1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
  6. Jul 21, 2012 #5


    User Avatar
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your original result is actually correct, but you made a sign error in your work.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook