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I need to prove that $$\lim_{r \to 0} \frac 1 {\operatorname{vol}(B_r(p))} \iint_{\partial B_r(p) } f \, dA = \operatorname{div}(f) $$

We have by the Divergence Theorem that $\dfrac 1 {\operatorname{vol}(B_r(p))} \iint_{\partial B_r(p)} f \, dA =\dfrac 1 {\operatorname{vol}(B_r(p)} \iiint_{B_r(p)} \nabla f \, dV $

where $\nabla f = \dfrac{\partial f}{\partial x} +\dfrac{\partial f}{\partial y} +\dfrac{\partial f}{\partial z}$.

Im not sure how to continue from here, I tried working with Spherical Coordinates but I can't compute this triple integral.

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You don't have to (compute the integral). What you need to know (or show) is that, for a (say) continuous function $h$ the identity $$ \lim_{r\rightarrow 0} \frac 1 {\operatorname{vol}(B_r(x)}\int_{B_r(x)}\, h(y) \, dy = h(x)$$ holds.

For continuous $h$ this is almost trivial, since for $\varepsilon >0$ you can finde $\delta >0$ such that $\|\bar y -y \|< \delta \Rightarrow \|h(\bar y)- h(y) \| < \varepsilon$. Now note that then for $r<\delta$

\begin{eqnarray} \left|\frac{1}{\operatorname{vol}(B_r(x))}\int_{B_r(x)}\, h(y) \, dy - h(x)\right| & =& \left|\frac{1}{\operatorname{vol}(B_r(x))}\int_{B_r(x)}\, (h(y) -h(x)) \,dy\right| \\ &\le& \frac{1}{\operatorname{vol}(B_r(x))}\int_{B_r(x)}\, |h(y) -h(x)| \,dy \\ & < & \frac{1}{\operatorname{vol}(B_r(x))}\int_{B_r(x)}\varepsilon \, dy = \varepsilon \end{eqnarray}

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  • $\begingroup$ $$ \lim_{r\rightarrow 0}\frac{1}{vol(B_r(x)}\int_{B_r(x))}\, h(y) dy = h(x)$$ -- this is not what i need to show . i need to show that $$ \lim_{r\rightarrow 0}\frac{1}{vol(B_r(x)}\int_{\partial B_r(x))}\, h(y) dy = \nabla h(x)$$ , dont i ? @Thomas $\endgroup$ – user335501 Jun 17 '17 at 15:38
  • $\begingroup$ Well, if you apply the divergence theorem $\dfrac 1 {\operatorname{vol}(B_r(p))} \iint_{\partial B_r(p)} f \, dA =\dfrac 1 {\operatorname{vol}(B_r(p))} \iiint_{B_r(p)} \nabla\cdot f \, dV$. What @Thomas wrote is true $\forall h$ continuous, in particular for $\nabla\cdot f$, thus the result. $\endgroup$ – Uskebasi Jun 17 '17 at 16:12
  • $\begingroup$ @Uskebasi right! i was confusing with something else. thanks $\endgroup$ – user335501 Jun 17 '17 at 18:01

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