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Math Help - Green's thm.

  1. #1
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    Green's thm.

    1) \oint_C x^3dx-y^3dy, where C is the circle x^2+y^2=1 in an anticlockwise direction.

    I got 0. Is that right? (Since Q_x-P_y = 0.)

    2) Now this one I'm sort of stuck on: EDIT: Oops! Corrected sign.
    \oint_C (x-y)dx {\color{red}+} (x+2y)dy where C is closed curve \frac{x^2}{4}+y^2=1 in an anticlockwise direction.

    So that equals:
    2\iint_D dA where D is the region bounded by C (the ellipse with radii 2 and 1).

    Now the problem here is computation. I opted to go to polar coords (maybe that was a bad idea?), so I got:
    x=2\cos\theta, ~y=\sin\theta

    \therefore r=\sqrt{4\cos^2\theta+sin^2\theta}. So far so good?

    So then I did:
    2\int_0^{2\pi}\int_0^{\sqrt{4\cos^2\theta+sin^2\th  eta}} r dr d\theta

    =\int_0^{2\pi} (4\cos^2\theta+sin^2\theta)d\theta

    Now.. do I really have to integrate that? Or is there a simpler way?
    Last edited by scorpion007; May 22nd 2009 at 03:34 AM.
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  2. #2
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    1) Looks correct.

    2) First of all, Q_x - P_y=0, so this one is zero as well.

    As for your parametrization and the limits:
    x=2r\cos\theta,\ y=r\sin\theta, 0 \leq r \leq 1 and the Jacobian is 2r

    \int_0^{2\pi}\int_0^1 (Q_x-P_y)\cdot 2r \cdot drd\theta
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  3. #3
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    Oops, I accidently wrote the wrong sign in the second question. My bad.

    Thanks very much for the explanation though! My parametrization is completely wrong, heh
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  4. #4
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    So the answer is 4\pi for the second one then?
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