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Polar coords

  1. Jan 18, 2005 #1
    I need help with finding areas. I'm having trouble picking the correct limits for my integration.

    Say for example we had r=2cos2t and a half-line t=pi/6, and I want to find the small area bounded between them.

    [tex]\frac{1}{2}\int (2\cos (2\theta))^{2}\,d\theta[/tex]

    I can do the integration, but what do I choose as its limits?

    I'm not particularly interested in the answer to this question; I'm looking for a good explanation or maybe a couple of pointers.
  2. jcsd
  3. Jan 18, 2005 #2


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    At what value(s) of theta does [itex]r[/itex] become zero ? Set the r(theta) equation to zero and solve for theta. Use the 1st quadrant value of theta obtained as the upper bound with [itex]\frac{\pi}{6}[/itex] as the lower.
  4. Jan 18, 2005 #3
    Why do we want the value of theta at r=0?
  5. Jan 18, 2005 #4


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    Have you even sketched the graph yet ? It will become clear as day if you do.
  6. Jan 18, 2005 #5
    Can you please check my working?

    (a) Sketch the curve with polar equation

    [tex]r=3\cos 2\theta, \, -\frac{\pi}{4}\leq\theta<\frac{\pi}{4}[/tex].

    The curve looks like 1 rose petal.

    (b) Find the area of the smaller finite region enclosed between the curve and the half-line [itex]\theta=\frac{\pi}{6}[/itex].

    [tex]Area = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (3\cos 2\theta)^2\,d\theta = \frac{9}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (1+\cos 4\theta)\,d\theta = \frac{9}{4}\left[\theta+\frac{1}{4}\sin 4\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = \frac{3\pi}{16}- \frac{9\sqrt{3}}{32}[/tex]

    A couple of arithmetic slips probably made it through, since I did all of this using my keyboard & notepad.exe. :tongue2: So I'm just wondering if my method was correct.
    Last edited by a moderator: Jan 18, 2005
  7. Jan 18, 2005 #6


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    The coefficient is 3 ? Not 2 ? If that's the case then your working in the last post is correct. But why did you give a different equation in your first post ? :confused:
  8. Jan 18, 2005 #7
    In my first post I made up an equation and a line, and when I went to do a question from my textbook, it happened to be almost like the one I made up. :tongue2:

    Anyway, thanks for your help. I appreciate it. :smile:
  9. Jan 18, 2005 #8


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