find the area inside one of the small petals of r=1-2sin3theta

Question

i need to help i dont know how to find the min and max values, and please answer this step by step

Answer 1

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=½_θ=π/18∫^θ=5π/18r²dθ=½_θ=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/18^5π/18,

A=½(⅚π-⅔√3+⅙√3-⅙π-⅔√3+⅙√3),

A=½(⅔π-√3)=⅓π-½√3=0.1812 approx.