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Area between Polar curves

  1. Sep 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area inside r = 9sinθ but outside r = 2




    2. Relevant equations
    Area = 1/2(Integral of (f(θ)^2 - g(θ)^2)dθ



    3. The attempt at a solution
    f(θ)^2 =
    81sin^2θ = 81((1-cos(2θ))/2)
    g(θ)^2 = 4

    f(θ)^2 - g(θ)^2 = 36.5 - cos(2θ)/2
    integral of (36.5 - cos(2θ)/2)
    [36.5θ - sin(2θ)/4]

    Area = 1/2[36.5θ - sin(2θ)/4]
    If I integrate from 0 to ∏/2 then multiply that area times two I get 114.6681319
    Which is the wrong answer.
    Where did I go wrong ;-(
     
  2. jcsd
  3. Sep 9, 2013 #2

    CAF123

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    How did you get the limits 0 to π/2? Have you sketched the two circles?
     
  4. Sep 9, 2013 #3
    Should I integrate from the intersection point to pi/2?
    9sin(theta) = 2
    theta = arcsin(2/9)
    theta = .224093
    If I do that that is half of the area and then I can multiply that by two to get the full area. Does that look better to you?
     
  5. Sep 9, 2013 #4

    CAF123

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    Yes, that is valid because the shaded region is symmetric wrt the y axis.
    I would suggest subbing in arcsin(2/9). The range of this function is [-pi/2,pi/2] so subbing this in will give you the required theta.

    Yes.
     
  6. Sep 10, 2013 #5
    I plug these in for theta and then get a required theta? I'm confused by what you mean here. Do you mean that the limits of integration are [-∏/2,∏/2]?
     
  7. Sep 10, 2013 #6

    CAF123

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    No, just sub in arcsin(2/9) instead of subbing in 0.224... to avoid rounding errors. The principal solution of arcsin(2/9) is the one that appears in [-pi/2,pi/2] which is the theta that the two curves intersect at in the first quadrant.
    You could find the other theta corresponding to the other intersection and integrate between these two values, but your method is also good, perhaps more elegant given you have noticed the symmetry.
     
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