The main results of the previous section were formulas for the surface elements $d\SS$ and $\dS$ of a surface in terms of two infinitesimal displacements in the surface, that is $$d\SS = d\rr_1 \times d\rr_2$$ and $$\dS = |d\SS| = |d\rr_1\times d\rr_2|$$ which are (1) and (2) of Section 8, respectively. This allows us to integrate any function over a surface, by considering an integral of the form $$\Sint f \, \dS$$ The surface area of $S$ is obtained as a special case by setting $f=1$. Another example is total mass, which is obtained by setting $f=\sigma$, where $\sigma$ denotes the surface mass density (mass per unit area). The center of mass can be obtained from $$\bar{x} = {1\over m} \Sint x\sigma \, \dS$$ and similar formulas for $\bar{y}$ and $\bar{z}$; compare the analogous formulas for the mass and center of mass of a wire in terms of line integrals.

The flux of a vector field $\FF$ across a surface $S$ is the total amount of $\FF$ pointing through $S$. An example which is easy to visualize is $\FF=\rho\,\vv$, where $\vv$ is the velocity field of a fluid with (volume) mass density $\rho$ (mass per unit volume). In this case, the flux measures how much of the fluid crosses the surface.

“Total amount” of course means to integrate, so we chop up our surface into small pieces. But how much of $\FF$ “points through” a small piece of the surface? We only care about the component of $\FF$ normal to the surface, so we need a dot product with a vector perpendicular to $S$. The flux through a small piece surely depends both on the size of $\FF$ and the area of the piece. Putting this all together, the flux takes the form $$\hbox{flux} = \Sint \FF\cdot d\SS$$ Again we see the fundamental role played by the vector surface element. Be careful to choose the correct orientation for $S$, depending on which flux is being computed: the flux up, the flux out, etc.

Another way to think about this is to first find the unit normal vector $\nn$ to the surface, then use the dot product to find the normal component of $\FF$, then integrate the resulting scalar field over the surface. These two viewpoints are related by $$\nn \, \dS = d\SS$$ If desired, you can compute $\nn$ itself from $$\nn = {d\rr_1\times d\rr_2 \over |d\rr_1\times d\rr_2|} = {\Partial{\rr}{u}\times\Partial{\rr}{v} \over \left| \Partial{\rr}{u}\times\Partial{\rr}{v} \right|} $$ but this is not usually necessary. You should compare this formula with $\TT = {d\rr\over ds} = {d\rr\over du}\Big/{ds\over du}$.

GOALS