r=1-2sin(3θ), dr/dθ=-6cos(3θ).
r=0 when sin(3θ)=½, 3θ=π/6 or 3θ=5π/6, that is, θ=π/18 and 5π/18.
Between these two values of θ one small petal forms. Between θ=π/18 and π/6, the upper half of the petal forms, and between θ=π/6 and 5π/18 the symmetrical lower half forms. [π/18,5π/18] is the range of values for calculating the required area.
r=0 (at the origin) when θ=π/18. At θ=π/9, r=1-2sin(π/3)=1-√3=-0.73 approximately, which means that r now extends "backwards" into quadrant 3. At θ=π/6, r=1-2sin(π/2)=1-2=-1. At θ=2π/9, r=1-2sin(2π/3)=1-√3 as it was when θ=π/9. So the petal stays in quadrant 3. Take a point on the petal (r,θ). As the petal progresses it sweeps out a sector with infinitesimal area=½r2dθ. The sum of these infinitesimals between the limits is the area of the small petal: A=½θ=π/18∫θ=5π/18r2dθ=½θ=π/18∫θ=5π/18(1-2sin(3θ))2dθ.
(1-2sin(3θ))2=1-4sin(3θ)+4sin2(3θ). cos(6θ)=1-2sin2(3θ), sin2(3θ)=½(1-cos(6θ)).
(1-2sin(3θ))2=1-4sin(3θ)+2-2cos(6θ)=3-4sin(3θ)-2cos(6θ).
A=½[3θ+4cos(3θ)/3-sin(6θ)/3]π/185π/18,
A=½(⅚π-⅔√3+⅙√3-⅙π-⅔√3+⅙√3),
A=½(⅔π-√3)=⅓π-½√3=0.1812 approx.