Integral of a function over the 2-sphere (Gauss theorem-explanation needed)

Question

Let $\mathcal S^2:=\{(x,y,z)\in \mathbb R^3 \vert\; x^2+y^2+z^2=1\}$ be the standard 2-sphere. For some $p=(x,y,z)\in \mathcal S^2$, the outer unit normal field of $B(0, 1)$ on $\mathcal S^2$ is given by $\nu(p) = p$ and can be extended to $\mathbb R^3$ by $\hat \nu(p) = p$. We also assume that $f:\mathcal S^2 \to \mathbb R$ is a function such that $f(x,y,z)=\hat f(x)$ where $\hat f$ denotes its extension in $\mathbb R^3$.

I am trying to get my head around the following estimate which is a part of some lecture notes:

QUESTIONS:

1. I see that in the first equality the Gauss theorem is used. However, I do not completely agree with this formulation. Why is the divergence of the vector field $f(p)p$ instead of simply $\nabla f(p)$ in the right-hand side? Moreover, in the left-hand side of the same equality, shouldn't we have a product of the form $f\cdot p$?
2. For the second equality I understand that $-1<x<1$ while $(y,z) \in B(0,\sqrt{1-x^2})$. But isn't some $dydz$ missing from here? Where did this go?
3. What is more, isn't $f(p)p=\hat f(x)\begin{pmatrix} x\\ y\\ z\end{pmatrix}$? If so, how come the substitution inside the integral appears?

I am really confused here and I would appreciate it if someone could break it down into pieces so I could understand what it's going on.

Many thanks in advance!

Answer 1

1. The vector field $\nu(p)=p$ has unit norm on the unit sphere, so it is the outward unit normal. So, on $S^2$, we have $f= f\cdot 1 = f\cdot \|\nu\|^2=f\langle\nu,\nu\rangle=\langle f\nu,\nu\rangle$. Therefore, we can apply the divergence theorem to the vector field $F=f\nu$. There is nothing special about $\nu(p)=p$; in general if you have a boundary surface $\partial\Omega$ with outward unit normal $\nu$ and a smooth function $f:\Bbb{R}^n\to\Bbb{R}$. then you can say $\int_{\partial\Omega}f\,dA=\int_{\Omega} \text{div}(f\nu)\,dV$.

2. I would just write it as \begin{align} \int_{S^2}f\,dA&=\int_{B(0,1)}\text{div}(f\nu)\,dV=\int_{-1}^1\int_{B(0,\sqrt{1-x^2})}\text{div}(f\nu)\,dA(y,z)\,dx, \end{align} where I use the notation $dA(y,z)$ to mean two dimensional Lebesgue measure, where the integral is taken with respect to the variables $y,z$.

3. You literally just compute the divergence straight from the definition: \begin{align} \text{div}(f\nu)= x\hat{f}'(x)+3\hat{f}(x)=\frac{1}{x^2}\frac{d}{dx}\left(x^3\hat{f}(x)\right), \end{align} where the final equality of course assumes $x\neq 0$.