referring to the attached pic:
both triangles ABC and DBC can be proved equilateral and congruent to each other.
the area of sector ABC
= (1/6)(pi)(1^2)
= pi/6
and the area of triangle ABC
= (1/2)(AB)(BC)(sin60)
= (1/2)(1)(1)(3^0.5 / 2)
= 3^0.5 / 4
the difference between these two areas is the area of each one red region (in attachment), let A be it.
hence, the total area of overlapped region of the two circles
= area of triangle ABC x 2 + 4A
= 3^0.5 /4 x 2 + 4(pi/6 - 3^0.5 / 4)
= 1.2284 sq. units