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Polar arc length

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the arc length of one of the leaves of the polar curve r= 6 cos 6θ.


    2. Relevant equations
    L = ∫sqrt(r^2 + (dr/dθ)^2) dθ
    (I use twice that since the length from 0 to π/12 is only half the petal)

    3. The attempt at a solution
    I seem to get an integral that can't be solved:

    L = 2∫sqrt((6 cos 6θ)^2 + (-36 sin 6θ)^2) dθ
    = 2∫sqrt(36 cos^2 6θ + 1296 sin^2 6θ) dθ

    I simplify the cos^2 and sin^2 to get

    L = 2∫sqrt(36 + 1260 sin^6θ) dθ
    = 12∫sqrt(1+35 sin^2 6θ) dθ

    but that's where I'm stuck. I have no idea how to do that integral. Any help would be sincerely appreciated!
     
  2. jcsd
  3. Sep 11, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

  4. Sep 11, 2014 #3
    But there's got to be some sort of trick to make this easy to work with - nothing in the notes or textbook has this Elliptic Integral thing in it.

    Any ideas?
     
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