Area of shaded region circle help 
Find the area of the shaded region


Area of the sector is $240^\circ$ or $\frac{4\pi}{3}$ 
Next find $\frac{b\cdot h}{2}$ which is $\frac{2\cdot2}{2}$ which is $2$.
Then subtract the former from the latter: $\frac{4\pi}{3} - 2$
Therefore the answer is $~2.189$?
Is this correct?

Here, aperture angle $\theta=120^\circ=\frac{2\pi}{3}$
Area of shaded portion  $$=\text{(area of the sector)}-\text{(area of isosceles triangle)}$$
$$=\frac{1}{2}(\theta)(r^2)-\frac{1}{2}(r^2)\sin\theta$$
$$=\frac{1}{2}\frac{2\pi}{3}(2)^2-\frac{1}{2}(2)^2\sin\frac{2\pi}{3}$$
$$=2.456739397$$
Edit:

Note: Area of an isosceles triangle having each of the equal sides $a$ & an angle $\theta$ included between them then the area of the triangle $$=\frac{1}{2}(a)(a)\sin\theta=\frac{1}{2}a^2\sin\theta$$

