^{2}+(dr/dθ)

^{2}=36sin

^{2}(3θ)+9*36cos

^{2}(3θ)=36(1-cos

^{2}(3θ)+9cos

^{2}(3θ))

^{2}(3θ)

_{0}

^{π/3}6√[1+8cos

^{2}(3θ)]dθ=10π

_{0}

^{π}6√[1+8cos

^{2}(3θ)]dθ=30π

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?

Tutors, sign in to answer this question.

r=6sin(3θ)

r^{2}+(dr/dθ)^{2}=36sin^{2}(3θ)+9*36cos^{2}(3θ)=36(1-cos^{2}(3θ)+9cos^{2}(3θ))

=36(1+8cos^{2}(3θ)

∫_{0}^{π/3}6√[1+8cos^{2}(3θ)]dθ=10π

∫_{0}^{π}6√[1+8cos^{2}(3θ)]dθ=30π

Probably somewhere a mistake with integration.

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