Integral - Length of cardioid arc question

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  • #1
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Homework Statement


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.




Homework Equations


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



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:

Answers and Replies

  • #2
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[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:
  • #3
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[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.


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?
 
  • #4
Dick
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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?

[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.
 
  • #5
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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?
 
  • #6
Dick
Science Advisor
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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?

Sure.
 
  • #7
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