Hannah O.

# 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?

By:

Tutor
4.6 (10)

PhD in Math, MS's in Calulus

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.