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Math Help - Topology. Computational exercise.

  1. #1
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    Topology. Computational exercise.

    Let U = \mathbb{R}^2     \setminus \{(0,0)\}; consider the 1-form in U defined by


    \omega = \dfrac{x dx + ydy}{x^2+y^2}


    (a) Show that d\omega =0


    (b) Show that there is a 0-form \theta on U, such that d\theta = \omega





    Please help!!!
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  2. #2
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    Re: Topology. Computational exercise.

    (a) d\omega=\frac{2xy dy\wedge dx + 2xy dx \wedge dy}{x^2+y^2}=0
    (b) since d\omega=0, for any close loop \gamma not enclosing 0, we have \oint_{\gamma} w ds = \int_D dw=0
    so we can define \theta(p) = \int_{p_0}^{p} w, where p_0 \ne 0 is fixed and the line integral goes on any curve that not crossing 0.
    Choose p_0=(1,0), and the curve from (1, 0) to p(x0,y0) be from (1,0) to (x0,0), then from (x0, 0) to (x0,y0)
    we get \theta(x_0,y_0)=\int_{(1,0)}^{(x_0, y_0)} w = \int_1^{x_0}\frac{xdx}{x^2+y^2} + \int_0^{y_0} \frac{ydy}{x^2+y^2}
    =  \frac{1}{2}log(x_0^2+y_0^2)
    Last edited by xxp9; December 6th 2012 at 08:19 AM.
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