Stranger region

Define $u = x y$ and $v = x^2-y^2$ . Then, $\text{du} = y \text{dx} + x \text{dy}$ and $\text{dv} = 2x \text{dx} - 2y \text{dy}$ . Solve this system of equations for $\text{dx}$ and $\text{dy}$ to get $\text{dx} = \frac{y}{x^2+y^2} \text{du} + \frac{x}{2x^2+2y^2} \text{dv}$ and $\text{dy} = \frac{x}{x^2+y^2} \text{du} - \frac{y}{2x^2+2y^2} \text{dv}$ . Thus, $\text{dA} = \text{dx} \wedge \text{dy} = \left(\frac{y}{x^2+y^2} \text{du} + \frac{x}{2x^2+2y^2} \text{dv}\right) \wedge \left(\frac{x}{x^2+y^2} \text{du} - \frac{y}{2x^2+2y^2} \text{dv}\right) = \left(-\frac{x^2}{2 \left(x^2+y^2\right)^2}\right) \text{du} \wedge \text{dv} + \left(\frac{y^2}{2 \left(x^2+y^2\right)^2}\right) \text{dv} \wedge \text{du} = \left(-\frac{x^2}{2 \left(x^2+y^2\right)^2}-\frac{y^2}{2 \left(x^2+y^2\right)^2}\right) \text{du} \wedge \text{dv} = \left(-\frac{1}{2 \left(x^2+y^2\right)}\right) \text{du} \wedge \text{dv}$ . 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 the $x$ -axis. So, our area is $\displaystyle \iint \text{dA} = \iint \text{dx} \wedge \text{dy} = 2 \int _{-1}^1\int _{-1}^1\frac{1}{\sqrt{16 u^2+4 v^2}} \text{du} \, \text{dv} = \frac{1}{4} \int_{-2}^{2} \int_{-4}^{4} \frac{1}{\sqrt{a^2+b^2}} \text{da} \, \text{db} = \log \left(\frac{123+55 \sqrt{5}}{2}\right) \approx 4.812 \Rightarrow \left\lfloor 100A \right\rfloor = \boxed{481}$ .

I think the best way to approach this is through polar coordinates.

We need to find $\angle AOB$ and $\angle AOC$ .

To find it let us find the coordinates where both the curves meet.

We have $xy=1$

So $y=\frac{1}{x}$

So putting it in $x^2-y^2 = 1$

we have

$x^4-x^2-1 = 0$

So $x=\sqrt{\frac{1+\sqrt{5}}{2}} =\sqrt{a}(say)$

So $y=\sqrt{\frac{2}{1+\sqrt{5}}} = \frac{1}{\sqrt{a}}$

So $\angle AOB = \arctan(\frac{1}{a})$

And $\angle AOC = \arctan(a)$

There are $8$ total regions like $AOB$ and 4 total regions which are like $BOC$

For $x^2-y^2 = 1$ using polar coordinates we have

$r^{2} = \frac{1}{cos(2\theta)}$

For $xy = 1$ using polar coordinates we have:-

$r^{2} = 2\cosec(2\theta)$

Area of $AOB$ = $\Large \int_{0}^{\arctan(\frac{1}{a})} \frac{1}{2cos(2\theta)}$

There are 8 such regions so total area of them is $\Large \int_{0}^{\arctan(\frac{1}{a})} 4sec(2\theta) d\theta$

Area of $BOC$ = $\Large \int_{\arctan(\frac{1}{a})}^{\arctan(a)} \cosec(2\theta) d\theta$

There are 4 such regions so their area is given by $\Large 4\int_{\arctan(\frac{1}{a})}^{\arctan(a)} \cosec(2\theta) d\theta$

Now indefinite integrals of $\cosec(x)$ and $\sec(x)$ are well known.

Using them and plugging in the limits we get the total area as

$\Large \int_{0}^{\arctan(\frac{1}{a})} 4sec(2\theta) d\theta + \Large 4\int_{\arctan(\frac{1}{a})}^{\arctan(a)} \cosec(2\theta) d\theta$

We get our answer as $\ln(\frac{123 + 55\sqrt{5}}{2})$

(Note that I have used the formula for area in polar coordinates : Area = $\int \frac{r^{2}}{2} d\theta$