Why do we need the vector normal to the surface in surface integration?

Question

I've been struggling with understanding how surface integrals come to be. Can someone please explain why they have such a format with a normal vector to the surface? thanks!

Answer 1

Part of it is because of the fact that these methods were first used to understand physics, and there are clear physical interpretations. The vector coming out of the surface is the "flow", and so a lot of physical laws like Guass' law and Fick's ask about balancing flows (i.e. for Gauss' law, if a ball contains electrical charge, then the total flow of "electric force" out of the ball is proportional to the total charge, since any electrical charge outside the ball has a force which flows inward and one that flows outward and together those cancel).

Geometrically, it's because you have to scale the quantity by the "infinitesimal area", and the normal vector comes from the cross product of vectors on the surface and is thus a measure of area. If you need to dot the vector with something to get a scalar, then this is the "natural" quantity.

More mathematically, it's because the gradient (or differential form) is a covector, and thus is a function that takes in a vector and spits out a number (this is something that you go into depth in differential geometry). This is because of how it has to handle coordinate transformations.

Answer 2

Let $F$ be a vector field, $n$ a normal vector to the surface, with $u=\frac{n}{|n|}$ the unit normal vector.

The integral of the flux of a vector field through a surface requires an orientation (direction) that defines an "inside" and an "outside" relative to the surface that is the boundary between the "inside" and the "outside" .

We want to ask, for example, how much of the vector field $F$ is "escaping" from the the inside to the outside. To do this, we decompose the vector field into two orthogonal components: (1) the part normal to the surface; (2) the part tangent to the surface. At any given point on the surface, only the part normal to the surface escapes. We find this part by taking the dot product with the unit normal vector, which gives the orthogonal projection of the vector field onto the normal line: $F\cdot u=|F|\cos\theta,$ where $\theta$ is the angle between the two vectors.

The local surface area about a point on the surface is given by the magnitude of the cross product of two vectors tangent to the surface at that point. This cross product depends on orientation, $u\times v=-v\times u,$ but both of these are normal vectors to the surface (they point in opposite directions). Depending on our choice of orientation, we choose one of these to be $n.$ Then $n=u|n|$, where $u$ is our unit normal vector, and $n=dS$ is the infinitesimal area.

At each point on the surface, we see that the total of the vector field escaping through the surface is the normal component $F\cdot u$ multiplied by the local (infinitesimal) surface area $|n|=dS.$ Summing over all points on the surface we get $$ \int_S F\cdot u\ dS. $$