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Integral - Length of cardioid arc question

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey all, I'm trying to calculate the length of the cardioid r(θ)=1+cosθ (polar coordinates) and I figured I'd try to do it in one integral from 0 to 2Pi.




    2. Relevant equations
    So the integral is [tex]\int_{0}^{2\pi} \sqrt{r^2 + (\frac{dr}{d\theta})^2}d\theta[/tex]



    3. The attempt at a solution
    and it becomes [tex]\int_{0}^{2\pi} \sqrt{1+cos\theta} d\theta[/tex] which becomes I've found
    [tex]\int_{0}^{2\pi} \sqrt{1+cos\theta} d\theta = \int_{0}^{2\pi} \frac{sin\theta}{\sqrt{1-cos\theta}} d\theta = 2\sqrt{1-cos\theta} |_{0}^{2\pi}[/tex]

    but that gives me 0 , although I run it through mathematica and it gives me [tex]4sqrt{2}[/tex]. Why?
     
    Last edited: Jan 6, 2012
  2. jcsd
  3. Jan 6, 2012 #2
    [tex]\int_0^{2\pi} \sqrt{r^2 + (\frac{dr}{d\theta})^2}d\theta = \int_0^{2\pi} \sqrt{(1+cos\theta)^2 + sin^2\theta} d\theta = \int_0^{2\pi} \sqrt{2+2cos\theta}d\theta[/tex]

    Edit: Although this oversight is not enough to fix the problem. The integral is clearly always positive thus it can't be zero. You can note that the integral between 0 and 2pi is symmetric around pi and thus integrate [itex]2\int_0^{\pi} \sqrt{2+2cos\theta}d\theta[/itex]. Something generally goes wrong when you try to do these integrals of trig functions between k and k+2pi, although I'm not sure exactly why that is; maybe somebody else more knowledgable can help explain why.
     
    Last edited: Jan 6, 2012
  4. Jan 6, 2012 #3

    Thanks for answering,
    yep I left the sqrt2 factor out because the real problem is the [tex]\int_{0}^{2\pi} \sqrt{1+cos\theta} d\theta [/tex]. I know, and by taking limits as 0 and pi/2 I can find a non-zero answer, but does mathematica do the same thing or am I missing something?
     
  5. Jan 6, 2012 #4

    Dick

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    [tex]
    \int_{0}^{2\pi} \sqrt{1+cos\theta} d\theta = \int_{0}^{2\pi} \frac{|sin\theta|}{\sqrt{1-cos\theta}} d\theta[/tex]

    You took a square root of [itex]sin^2\theta[/itex] and left out the absolute value. That's what's going wrong.
     
  6. Jan 6, 2012 #5
    Ah I tend to ignore these restrictions because I've been heavily working with indefinite integrals. So I break the integral into two others depending on where sin(theta) is positive or not and add them?
     
  7. Jan 6, 2012 #6

    Dick

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    Sure.
     
  8. Jan 6, 2012 #7
    Ok thanks, rock on!
     
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