Find the area between the loops of the limacon r=2(1+2cosθ)

Answer 1

since cos 2π/3 = -1/2, r(2π/3) = 0

So, the graph loops back to r=0 at θ=2π/3

Using symmetry, we need consider only the top half of the figure, so we just need to subtract the inner loop from the outer loop, giving us

A/2 = ∫[0,2π/3] 1/2 r^2 dθ - ∫[2π/3,4π/3] 1/2 r^2 dθ

A = ∫[0,2π/3] (2(1+2cosθ))^2 dθ - ∫[2π/3,4π/3] (2(1+2cosθ))^2 dθ

A/4 = ∫[0,2π/3] (1+2cosθ)^2 dθ - ∫[2π/3,4π/3] (1+2cosθ)^2 dθ

A/4 = 3√3/2+2π - (2π-3√3)
A = 18√3