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Homework Help: Find the Exact length of the Polar Curve

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Exact length of the Polar Curve for

    No limits of Integration were given which I found to be odd.

    2. Relevant equations

    L= ∫√(r^2+(dr/dθ)^2)dθ

    3. The attempt at a solution



    Got to this point and figured I did something incorrectly.
    Last edited: Apr 29, 2012
  2. jcsd
  3. Apr 29, 2012 #2


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    It's not a good idea to put the statement of the problem only in the title! I started to say "no that is not a problem statement" until I looked up at your title!

    Perhaps they expect you to be able to get that yourself. When [itex]\theta= 0[/itex], [itex]cos(\theta)= 1[/itex] so r= 4; when [itex]\theta= \pi/2[/itex], [itex]cos(\theta)= 0[/itex] so r= 2; when [itex]\theta= pi[/itex], [itex]cos(\theta)= -1[/itex] so r= 0; when [itex]\theta= 3\pi/2[/itex], [itex]cos(\theta)= 0[/itex] so r= 2 and when [itex]\theta= 2\pi[/itex], [itex]cos(\theta)= 1[/itex] so r= 4 again. We make one complete loop around the figure (a "cardiod" since it looks roughly like the "valentine" heart) as [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex]. That shouldn't be too surprizing since cosine has period [itex]2\pi[/itex].

    No, that's correct. Now integrate from 0 to [itex]2\pi[/itex].
  4. Apr 29, 2012 #3


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    Like HallsofIvy said, you should plot the polar curve: r=2(1+cosθ) to find the limits of integration. I have attached the graph to this post.

    Attached Files:

  5. Apr 29, 2012 #4
    Sorry for not putting enough info originally, new to the forums.
    Also when I integrate, I get stuck at
    Only step I could think of next is to u-sub what's inside the radical.
  6. Apr 29, 2012 #5


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    Try trigonometric substitution. Let [itex]\sqrt{\cos \theta}=\tan \phi[/itex]

    After integrating, the answer should be: [tex]2\sqrt{8}.\sqrt{\cos \theta +1}.\tan \frac{\theta}{2}[/tex] Now, you have to put the limits and evaluate.
    The problem is that for [itex]0 \leq \theta \leq 2\pi[/itex], the evaluation gives 0.

    Are you sure the integral for L is correct in your first post?
    Last edited: Apr 29, 2012
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