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I want to find a continuously differentiable and one-to-one mapping from the first quadrant of $\Bbb{R}^2$ to itself such that the region bounded by $x^2\le y\le 2x^2$ and $1\le xy \le 3$ is mapped to a rectangle. Is there some systematic way to find such mappings? Right now I am just guessing and it is not very effective.

In other words what change of variables $(x,y)\to(u,v)$ will change the curve of $y=ax^2$ to $y= C_1$ and $y=b/x$ to $x=C_2$ for some constants $C_1,C_2$. Is the best way just by inspection?

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The change of variables $u=\frac{y}{x^2}$ and $v=xy$ maps the region to the rectangle $1\leq u\leq 2$ and $1\leq v\leq 3$ in $uv$-plane.

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    $\begingroup$ $u=\dfrac{y}{x^2}$ $\endgroup$ – Nosrati Dec 8 '17 at 1:34
  • $\begingroup$ yes you right thanks $\endgroup$ – daulomb Dec 8 '17 at 1:34
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I got the same answer as user @daulomb, but I wanted to explain how I got there:

We want to change the variables $x,y$ using a transformation $(x,y)\rightarrow(u,v)$ such that the curves that limit your region are linear in $(u,v)$.

We try with transformations in the form $x=ku^av^b$, $y=lu^cv^d$, where $a,b,c,d,k,l$ are constants. Clearly, constants $k,l$ will not affect the linearity of our transformed curves, so we just assume $k=l=1$ and focus on finding appropriate $a,b,c,d$. This transformation maps one-to-one the first quadrant to itself, as long as it is not in the form $c=d=0$ or $a=b=0$.

With this transformation, $xy=u^{a+c}v^{b+d}$. So the inequalities $1\leq xy\leq3$ become $1\leq u^{a+c}v^{b+d}\leq 3.$ For this to be the boundary of a rectangle in $(u,v)$, we need either $a+c=1$ and $b+d=0$, or $a+c=0$ and $b+d=1$.

Similarly, $x^2=u^{2a}v^{2b}$, so inequality $x^2\leq y\leq 2x^2$ becomes $u^{2a}v^{2b}\leq u^{c}v^{d}\leq 2 u^{2a}v^{2b}$. Rearranging terms, $1\leq u^{c-2a}v^{d-2b}\leq 2$. Then, for this to be a rectangle, either $c-2a=1$ and $d-2b=0$, or $c-2a=0$ and $d-2b=1$.

With these conditions on $a,b,c,d$ we can build four systems of linear equations on $a,b,c,d$ and, solving each one, we find two sets of solutions in which our condition for one-to-one mapping is not true, namely solutions $(a,b,c,d)=(0,0,1,0)$ and $(a,b,c,d)=(0,0,0,1)$. We find two solutions that lead to a one-to-one transformation, namely $(a,b,c,d)=\frac{1}{3}(1,-1,2,1)$ and $(a,b,c,d)=\frac{1}{3}(-1,1,1,2)$.

These are the transformations $(x,y)=(u^{1/3}v^{-1/3},u^{2/3}v^{1/3})$, whose inverse is $(u,v)=(xy,y/x^2)$, and $(x,y)=(u^{-1/3}v^{1/3},u^{1/3}v^{2/3})$, whose inverse is $(u,v)=(y/x^2,xy)$. The first maps your region into the rectangle $1\leq u\leq 3$, $1\leq v\leq 2$, and the second transformation maps it to $1\leq u\leq 2$, $1\leq v\leq 3$.

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