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Finding arc length of polar Curve

  1. Sep 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the arc length of polar curve 9+9cosθ


    2. Relevant equations

    L = integral of sqrt(r^2 + (dr/dθ)^2 dθ
    dr/dθ = -9sinθ
    r = 9+9cosθ

    )3. The attempt at a solution

    1. Simplifying the integral
    r^2 = (9+9cosθ^2) = 81 +162cosθ + 81cos^2(θ)
    (dr/dθ)^2 = 81sin^2(θ)

    r^2 + (dr/dθ)^2 = 81 + 162cosθ + 81cos^2(θ) + 81sin^2(θ)
    81sin^2(θ) + 81cos^2(θ) = 81

    162 + 162cosθ = r^2 + (dr/dθ)^2

    now I have to take the integral of the squareroot...

    Integral of sqrt(162 + 162cosθ)dθ
    chain rule..?
    (2/3)(162+162cosθ)^3/2*(162θ + 162sinθ)

    Integrated between 0 and 2pi...?
    which would lead to a crazy high number that I got as 3957501.966.
    Anyone know where I went wrong?
     
  2. jcsd
  3. Sep 9, 2013 #2

    SteamKing

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    Science Advisor
    Homework Helper

    Well, for one thing, your indefinite integration is wrong. You can't use the power formula because you don't have the proper du outside the square root. You need to use a trig substitution instead.
     
  4. Sep 9, 2013 #3
    For the other thing please use LaTeX...just a suggestion.
     
  5. Sep 9, 2013 #4
    Sorry i'll learn LaTex soon, in a bit of a pinch at the moment.

    is the final integral end up being
    integral of(162 + 162cos(theta) ???
     
  6. Sep 9, 2013 #5

    CAF123

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    Gold Member

    Yes, a half angle sub will help to evaluate it.
     
  7. Sep 9, 2013 #6
    Why use a half angle sub? Doesn't it just come out to 162θ+162sinθ between 0 and 2∏?
     
  8. Sep 10, 2013 #7

    CAF123

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    Gold Member

    Sorry,I misread what you wrote: it should have been $$\int_0^{2\pi} \sqrt{162 + 162\cos\theta}\,d\theta,$$ which you will need to use a trig sub to evaluate. Factor out the 162 from the sqrt first.
     
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