Flux of constant vector field $F(x,y,z) = C$ is just $|C|$ times the area of the surface? Suppose $S$ is a surface of finite area and $F(x,y,z) = C$ is a constant vector field over $S$. Is the flux over $F$ across S just $|C|$ times the area of $S$?
I'm not finding a way to argument on this. Can someone answer it? Thanks!
 A: No. For instance, the constant vector field $F(x, y, z) = (1,0,0) $ has flux zero through the unit square in the $xy$-plane.
The right answer is $|C| ~ A ~ (u \cdot n)$, where $A$ is the area of $S$, $n$ is the normal to your bit of surface, and $u$ is a unit vector in the directions of $C$, i.e., $u = \frac{1}{|C|} C$. (But this assumes that the surface is in fact planar, i.e., that the normal vector is constant over the surface!)
Because of this, a simpler formula (in the case of a planar surface piece) is that the flux is
$$
A (C \cdot n).
$$
A: Project your surface $S$ to the plane perpendicular to the direction of $C$; compute the area of the projection, counting various pieces of multiple projection with signs (based on orientation of your surface $S$) and multiplicities. Multiply by $|C|$. This is the flux of $C$ through $S$.
You will get the same result using any surface with the same boundary instead of $S$ (by Stokes's theorem or by Gauss's theorem). So in a sense (which can be made precise with a bit of effort), the answer is the area "enclosed" by the projection of the boundary (times $|C|$).
This also gives a way of seeing why this is true: replace $S$ by a "cylinder" with sides parallel to $|C|$ plus a flat "cap" in a plane perpendicular to $C$. The total flux through this new surface is the same as through $S$. The flux through the sides of the cylinder is $0$ since the surface is parallel to the field, while the flux through the "base" is $|C|$ time the area of the base.
A: In general, no. Remember that the surface integral can be calculated as follows:
$$\int_S F\text{d}\mathbf{S} = \int_S F_n\text{d}{S} = \int_S \langle F,\mathbf{n}\rangle\text{d}{S},$$
where the first term is a $\textit{field}$ integral over $S$ and the second and third terms are the $\textit{scalar}$ integral over $S$ of the normal component of the field ($F_n=\langle F,\mathbf{n}\rangle$ denotes the normal component of the field, $\mathbf{n}$ is the vector normal to the surface and has the same sense as the chosen orientation). In the general case, $\langle C,\mathbf{n}\rangle$ needn't be equal to $|C|$ (this just happens when $C$ is perpendicular to the surface in every point). In fact, if $C$ is tangent to the surface in every point (e.g., if $S$ is a plane and $C$ is parallel to the plane), then $\langle C,\mathbf{n}\rangle$ is identically $0$ and the integral is also $0$.
