Which limits do you use when finding arclength of a rose?

Question

What is the arclength of r=6sin(3θ)?

Given the formula Lpolar= ∫√[r^2+(dr/dθ)^2]dθ I can find the arc length using the r stated above and r'=18cos(3θ)

Given that one petal is between θ of 0 and pi/3 (found by setting r=0 and choosing the first consecutive numbers), I would think that multiplying the integral with these limits by 3 (the number of petals) would give me the same answer as the integral from 0 to 2pi

However the integral from 0 to 2pi is ~80 and 3*the integral from 0 to pi/3 ~40

 

Why is this? Have I made a mistake?

 

Answer 1

r=6sin(3θ)

r²+(dr/dθ)²=36sin²(3θ)+9*36cos²(3θ)=36(1-cos²(3θ)+9cos²(3θ))

=36(1+8cos²(3θ)

∫₀^π/36√[1+8cos²(3θ)]dθ=10π

∫₀^π6√[1+8cos²(3θ)]dθ=30π

Probably somewhere a mistake with integration.