In a general flux integral of a level surface, of the form
$$\iint{\mathbf{F}\bullet d\mathbf{S}}$$
what exactly does $d\mathbf{S}$ represent? I have seen both
$d\mathbf{S} = \mathbf{\hat N}dS = \pm (\mathbf{\frac {n}{|n|}})(\mathbf{|n|}) dudv$
(for parametric surfaces), and
$d\mathbf{S} = \mathbf{\hat N}dS = \pm \frac{\nabla G(x,y,z)}{G3(x,y,z)}dxdy$
for level surfaces. At a glance, I feel like I get them, but whenever I sit down to actually solve any problems I get confused about what exactly it represents in the integrals. Finding the normal is usually not a problem, nor is calculating $\frac{\nabla G}{G3}$, but then I get stuck on what to put for dS.
Like the following example, in calculating the flux of $\mathbf{F} = xi + zj $ out of the surface x+2y+3z = 6. The textbook calculates $\mathbf{\hat N}$ to be $\frac{i+2j+3k}{\sqrt{14}}$ (and I agree), but then it goes on to calculate $dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} = \frac{\sqrt{14}}{2} dxdz$ . I'm not entirely sure why they did that, or why they set it up the way they did. How do you find/choose dS and what does it mean to the integral?
Thanks!
Best Answer
$dS$ is a surface element, a differential sized part of the surface $S$. It is usually oriented, positive if its normal $n$ is outward pointing (e.g. if $S$ is the boundary of a volume). $$ dS = n \lVert dS \rVert $$
For those examples $\lVert dS \rVert = du \, dv$ and $\lVert dS \rVert = dx \, dy$. The other parts are the more or less complicated normal vectors of those surface elements.
The integration is along the $x$-$z$ plane, while the surface, $$ S: x+2y+3z = 6 \quad n = (1,2,3)^t/\sqrt{14} $$ which is a plane as well, is not parallel to the $x$-$z$ plane.
The area of the projection $P_y S$ has to be adjusted, to give the correct area $\lVert S \rVert$ for $S$. We want $$ \lVert S \rVert = \int\limits_S \lVert dS \rVert = \int\limits_S \lVert n\,du\,dv \rVert = f \lVert P_y S \rVert = f \int\limits_{P_y S} \lVert dx \, dz \rVert $$ In your example they simply take $f = 1/\lVert n \cdot e_y\rVert$.
Let us check this: First we look for unit vectors $u$ and $v$ orthogonal to $n$ and each other. $$ 0 = n \cdot a = (1, 2, 3)^t / \sqrt{14} \cdot (2, -1, 0)^t \quad e_u = (2, -1, 0)^t / \sqrt{5} \\ e_v = n \times e_u = (3, 6, -5)^t / \sqrt{70} $$ These are unit vectors, so the area of the square between $e_u$ and $e_v$ is 1. Now these unit vectors have the projections on the $x$-$z$ plane: $$ u_p = P_y e_u = (2, 0, 0)^t/\sqrt{5} \quad \lVert u_p \rVert = 2/\sqrt{5} \\ v_p = P_y e_v = (3, 0, -5)^t/\sqrt{70} \quad \lVert v_p \rVert = \sqrt{34/70} = \sqrt{17/35} \\ $$ where $P_y a = a - (a\cdot e_y) e_y$ for a vector $a$. The area of the projection is $$ \lVert u_p \times v_p \rVert = \lVert ((2, 0, 0)^t/\sqrt{5}) \times ((3, 0, -5)^t/\sqrt{70}) \rVert = \lVert (0, 10, 0)^t/\sqrt{350} \rVert = 2 /\sqrt{14} $$
This should explain the factor $\sqrt{14}/2$.
What is missing is a derivation for the shorter $$ \lVert P_y u \times P_y v \rVert = \lVert n \cdot e_y \rVert \, \lVert u \rVert \, \lVert v \rVert $$