Divergence theorem application.

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.

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 $r<\delta$
• $$\lim_{r\rightarrow 0}\frac{1}{vol(B_r(x)}\int_{B_r(x))}\, h(y) dy = h(x)$$ -- 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 – user335501 Jun 17 '17 at 15:38
• 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. – Uskebasi Jun 17 '17 at 16:12