Let $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a continuously differentiable function. Let $S_\varepsilon,B_\varepsilon$ be the sphere and ball of radius $\varepsilon$ around the origin, respectively. Show that $$\frac{1}{v(B_{\varepsilon})}\iint_{S_{\varepsilon}}zf(x,y,z){\rm d}S\underset{\varepsilon\rightarrow0}{\rightarrow}\frac{\partial f}{\partial z}(0,0,0).$$ Here $v$ signifies volume.
My strategy is to divide $S_\varepsilon$ into two hemispheres $S_\varepsilon^+$ and $S_\varepsilon^{-}$ according to the sign of $z$. Then we can write $$ \iint_{S_{\varepsilon}}zf(x,y,z){\rm d}S=\iint_{S_{\varepsilon}^{+}}z\left[f(x,y,z)-f(x,y,-z)\right]{\rm d}S. $$ Now I would like to substitute $$ f(x,y,z)-f(x,y,-z)=\int_{-z}^{z}\frac{\partial f}{\partial z}(x,y,u){\rm d}u, $$ then turn the entire expression into a triple integral over $B_\varepsilon$. Here things get a bit messy, mostly because of having both $z$ and $u$, and I wonder if there's a more elegant way.
As a side note, I realize the goal should be reaching something that looks like $$ \frac{1}{v(B_{\varepsilon})}\iiint_{B_{\varepsilon}}\frac{\partial f}{\partial z}(x,y,z){\rm d}V $$ since this is simply an average, and it approaches the desired value $\frac{\partial f}{\partial z}(0,0,0)$.
Any thoughts or suggestions?
Note that $$ zf(x,y,z)=\varepsilon n\cdot(0,0,f(x,y,z)), $$ where $n$ is the unit normal at each point of $S_\varepsilon$. It follows from the Divergence Theorem that $$\iint_{S_{\varepsilon}}zf(x,y,z)\,{\rm d}S= \varepsilon\iiint_{B_{\varepsilon}}\frac{\partial f}{\partial z}(x,y,z)\,{\rm d}V. $$ Finally, by the continuity of the derivative we obtain $$\frac{1}{{\color{red}\varepsilon}\, v(B_{\varepsilon})}\iint_{S_{\varepsilon}}zf(x,y,z){\rm d}S\underset{\varepsilon\rightarrow0}{\rightarrow}\frac{\partial f}{\partial z}(0,0,0). $$