Find area of region bounded by $xy = a^2$, $xy = b^2$, $x = py$, $x = qy$ I need to obtain the area of the region bounded by the curves 
$xy = a^2$, $xy = b^2$, $x = py$, $x = qy$, $0 < a < b$, $0 < p < q.$
I guess I should use Jacobian of the transformation. I tried $u = \sqrt{xy}$ and $v = \frac{x}{y}$, but it didn't work out. Can you suggest an appropriate substitution?
 A: If you take $u=x y$, $v=\tfrac xy$, then you have 
$$
a^2\leq u\leq b^2,\ \ \ p\leq v\leq q. 
$$
For the Jacobian we first get (no issues with the square roots as both $u$ and $v$ are positive)
$$
x=\sqrt{uv},\ \ \ y=\sqrt{\tfrac uv},
$$
and then 
$$
J=\begin{vmatrix} 
-\frac{\sqrt v}{2\sqrt u}&-\frac{\sqrt u}{2\sqrt v}\\
-\tfrac1{2\sqrt{uv}}&-\frac{\sqrt u}{2\sqrt{v^3}}
\end{vmatrix}
=\left|-\tfrac{1}{2v}\right|=\tfrac1{2v}.
$$
So you area is 
\begin{align}
A&=\int_{a^2}^{b^2}\int_p^q 1\,\left(\frac1{2v}\right)\,dv\,du=(b^2-a^2)\log\sqrt \frac qp.
\end{align}
A: Here is to integrate in polar coordinates. The boundaries are
$a^2\le r^2\sin\theta\cos\theta\le b^2$ and $\frac1q\le \tan\theta \le \frac1p$, or, the corresponding integral limits,
$$\theta_1= \tan^{-1}\frac1q,\>\>\>\>\> \theta_2= \tan^{-1}\frac1p,
\>\>\>\>\>r_1^2(\theta)=\frac{a^2}{\sin\theta\cos\theta},
\>\>\>\>\>r_2^2(\theta)=\frac{b^2}{\sin\theta\cos\theta}$$
Thus, the integral is
$$\int_{\theta_1}^{\theta_2} 
\int_{r_1(\theta)}^{r_2(\theta)}  rdrd\theta 
=\frac{b^2-a^2}2\int_{\theta_1}^{\theta_2} \frac{d\theta}{\sin\theta\cos\theta}
=\frac{b^2-a^2}2\ln(\tan\theta)\bigg|_{\theta_1}^{\theta_2}=\frac{b^2-a^2}2\ln\frac qp$$
