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Math Help - flux and integrals over a closed surface

  1. #1
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    flux and integrals over a closed surface

    let C be a smooth simple closed curve in the plane and n be a unit vector normal to C.
    prove that the integral of n.ds over C is 0.
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  2. #2
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    Quote Originally Posted by lauren2988 View Post
    let C be a smooth simple closed curve in the plane and n be a unit vector normal to C.
    prove that the integral of n.ds over C is 0.
    You cannot compute a line integral over a vector field.
    Therefore the integral of n.ds over just makes no sense.
    I think you might mean the following:

    Define the vector-field \bold{F}: \mathbb{R}^2 \to \mathbb{R}^2 by \bold{F}(x,y) = (1,1).
    From here we see that \text{div}(\bold{F}) = 0 and \bold{F}\cdot \bold{n} = n_1 + n_2 (sum of components of the normal vector).
    Thus, by divergence theorem,
    0=\iint_S \text{div}(\bold{F}) dA = \oint_C (n_1+n_2) ds
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