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Area of polar curve

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area inside the larger loop and outside the smaller loop of the limacon r=.5+cosθ
    Picture here http://www.wolframalpha.com/input/?i=r=.5+costheta

    2. Relevant equations
    Area = .5∫r^2


    3. The attempt at a solution
    To get the area of the outer loop, you just get the value of the entire area, and then just subtract the area of the inner loop. I use two integrals for this

    To start, i get the area of the inner loop.
    r=.5+cosθ
    -.5=cosθ
    θ=120 degrees or 2pi/3. and 240 degrees, or 1.33pi.
    .5∫(.5+cosθ)^2 evaluated at 2pi/3(lower bound on integral) and 1.33pi(upper bound on integral).
    I get .135879

    Now, I have to get the area of the entire limacon. I use the bounds zero and 2pi

    .5∫(.5+cos)^2 which 2pi as upper limit and 0 as lower limit.

    I get 2.35619

    Now, 2.35619-.135879 i get 2.2203.

    This is the wrong answer though, the correct answer is 2.08. If i was to subtract another .135879 from 2.2203, that would give me the right answer. But why do i have to subtract another .135879? I should be getting the correct answer with the set up i have.
     
    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 3, 2013 #2

    haruspex

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    The integral right around will include the area in the inner loop twice: once as part of the outer loop and again as the inner loop part of the range.
     
  4. Apr 3, 2013 #3
    So your saying that it takes longer than 2pi to complete before it starts tracing over itself again? and idea how i can find the time it takes a polar curve to complete 1 cycle?
     
  5. Apr 3, 2013 #4

    SammyS

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    This integral needs to have dθ with it.

    The inner loop is for the case where r is negative. That's for values of θ satisfying [itex]\displaystyle \ \frac{1}{2}+\cos(\theta)<0\ .[/itex]
     
  6. Apr 3, 2013 #5

    haruspex

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    No, I'm not saying that. The outer loop range is θ from -2π/3 to 2π/3, say, and the inner loop from 2π/3 to 4π/3. Suppose the two loops have areas A and B. The outer loop contains the inner loop, so the area betwen them is A-B. You computed the integral from 0 to 2π, which is A+B. A-B = A+B -2B.
     
  7. Apr 3, 2013 #6
    Hmm. you would think A+B-B, the B's just cancel and your left with A. but i guess not though
     
  8. Apr 3, 2013 #7

    haruspex

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    But you don't want A. A is the area within the outer loop. That includes the area of the inner loop. you want A-B, the area between the two loops.
     
  9. Apr 3, 2013 #8
    Hmm, ok thanks for explaining. Also, 1 small question. why is the inner loop at .5 at 180 degrees, when cos(pi)+.5 is actually -.5?
     
  10. Apr 3, 2013 #9

    haruspex

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    When theta = pi, r = -.5 in the direction pi, which is the same as +.5 in the direction 0.
     
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