Find the area of the region that lies within thecurve r=2cosθ but is outside the curve r=1.To get full credit you need to SHOW WORK.

Answer 1

I don't need full credit, but you can supply any steps you think necessary.
since the region is symmetric, just double the area in the first quadrant,
The two circles intersect where
2cosθ = 1
θ = π/3
so the area is
A = 2∫[0,π/3] 1/2 (R^2-r^2) dθ
= 2∫[0,π/3] 1/2 ((2cosθ)^2-1^2) dθ = 2(π/6 + √3/4)