imc

You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is one-fourth of the area of the circle with radius $2,$ and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is $$\frac{1}{4}4\pi -\frac{1}{2}(2)(2)=\pi -2$$ That means the area of the whole intersection is $\boxed{2(\pi -2)}$