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Arc Length.

  1. Feb 5, 2008 #1
    It's easy question,but I don't know whether I solved it correctly.
    1. The problem statement, all variables and given/known data
    Calculate the length of the curve given by
    [tex]r=a\sin^3 \frac{\theta}{3}[/tex]
    in polar coordinates. Here, a > 0 is some number.

    2. Relevant equations

    [tex]l=\int \sqrt{r^2(\theta)+(\frac{dr}{d\theta})^2}d\theta[/tex]

    3. The attempt at a solution

    [tex]l=\int \sqrt{a^2 \sin^6\frac{\theta}{3}+a^2\sin^4\frac{\theta}{3}\cos^2\frac{\theta}{3}}\theta[/tex]

    [tex]l=a\int \sin^2\frac{\theta}{3}d\theta[/tex]
    for [tex]0<\frac{2\theta}{3}<2\pi[/tex]


  2. jcsd
  3. Feb 5, 2008 #2
    It's all correct except for the very last line. You forgot to include [tex]a[/tex]. Your answer should be:

    [tex] s = \frac{3a\pi}{2}[/tex]

    p.s. Use [tex]s[/tex] for arclength--it's more widely used and recognized. Also, you can include limits of integration like this: \int^b_a Always put the ^ first, though. Otherwise it doesn't work right.
    Last edited: Feb 5, 2008
  4. Feb 5, 2008 #3


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    Staff Emeritus
    Science Advisor

    It doesn't? What the difference between
    [tex]\int_0^1 f(x)dx[/tex]
    [tex]\int^1_0 f(x)dx[/tex]
  5. Feb 5, 2008 #4
    hmm. That's odd. I guess it just didn't work right when I tried it. Ah, well. Not a very scientific conclusion, eh?
  6. Feb 6, 2008 #5
    Thanks a lot!!!
  7. Feb 6, 2008 #6


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    Staff Emeritus
    Science Advisor

    That's alright. There are millions of thing that work for everyone except me!
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