What is the area of shaded region which is lies between outer and inner circle. There is a outer circle with radius 2r and another inner circle with radius r whose center is the middle of big circle.As depicted in the following figure.



Foo graph Image

There is a sector of 120 degree in inner circle which leads to shaded part between outer and inner circle. What is the area of shaded circle in terms of r ?

i.e. What is the value of K ?
WLOG let $r=1$.

Let us use a coordinate system centered on the large circle. Then the oblique line on the right has equation $x=\sqrt3(y-1)$ and meets the circle $x^2+y^2=4$.

Solving the quadratic equation, the intersection point is $(\dfrac{\sqrt{39}-\sqrt3}4,\dfrac{\sqrt{13}+3}4)$.

Hence, the aperture angle of the large sector

$$\theta=2\arctan\left(\frac xy\right)=1.19873\cdots \text{rad}=68.6821\cdots°.$$

The requested area is the difference between the large and the small sector and two triangles having base $1$ and height $x$.

$$K=\frac\theta2\,2^2-\frac\pi31^2-\frac22x=4\arctan\left(\frac{\sqrt{39}-\sqrt3}{\sqrt{13}+3}\right)-\frac\pi3-\dfrac{\sqrt{39}-\sqrt3}4=0.222026268\cdots.$$
something like this:
$$
2r^2\left(\int_0^{\frac{\sqrt{3}}2} \left(\sqrt{4-x^2} - 1 - \sqrt{1-x^2} \right)dx + \int_{\frac{\sqrt{3}}2}^{\frac{\sqrt{13}-1}4} \left(\sqrt{4-x^2} - 1-\frac{\sqrt{3}x}2 \right)dx \right)
$$