Find the area of the region that is inside the circle r = 6cos(theta) but outside the cardoid r = 2 + 2cos(theta)
r = 6·cosθ
r = 2 + 2·cosθ
The Attempt at a Solution
intersections of the two curves.
6·cosθ = 2 + 2·cosθ → 4·cosθ = 2
cosθ = 1/2 → θ = ±π/3
Can someone finish the integral for me, I'm not good at integrals, this is as far as i can get. thanks for any help.
A = 2 x (1/2) ∫ [(6·cosθ)² - (2 + 2·cosθ)²] dθ