Answer :
r = 4 + 4sinθ
y = rsinθ > 3 when π/6 < θ < 5π/6
So to find the shaded area we need to integrate
dA = rdr dθ
between the limits
r = 3/sinθ to 4 + 4sinθ
and θ = π/6 to 5π/6
You need to integrate r first, and you'll get
∫ rdrdθ = ∫ [r^2 /2] dθ
= ∫ [(4+4sinθ)^2 - (3cscθ)^2] dθ
= ∫ [16 + 32sinθ + 16sin^2 θ - 9csc^2 θ] dθ
= [24θ - 32cosθ + 9cotθ - 4sin(2θ)]
Apply the limits to get
16π + 18√3
y = rsinθ > 3 when π/6 < θ < 5π/6
So to find the shaded area we need to integrate
dA = rdr dθ
between the limits
r = 3/sinθ to 4 + 4sinθ
and θ = π/6 to 5π/6
You need to integrate r first, and you'll get
∫ rdrdθ = ∫ [r^2 /2] dθ
= ∫ [(4+4sinθ)^2 - (3cscθ)^2] dθ
= ∫ [16 + 32sinθ + 16sin^2 θ - 9csc^2 θ] dθ
= [24θ - 32cosθ + 9cotθ - 4sin(2θ)]
Apply the limits to get
16π + 18√3