Overlapping area of 2 circles

[Guest]

A, B and E are 3 circles all with radius 1 unit.
A and E touch at P while B and E touch at Q.
∠P O Q = x degrees where O is the center of E.
Find the area of the overlapping portion of A and B
if 0 ≦ x ≦ 60

[Guest]

Let $\varphi=arccos\left(\sqrt{2-2cosx}\right)$. Then $S_{A\cap B}=2\varphi-sin2\varphi$.

[Guest]

Thanks much for your reply !

Let us check your answer :

(I) For x = 60 , cos 60 = 1/2 , thus 2cos 60 = 1 ,
then φ = arccos 1 = 0 degree .
Thus SA∩B = 0 degree - 0 = 0 degree ?

(II) For x = 0 , cos 0 = 1 , thus 2 cos 0 = 2 ,
then φ = arccos 0 = 90 degree .
Thus SA∩B = 2φ - sin 180 = 2φ - 0 = 180 degree ?

Should there be something missed ?

[Guest]

$\varphi$ is in radians.

[Guest]

$x=0\Rightarrow\varphi=arccos0=\frac{\pi}{2}\Rightarrow S=\pi-0=\pi$

$x=60^\circ\Rightarrow\varphi=arccos1=0\Rightarrow S=0$

[Guest]

Sorry ! I mis-comprehended your answer . Thanks again !

What will be the expectation of SA∩B if x varies randomly
from 0 to 360 ?

[Guest]

$\bar{S}=\frac{\int_{0^\circ}^{60^\circ}(2arccos\sqrt{2-2cosx}-sin(2arccos\sqrt{2-2cosx}))dx}{\frac{\pi}{3}-0}=...$

$...=\frac{24}{\pi}\int_0^{\frac{\pi}{2}}sin^2\varphi*arcsin\left(\frac{cos{\varphi}}{2}\right)d{\varphi}\approx$

$\approx\frac{24}{\pi}*0.169\ 643\approx 1.295\ 977$

[Guest]

If $x>60^\circ\Rightarrow \sqrt{2-2cosx}>1\Rightarrow arccos\sqrt{2-2cosx}\notin\mathbb{R}$

[Guest]

Thanks much again !

The formula is really complicated !