<s>[INST] 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?
 [/INST] 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
$$\int_{\theta_1}^{\theta_2} 
\int_{r_1(\theta)}^{r_2(\theta)}  rdrd\theta 
=\frac{b^2-a^2}2\int_{\theta_1}^{\theta_2} \frac{d\theta}{\sin\theta\cos\theta}
=\frac{b^2-a^2}2\ln(\tan\theta)\bigg|_{\theta_1}^{\theta_2}=\frac{b^2-a^2}2\ln\frac qp$$
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