How to prove that $\int_{\partial A}\frac{x}{\lVert x\rVert^n}\cdot \nu(x)d\sigma = \int_{\partial B_1(0)}1d\sigma$? I am having some difficulty finding a proof idea for this question I found in an old exam:
Suppose $A\subset\mathbb R^n$ is a bounded $C^1$ set with outer unit normal $\nu_A(x)$ and $0\in A$. Prove that $$\int_{\partial A}\frac{x}{\lVert x\rVert^n}\cdot \nu_A(x)d\sigma = \int_{\partial B_1(0)}1d\sigma.$$
Hint: Consider the set $A_\epsilon:=A\setminus B_\epsilon(0)$ for sufficiently small $\epsilon$.
 A: Thanks @TedShifrin
Following the hint and comment recommendations, consider $A_\epsilon\subset A$ and note that $B_\epsilon(0)\subset A$ for sufficiently small $\epsilon$, using the quotient rule
$$\begin{aligned}\int_{A_\epsilon}\text{div} \frac{x}{\lVert x\rVert^n}dx 
&= \int_{A_\epsilon}\sum_{i=1}^n \frac{\partial_i}{\partial x_i} \frac{x}{\lVert x\rVert^n}dx\\
&= \int_{A_\epsilon}\sum_{i=1}^n \frac{\lVert x\rVert^n - x_in\lVert x\rVert^{n-1}x_i\lVert x\rVert^{-1}}{\lVert x\rVert^{2n}}dx\\
&= \int_{A_\epsilon}\sum_{i=1}^n\frac{\lVert x\rVert ^2 - x_i^2 n}{\lVert x\rVert^{n+2}}dx\\
&=n\int_{A_\epsilon}\frac 1 {\lVert x\rVert^n}(1 -\sum_{i=1}^n\frac{x_i^2}{\lVert x\rVert^{2}})dx\\
&= 0.
\end{aligned}$$
So by the divergence theorem, keeping in mind the orientation of the outer normals, we have
$$0 = \int_{\partial A} \frac{x}{\lVert x\rVert^n}\cdot \nu_A(x)d\sigma - \int_{\partial B_\epsilon(0)}\frac{x}{\epsilon^n}\cdot \frac x {\epsilon}d\sigma.$$
Taking a closer look at the second term,
$$\int_{\partial B_\epsilon(0)}\frac{x}{\epsilon^n}\cdot \frac x {\epsilon}d\sigma = \int_{\partial B_\epsilon(0)}\frac{\lVert x\rVert^2}{\epsilon^{n+1}}d\sigma = \int_{\partial B_\epsilon(0)}\frac{1}{\epsilon^{n-1}}d\sigma = \int_{\partial B_1(0)}1d\sigma$$
