Significance of the dot product in surface integral The general definition of a surface integral is given by
\begin{equation}
\iint_S \mathbf{F}\cdot \mathbf{\hat{n}}\ dS
\end{equation}

In this explanation, the author finds the normal and dots it with the position vector of that point. Why did he dot the two vectors? And what's the overall significance of the dot product in surface integrals?
 A: To integrate a vector field over a surface is to get how much the vector field is flowing through the surface (the flux: https://en.wikipedia.org/wiki/Flux).
First, if you don't take the inner product, then you are integrating a vector function and the answer should be a vector, not a number, but we want a number to denote the total flux, not a vector. Now the question is why not just to integrate the length of the vector over the surface?
To give a simple example, let's say we have a vector field that hits a plane with an angle like this: ->/
To give an interpretation of this problem, if you are driving a car toward a wall, would you rather the wall like "/" or "|"? It should be the former, as you are not hitting the wall as directly as the second case, or you are passing through the wall lighter. Why the difference? The dot products of the normal vectors of the two planes with the velocity of you and your car are different.
More rigorously, if we decompose the vector $\vec v$ into $\vec v_n + \vec v_t$ where $\vec v_n$ is along the direction of the normal vector of the plane and $\vec v_t$ is on the tangent plane. Then only the $\vec v_n$ is contributing to the flux, and $\vec v_t$ is basically trying to stay inside the surface. If $\vec n$ is the unit normal vector of the surface at a given point, then $\vec v \cdot \vec n = (\vec v_n + \vec v_t)\cdot \vec n = \vec v_n \cdot \vec n$.  $|\vec v \cdot \vec n| = |\vec v_n|$ is exactly the "amount" of vector that is passing through the surface, while the sign of $\vec v \cdot \vec n$ indicates which direction it goes (e.g. from the left of the plane to right or the opposite, depending on the choice of $\vec n$).
