1. In Christ
    Joined
    30 Apr '07
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    04 Jun '07 17:16
    Let P = -y/(x^2 + y^2), Q = x/(x^2 + y^2), and D be the unit disk.

    I'll use S for the integral sign.

    Parameterize: x(t) = cos(t), y(t) = sin(t), 0
  2. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
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    20443
    04 Jun '07 21:58
    Originally posted by Jirakon
    Let P = -y/(x^2 + y^2), Q = x/(x^2 + y^2), and D be the unit disk.

    I'll use S for the integral sign.

    Parameterize: x(t) = cos(t), y(t) = sin(t), 0
    So P = -sin(t) and Q = cos(t)

    Then what?
  3. In Christ
    Joined
    30 Apr '07
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    172
    04 Jun '07 22:101 edit
    Wait...what? I had just typed a whole proof of how 2pi = 0, but I used a less than sign after that 0 in the first post, which was probably misinterpreted as an incomplete tag. Let me try again:

    Let P = -y/(x^2 + y^2), Q = x/(x^2 + y^2), and D be the unit disk.

    I'll use S for the integral sign.

    Parameterize: x(t) = cos(t), y(t) = sin(t), t is between 0 and 2pi inclusive

    S Pdx + Qdy
    S -sin(t) x -sin(t) + cos(t) x cos(t) dt
    S sin^2(t) + cos^2(t) dt
    S dt from 0 to 2pi = 2pi

    SS dQ/dx - dP/dy dx dy
    SS (y^2 - x^2)/(x^2 + y^2)^2 - (y^2 - x^2)/(x^2 + y^2)^2 dx dy
    SS 0 dx dy = 0

    Therefore 2pi = 0