In class we were given this definition: $Div (\vec{F}):= lim_{r \rightarrow0} \oint_{C} \vec{F} \cdot \vec{n} \:ds$ (where r is the radius of the circle C and $\vec{n}$ is the outward pointing normal of the curve C).
How do i show that defining divergence in this way also implies that
$Div(\vec{F}) = \partial_1 F_1 + \partial_2 F_2$?
And so on in higher dimensions
Edit: here are my results so far:
$\vec{T} = \frac{dx}{ds} e_x + \frac{dy}{ds} e_y $ which implies for a counter-clockwise (positive?) arc-length-parametrisation $\vec{n} = \frac{dy}{ds} e_x - \frac{dx}{ds} e_y$ such that
$\operatorname{Div}(\vec{F}) = \lim_{r\to0} \frac{1}{\pi r^2}\oint_{C}\vec{F} \cdot \vec{n} \: ds = \lim_{r\to0} \frac{1}{\pi r^2}\oint_{C} \begin{bmatrix} F_1 \\ F_2 \end{bmatrix} \cdot \begin{bmatrix} \frac{dy}{ds} \\ \frac{dx}{ds} \end{bmatrix} \: ds = \lim_{r\to0} \frac{1}{\pi r^2}\oint_{C} F_1\frac{dy}{ds} + F_2\frac{dx}{ds} \: ds$
But this still isn't quite right edit2: meant to say this still quite doesn't show what I'm looking for
Something is wrong. The divergence of $F(x,y)=({x\atop y})$ is two but according to your limit formula $$ \lim_{r\to 0}\int\limits_{\sqrt{x^2+y^2}=r}F\cdot\hat{\mathbf{n}}\,ds= \lim_{r\to 0}\;\int_0^{2\pi}r^2\,\,d\theta =\lim_{r\to 0} \,2\,\pi\,r^2=0\,. $$ Hint:
The correct limit formula is $$ \operatorname{div}F=\lim_{r\to 0}\;\frac{1}{\pi r^2}\int\limits_{\sqrt{x^2+y^2}=r}F\cdot\hat{\mathbf{n}}\,ds $$ which we often see as $$ \operatorname{div}F(x,y)=\lim_{|V|\to 0}\frac{1}{|V|}\int_{C}F\cdot\hat{\mathbf{n}}\,ds $$ where $|V|$ is the area ("volume") enclosed by $C\,.$ That volume has to shrink to the point $(x,y)\,.$ Hint for proof: Gauss' or Green's theorem.