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Greens Theorem Area of ellipse

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the area swept out by the line from the origin to the ellipse x=acos(t) y=asin(t) as t varies from 0 to [itex]t_{0}[/itex] where [itex]t_{0}[/itex] is a constant between 0 and 2[itex]\pi[/itex]

    2. Relevant equations



    3. The attempt at a solution

    so using Greens Theorem in reverse i get A=[itex]\frac{1}{2}\oint_{c} ydx-xdy[/itex]

    x=acos(t) dx=-asin(t)
    y=asin(t) dy=cos(t)

    so sub into my equation i get [itex]\frac{1}{2}\int^{t_0}_{0} a(sin^2 (t) - cos^2(t)) dt[/itex]

    [itex]-\frac{1}{2}\int^{t_0}_{0} a(1) dt[/itex]

    I think im good up to here, i then integrate and get [itex]-\frac{1}{2} at_0[/itex] but im not sure about how to use the information that [itex]t_0[/itex] varies from 0 to 2[itex]\pi[/itex]
     
  2. jcsd
  3. Jul 21, 2012 #2
    Are you sure? I don't think sin^2 (t) - cos^2(t) is equal to -1 ...
     
  4. Jul 21, 2012 #3
    ahh yes, what a mistake a ta make a

    after i integrate i should get -sin2t[itex]_0[/itex] but my problem is still how to use the information that [itex]t_o[/itex] varies from 0 to 2[itex]\pi[/itex]
    ...
     
  5. Jul 21, 2012 #4
    .. I don't think you have to worry about the range of t_0. All this means is that the total area will be less or equal to that of the ellipse, it doesn't change the equations.
    Wakabaloola
     
  6. Jul 21, 2012 #5

    uart

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    Are you sure that's the problem statement. It's not a very general ellipse, seeing as it is a circle. In this case of course the area is simply that of a sector of angle t0, hence A = (1/2) a^2 t0.
     
  7. Jul 21, 2012 #6

    vela

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    Your original result is actually correct, but you made a sign error in your work.
     
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