To find this, we would have to use how to find the area of a section of the circle and a triangle.
We will lable the circles circle A on the left and circle D on the right.
First. consider the figure on the attachment. All the red lines are equal length since they are all radii. Since the right circle has the same radius as the left circle, the blue lines have equal length to the red lines as well. This makes ABDC is a rhombus, with AE = ED = 1.
Then, we consider triangle ABE, where AB = 2, and AE = 1. Then, BE = 2*(sqrt3)/2 = sqrt3 and BC = 2*sqrt3.
Given this, we can find angle BAC with the law of cosines to be 120 degrees.
So, the area of the sector BDC would be A = pi*2^2 *120/360 = 4pi/3.
We find the area of triangle ABC with A = 1/2 A*B*sinC = 1/2*2*2*sin120 = sqrt3.
The difference of this area is 4pi/3 - sqrt3. The common area for the circles is double this, so:
A = 2(4pi/3 - sqrt3) = 8pi/3 - 2sqrt3, approx. 4.91 square units.
Good luck. I hope this helps.
Thank you very much, but I need the answer in improper integral, not in geometry. Could you help me, please?