Use polar coordinates to set up the double integral x dA, where the bounds are given by the region lying in the intersection of the 2 circles r = 2 sin (theta) and r = 2 cos theta.

Thanks for any help!

recall that

dA = r dr dθ

so, since
x = r cosθ

you have

∫[0,π/2] r cosθ r dr dθ
= ∫[0,π/2] r^2 cosθ dr dθ
But, r changes at π/4, so that gives

∫[0,π/4]∫[0,2sinθ] r^2 cosθ dr dθ
+ ∫[π/4,π/2]∫[0,2cosθ] r^2 cosθ dr dθ
= 1/6 + (π/4 - 2/3)
= (π-2)/4

As always, check my math.