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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- Dec 1st 2008, 05:18 PMlauren2988flux 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. - Dec 1st 2008, 06:53 PMThePerfectHacker
You cannot compute a line integral over a vector field. (Surprised)

Therefore the integral of n.ds over just makes no sense.

I think you might mean the following:

Define the vector-field $\displaystyle \bold{F}: \mathbb{R}^2 \to \mathbb{R}^2$ by $\displaystyle \bold{F}(x,y) = (1,1)$.

From here we see that $\displaystyle \text{div}(\bold{F}) = 0$ and $\displaystyle \bold{F}\cdot \bold{n} = n_1 + n_2$ (sum of components of the normal vector).

Thus, by divergence theorem,

$\displaystyle 0=\iint_S \text{div}(\bold{F}) dA = \oint_C (n_1+n_2) ds$