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
 A: 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 $$\displaystyle \iint  {dA} = \iint  {dx} \wedge  {dy} = 2 \int _{-1}^1\int _{-1}^1\frac{1}{\sqrt{16 u^2+4 v^2}}  {du} \,  {dv} = \frac{1}{4} \int_{-2}^{2} \int_{-4}^{4} \frac{1}{\sqrt{a^2+b^2}}  {da} \,  {db} = \log \left(\frac{123+55 \sqrt{5}}{2}\right) \approx 4.812$$.
A: a) Can you visualize the region in a graph?

b) Can you subdivide the region into copies of two easier regions?
b) Can you rewrite the region in polar coordinates? (Hyperbolas can be rewritten this way.)
c) Can you find an area in polar coordinates?
