We consider the polar function r = 2 R Cos[q] and require that the polar region BAC be 1/3 Pi R^2. We integrate from angle -a to angle a (where a = q/2).

Input := 


cut[a_] = Integrate[1/2 (2 R Cos[t])^2,{t,-a,a}]

Output =


 2

R  (2 a + Sin[2 a])