Let $f: \mathbb{R^3} \rightarrow \mathbb{R}$ such that $f\in C^1$ assume that $\nabla(f) \neq 0$ at every point in the surface $S$ such that $S=${$f(x,y,z)=0$}. you may assume that $S$ is smooth and bounded. Show that $$ \iint_{S} \nabla(f) \cdot d \vec{S} \neq 0$$
I read the solution for this problem it was written :
By the assumptions the surface S is smooth at each point, and its normal is $\nabla(f)$ we can write $$\iint_{S} \nabla(f) \cdot d \vec{S}=\iint_{S} \nabla(f) \cdot \frac{\nabla(f)}{\|\nabla(f)\|} d S=\iint_{S}\|\nabla(f)\| \cdot d S$$
My questions is:
how the fact that $S$ is smooth relate to the fact that the normal vector to $S$ is $\nabla(f)$?
if the gradient is the normal vector to $S$, why is it possible to write this solution? is it usually $d\vec{s}=\hat n ds$ while n is parallel vector to the surface?
The surface is defined by $f(x,y,z)=0$. The normal direction is defined by the gradient since the most efficient way to increase from a level curve is along a direction orthogonal to it. Hence, $n=\frac{\nabla (f)}{||\nabla(f) ||}$ is the unit normal direction.
From there, the claim is true because you are integrating a magnitude over a region where the vector is never zero. Hence, the integral is with respect to a positive integrand.
The "smoothness" assumption just makes the normal calculation more direct, in terms of the gradient.