What is the area of the inner loop of r = 3+6sinΘ?

Answer 1

you need to find the angles θ which bound the inner loop.
3+6sinθ = 0 when sinθ = -1/2
So, the loop is traced out when 7π/6 <= θ <= 11π/6

So, the area inside the loop is

∫[7π/6,11π/6] 1/2 r^2 dθ
= ∫[7π/6,11π/6] 1/2 (3+6sinθ)^2 dθ = 9π-(27√3)/2