Let $S_r(x)$ the sphere of radius $r>0$ centered at the point $x\in\mathbb{R}^{n}$, that is
$$S_{r}(x)=\{y\in\mathbb{R}^n : |x −y| = r\} $$
Let $\sigma$ be the $(n-1)$-volume on $S_r(x)$, and note that $\sigma(S_{r} (x)) =\omega_{n}r^{n-1}$ where $\omega$ is the area of the unit $(n-1)$-sphere in $\mathbb{R}^n$. Prove that, for any $f\in C^{2}(\mathbb{R}^{n})$ and for all $x\in\mathbb{R}^{n}$,
$$\dfrac{1}{\omega_{n}r^{n-1}}\left(\int_{S_{r}(x)}{fd\sigma}\right)-f(x)=\Delta f(x) \dfrac{r^{2}}{2n}+O(r^{2})$$
My approach: Let $x=0$, we have, by Taylor’s expansion at $0$,
$$f(y)=f(0)+\sum_{i=1}^{n}{\dfrac{\partial f(0)}{\partial y_i}y_i}+\frac{1}{2}\sum_{i,j=1}^{n}{\dfrac{\partial^2 f(0)}{\partial y_i\partial y_j}y_{i}y_{j}}$$
So, I need prove that $$\int_{S_r}{y_{i}d\sigma(y)}=0 $$ $$\int_{S_r}{y_{i}y_{j}d\sigma(y)}=0\qquad \mbox{ if } i\neq j$$
Where $S_r=S_{r}(0)$, Can give me some hint to prove this, I think this happen 'cause a symmetri on the sphere. So, we can see that the function $f(y)=y_i$ is an odd function, so if we integrate this function in symmetric domain we have that the first integral is zero, i.e.,
$$I=\int_{S_r}{y_{i}d\sigma(y)}=0$$
So, $I=0$. But isn't clear to me why the second integral vanish.
Another quesion I have is, what happend if I take $f(x)$ as $\Delta f(x)$?, I mean, is correct that
$$\dfrac{1}{\omega_{n}r^{n-1}}\left(\int_{S_{r}(x)}{\Delta f d\sigma}\right)-\Delta f(x)=\Delta (\Delta f(x)) \dfrac{r^{2}}{2n}+O(r^{2})\qquad\quad (*)$$
Thanks!!!!!
