Transformation theorem: calculate picture of a set I have this function:
$T:(0,\infty)^2 \rightarrow T((0, \infty)^2), \quad T(x,y)=\left( \frac{y^2}{x},\frac{x^2}{y} \right)$
Now I try to estimate $T(M)$ with:
$0<p<q, \quad 0<a<b$
$M=\left\{ (x,y) \in \mathbb{R}^2 | ax < y^2 < bx \text{ and } py < x^2<qy \right\}$
Later I want to calculate the area of $M$ (relating to $\lambda_2$).
Until now I got:
$x,y > 0$ and $a,p,x,y < \infty \Rightarrow inf(M)=\left ( \begin{array}{c}
0\\
0\\
\end{array} \right ) \notin M$ and $sup(M) = \left ( \begin{array}{c}
\infty\\
\infty\\
\end{array} \right ) \notin M$
My Problem is that I don't have an explicit relation between $x$ and $y$ that I could use to calculate $T(M)$.
Has anyone an idea? I would appreciate it.
 A: (Originally this was a set of hints $\ldots$)
The set $M$ is bounded by four parabolic arcs. One could compute its area by cutting it up into simpler shapes, each of them bounded by segments parallel to the axes and at most one arc.
Instead another procedure is suggested: One can write the definition of $M$ in the form
$$M:=\left\{(x,y)\>\biggm|\>a<{y^2\over x}<b, \quad p<{x^2\over y}<q\right\}\ .$$
This shows that the set $Q:=T(M)$ in the $(u,v)$-plane is the rectangle $[a,b]\times[p,q]$. The area of $M$ can then be computed in the following way: Determine explicitly the inverse  $S:=T^{-1}$, which maps $Q$ bijectively back onto $M$. This $S$ is of the form
$$S:\quad(u,v)\mapsto (x,y):=\bigl(x(u,v),y(u,v)\bigr)\ ,$$
whereby the functions $x(u,v)$, $y(u,v)$ can be found algebraically by solving the system
$$u={y^2\over x}, \quad v={x^2\over y}$$
for $x$ and $y$. One obtains
$$x=u^{1/3}\>v^{2/3},\quad y=u^{2/3}v^{1/3}\ ,$$
and the Jacobian determinant of $S$ computes to
$$J_S(u,v)=x_uy_v-x_vy_u=-{1\over3}\ .$$
Therefore we get
$${\rm area}(M)=\int_Q \bigl|J_S(u,v)\bigr|\>{\rm d}(u,v)={1\over3}(q-p)(b-a)\ .$$
