Find the area between $y=x; y=2x; y=1/x; y=3/x$ Find the area between $y=x; y=2x; y=1/x; y=3/x$ using the substitution $(x,y)=(\frac{u}{v},uv)$
I made a sketch but don't see how I can make use of the substitution given, to help.
 A: it is easier to find the area using polar coordinates. the transformation between polar and cartesian are given by $$x = r \cos t, y = r \sin t, y = 1/x \to r\sin t = \frac1{r\cos t} \to r^2 = \frac 1{\sin t\cos t} .$$ the area is $\frac 12 \int_{t_1}^{t_2} (r_2^2 - r_1^2) \, dt$. that is 
$$\frac12\int_{\pi/4}^{\tan^{-1} 2} \left(\frac 3{\sin t\cos t} - \frac 1{\sin t \cos t}\right) \, dt  =2\int_{\pi/4}^{\tan^{-1} 2}\csc 2t\, dt=  \ln(\csc 2t + \cot 2t)\big|_{\pi/4}^{\tan^{-1} 2}
=\ln 2.$$
A: $x=\frac{u}{v}, y=uv\implies u^2=xy\;$ and $\;v^2=\frac{y}{x}$, so 
$1\le u^2\le3\;, \;1\le v^2\le2 \implies 1\le u\le\sqrt{3},\;1\le v\le\sqrt{2}$.
Since $\displaystyle\begin{vmatrix}x_u &x_v\\y_u & y_v\end{vmatrix}=\begin{vmatrix}\frac{1}{v} &-\frac{u}{v^2}\\v & u\end{vmatrix}=\frac{2u}{v}$, 
$\displaystyle A=\int_1^{\sqrt{2}}\int_1^{\sqrt{3}}\frac{2u}{v}dudv=\int_1^{\sqrt{2}}\frac{2}{v}dv=\big[2\ln v\big]_1^{\sqrt{2}}=2\ln\sqrt{2}=\ln 2$.

Alternatively, $xy=k\implies r^2\sin\theta\cos\theta=k\implies r^2=\frac{k}{\sin\theta\cos\theta}=2k\csc 2\theta$, 
so using polar coordinates gives $\displaystyle A=\int_{\pi/4}^{\tan^{-1}2}\frac{1}{2}\left(6\csc 2\theta-2\csc2\theta\right)d\theta=\int_{\pi/4}^{\tan^{-1}2}2\csc2\theta d\theta$
$\displaystyle=\int_{\pi/4}^{\tan^{-1}2} \frac{2}{2\sin\theta\cos\theta}d\theta=\int_{\pi/4}^{\tan^{-1}2}\frac{\sec^2\theta}{\tan\theta}d\theta=\big[\ln(\tan\theta)\big]_{\pi/4}^{\tan^{-1}2}=\ln 2-\ln1=\ln 2$.
