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Finding area in polar coordinates

  1. Jan 21, 2015 #1
    I've attached the solution to this post. The question is essentially just asking to find the area in one loop for r = cos[3(theta)].

    This seems like a fairly simple question (and answer). I've solved and understand the general integration, but I am just a little uncertain on why exactly different limits for integration for the variable theta do not work. For example, the limits used in the solution are -pi/6 to pi/6 which makes perfect sense. Since this is a sinusoidal function with the frequency 3, I was wondering why any interval 2pi/3 rad does not work. For example, the limits of integration I originally used was 0 to 2pi/3. But when I do this, my answer is incorrect. I don't quite seem to understand why this is exactly, but if I'm to wager a guess: is it b/c the new function has a frequency of 6 (i.e. the original function was only + values for r so it did not have its reflection)? If not, any explanation would be greatly appreciated!

    Also, is there any easy way to find the angle for the loop without drawing it? Is there a particular pattern for where the loops begin and end depending on the sinusoid and its frequency?
     

    Attached Files:

  2. jcsd
  3. Jan 21, 2015 #2

    HallsofIvy

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    Yes, as [itex]\theta[/itex] goes [itex]-\pi/6[/itex] to [itex]\pi/6[/itex] point goes exactly once around that loop. If, instead, you go from 0 to [itex]2\pi/3[/itex] the point goes from the right tip of the first loop to the tip of the loop on the bottom. That does NOT bound any area and I don't see why you expect it to give the same integral
     
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