# flux and integrals over a closed surface

• December 1st 2008, 05:18 PM
lauren2988
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.
• December 1st 2008, 06:53 PM
ThePerfectHacker
Quote:

Originally Posted by lauren2988
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. (Surprised)
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,
$0=\iint_S \text{div}(\bold{F}) dA = \oint_C (n_1+n_2) ds$