Let $\mathbf{f}:\mathbb{R}^3\to\mathbb{R}^3$ be a continuous vector field which is parallel to the tangent plane at each point of a piecewise-smooth simple surface S. What is the flux of $\mathbf{f}$ through S along its normal vector?
What I have done so far:
The flux of the integral is $\iint\limits_{S}\mathbf{f\cdot \mathbf{n}}dS$ and it's normal vector will be perpendicular to it at every point. Thus making $\mathbf{f}\cdot \mathbf{n}=0$ Hence I think the answer is $0$.
You are right, the dot product with $\mathbb{n}$ takes the normal component of your field, which is $0$. You can think that, at each point of your surface, there is a base of $\mathbb{R}^3$ with two vectors in the tangent plane at that point and the third vector, the normal $\mathbb{n}$. Hence, you can decompose your field $f$ at each point of the surface and the flux takes only the normal component.