)
\end{align}$$ 
Either way we are done. 

*Surjective: Trivial. Just look at the range of $(xy,\frac{x}{y})$ which we are working with. 


Now, you can easily conclude that the support $S_{u,v}$ is indeed given by $S$, and shade that region out (apologies for the rather ugly picture):

This gives the region of integration from which the limits that are taken should be fairly obvious. 
