Area of region inequality What is the area of the region defined by the following set of inequalities?
$$\begin{array}{cc} (1) &-1 < xy < 1 \\ (2) &-1 < x^2-y^2 < 1 \end{array}$$
I think we use integration here

The area that we want is:

Define $u = x y$ and $v = x^2-y^2$. Then, $ {du} = y  {dx} + x  {dy}$ and $ {dv} = 2x  {dx} - 2y  {dy}$. 
Solve this system of equations for $ {dx}$ and $ {dy}$ to get: 
$ {dx} = \frac{y}{x^2+y^2}  {du} + \frac{x}{2x^2+2y^2}  {dv}$ and $ {dy} = \frac{x}{x^2+y^2}  {du} - \frac{y}{2x^2+2y^2}  {dv}$. 
Thus, $$ {dA} =  {dx} \wedge  {dy} = \left(\frac{y}{x^2+y^2}  {du} + \frac{x}{2x^2+2y^2}  {dv}\right) \wedge \left(\frac{x}{x^2+y^2}  {du} - \frac{y}{2x^2+2y^2}  {dv}\right) = \left(-\frac{x^2}{2 \left(x^2+y^2\right)^2}\right)  {du} \wedge  {dv} + \left(\frac{y^2}{2 \left(x^2+y^2\right)^2}\right)  {dv} \wedge  {du} = \left(-\frac{x^2}{2 \left(x^2+y^2\right)^2}-\frac{y^2}{2 \left(x^2+y^2\right)^2}\right)  {du} \wedge  {dv} = \left(-\frac{1}{2 \left(x^2+y^2\right)}\right)  {du} \wedge  {dv}$$. 
Call $f = -\frac{1}{2 \left(x^2+y^2\right)}$, and recall that $u = xy$ and $v = x^2-y^2$. 
Observe then that $f^2 = \frac{1}{4 (x^4+2x^2 y^2 + y^4)} = \frac{1}{4 (4 u^2 + v^2)} = \frac{1}{16u^2+4v^2}$. Thus, $f = \pm \frac{1}{\sqrt{16u^2+4v^2}}$, where each solution for $f$ corresponds to half of the problem region, as the two are symmetric over $y=x$. 
So, our area is $$\displaystyle \iint  {dA} = \iint  {dx} \wedge  {dy} = 2 \int _{-1}^1\int _{-1}^1\frac{1}{\sqrt{16 u^2+4 v^2}}  {du} \,  {dv} = \frac{1}{4} \int_{-2}^{2} \int_{-4}^{4} \frac{1}{\sqrt{a^2+b^2}}  {da} \,  {db} = \log \left(\frac{123+55 \sqrt{5}}{2}\right) \approx 4.812$$.
