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

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?
 

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Bobosharif S. | Mathematics/Statistics TutorMathematics/Statistics Tutor
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r=6sin(3θ)
r2+(dr/dθ)2=36sin2(3θ)+9*36cos2(3θ)=36(1-cos2(3θ)+9cos2(3θ))
=36(1+8cos2(3θ)
0π/36√[1+8cos2(3θ)]dθ=10π
0π6√[1+8cos2(3θ)]dθ=30π
Probably somewhere a mistake with integration.