TITLE: Puzzling Double integral
QUESTION [5 upvotes]: I was asked to solve this double integral:
Compute the area between $y=2x^2$ and $y=x^2$ and the hyperbolae $xy=1$ and $xy=2$ in 
$$ \iint dx \,dy$$
I tried to solve it starting with considering that 
$$x^2 \leq y \leq 2x^2 $$
suitabile for integration interval in $y$, obtaining the incomplete form
$$ \int^{x^2}_{2x^2} \int_\ldots^\ldots dx \,dy$$
but I also have
$$1 \leq xy \leq 2$$ and I would obtain a result in which I still have one independent variabile. 
Please, can anyone help me? Thanks in advance.

REPLY [4 votes]: $y \in [x^2, 2x^2]$ can be interpreted as $\frac{y}{x^2} \in [1,2].$
Consider the change of variables

$u = \frac{y}{x^2} \in [1,2]$
$v = xy \in [1,2]$

The Jacobian is given by
$$\frac1J = \begin{vmatrix} u_x & u_y \\ v_x & v_y
\end{vmatrix} = \begin{vmatrix} -\frac{2y}{x^3} & \frac1{x^2} \\ y & x
\end{vmatrix} = -\frac{3y}{x^3} = -3u$$
so $$dx\,dy = |J| \,du\,dv = \frac1{3u} \, du\,dv$$ 
Therefore you need to calculate
$$\int_{[1,2]^2} \frac{1}{3u}\,du\,dv= \frac13\ln 2$$