Very strange shape (2)

Just a quick outline ...

Focussing on the 'shape' in the first quadrant, the points of intersection are the positive roots of $x^{4} - 4*x^{2} + 1 = 0$ , which come out to be $a = \sqrt{2 - \sqrt{3}}$ and $b = \sqrt{2 + \sqrt{3}}$ .

We then need to calculate

$\int_{a}^{b} (\sqrt{4 - x^{2}} - \frac{1}{x}) dx$ .

For the first term we'll need to make a trig substitution of $x = 2\sin(\theta)$ . Going through the motions, we find that the integral comes out to

$2*\arcsin(\frac{x}{2}) + \frac{x}{2}\sqrt{4 - x^{2}} - \ln{x}$ ,

evaluated from $a$ to $b$ . The area of this 'upper' shape is then

$\dfrac{2\pi}{3} - \ln(2 + \sqrt{3})$ ,

which by symmetry will also be the area of the 'lower' shape. To $3$ decimal places the combined area is then $1.555$ , leaving us with a final answer of $\lfloor \beta \rfloor = \boxed{1}$ .

Too much bashable, anyone with a little logic will bash this the same way as I did.

We know how the 2 graphs look, 1 is a hyperbola and other is a circle. They intersect as shown in the following figure img

Now, area of the circle is $\pi r^2 = 4\pi$ .

Hence the area of one quarter circle is $\pi$ .

And area of one part of the intersection (2 parts as you see) ,is even less than $\dfrac{1}{3}$ of the area of the quarter circle, so the area shaded is less than $\dfrac{2\pi }{3} = 2.0943951023931953$ , so answer will be either $2$ or $1$ .

But the area is too less than the $\frac{1}{3}$ of quarter circle, so it won't be greater than 2.

Answer $\boxed{1}$

Bashed badly, but waiting for elegant way. Sorry @Sharky Kesa .