1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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


    User Avatar
    Science Advisor

    That's alright. There are millions of thing that work for everyone except me!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook