Let $U=\{x:x\in \mathbb{R}^n, x\neq 0\}$ and $f\in C^2(U)$. Define, for $x\in U$, $$f_{\#}(x) = \int_{S_1(0)} f(|x|w)d\sigma(w)$$ Show that $\nabla(f_{\#}) = (\nabla f)_{\#}$
Hint: use the divergence theorem and the formula for polar integration
I know that I must use the divergence theorem:
$$\int_{\Omega}div(\vec{X})(x) dx = \int_{\partial \Omega}\vec{X}(x)\cdot n(x) dx$$
But I don't know what it means by polar integration formula, since the polar integration technique is only for $\mathbb{R}^2$. I don't see what to do in $n$ dimensions.
I tried seeing the integral as
$$f_{\#}(x) = \int_{S_1(0)}f(|x|w)d\sigma(w) = \frac{1}{|x|^{n-1}}\int_{S_{|x|}(0)}f(w)d\sigma(w) $$
but then I'd have to take the laplacian of that, which is unpratical.
I tried to integrate both sides to end with a ball and try to see some divergence in order to apply the divergence theorem:
$$\int_0^1\int_{S_1(0)}|x|^ {n-1}f(|x|w)d\sigma(w) = \int_{B_1(0)}f(|x|w)d\sigma(w)$$
But I don't see how it helps.
I think the best way is to begin with $(\nabla)_{\#}$:
$$(\nabla f)_{\#}(x) = \int_{S_1(0)}\nabla f(|x|w)d\sigma w $$
I think that if I see the function as a vector field times the normal I can use the divergence theorem to transform this integral to an integral of a ball, but I think I cannot do that.