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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!

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  • $\begingroup$ $\vec{n}$ is just the normal vector at the surface element dS. $\endgroup$ – tired Apr 29 '15 at 13:57
  • $\begingroup$ It seems you are confused about $dS$, not $d\bf S$, is that right? $\endgroup$ – Samuel Apr 29 '15 at 13:59
  • $\begingroup$ @Samuel Yes, I guess so! I mean the other terms can be computed in a straight forward fashion, but the role of dS is a bit more foggy. $\endgroup$ – Laplacinator Apr 29 '15 at 16:27
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$dS$ 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). $$ dS = n \lVert dS \rVert $$

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.

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.

$$ dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} = \frac{\sqrt{14}}{2} dxdz $$

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.

surface element and projection

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 $$ \lVert P_y u \times P_y v \rVert = \lVert n \cdot e_y \rVert \, \lVert u \rVert \, \lVert v \rVert $$

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  • $\begingroup$ Thanks! That really helps clear some of it up. I'm gonna read through that part on adjusting area elements that aren't parallell to any axis some more , that seems like a very important point. Follow up question, would that mean that the area element of a plane parallell to the xy-plane would always be dxdy? Is the $\frac{\sqrt{14}}{2}$ in fact just "scaling" the differential square to "fit" our particular plane? $\endgroup$ – Laplacinator Apr 29 '15 at 16:22
  • $\begingroup$ Exactly. The area of the projection of the surface $du dv$ on the $x$-$z$ plane is smaller than the area of the original $du dv$ , because its normal $n$ is not normal to that plane. So it must be upscaled. If the surface is part of a parallel plane, we do not have to do this, as surface element and its projection are same sized. $\endgroup$ – mvw Apr 29 '15 at 16:53
  • $\begingroup$ @mvw Would you care to explain what $\pm(\mathbf{\frac {n}{|n|}})(\mathbf{|n|})$ represents exactly? $\endgroup$ – Demosthene Apr 29 '15 at 18:26
  • $\begingroup$ @Demosthene From the looks it is about normal vectors $n$ which are not unit vectors, decomposing them into a product $\lVert n \rVert e_n$, where $e_n$ is the unit vector with the same direction as $n$. I would need to see an example to understand more, esp. why the signs show up that way. $\endgroup$ – mvw Apr 29 '15 at 18:33
  • $\begingroup$ @mvw Yes, I am familiar with the notations $\mathbf{\frac{n}{|n|}}$ or $\hat{n}$ to denote a normalized vector. Following that logic, I don't see the point of writing $\mathbf{\frac{n}{|n|}|n|}$. Isn't that just $\mathbf{n}$? $\endgroup$ – Demosthene Apr 29 '15 at 19:34
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$d\mathbf S = \hat {\mathbf N}dS$ -- where $\hat {\mathbf N}$ is the unit normal (outward if closed, otherwise you have to choose an orientation) to the surface and $dS$ is the differential area element -- is the definition. But you don't really need to worry about it. You should always parametrize your surface first by some $\mathbf r(s,t)$, $s_0 \le s \le s_1$ and $t_0 \le t\le t_1$. Then $$\iint_\Gamma \mathbf f \cdot d\mathbf S = \int_{t_0}^{t_1} \int_{s_0}^{s_1} \mathbf f(\mathbf r(s,t)) \cdot \left(\frac {\partial \mathbf r}{\partial s} \times \frac {\partial \mathbf r}{\partial t}\right)\,ds\,dt$$

Each of those terms on the right you should then be able to calculate.

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  • $\begingroup$ I see! Thanks a lot, but is it really necessary to always parametrize a surface first? $\endgroup$ – Laplacinator Apr 29 '15 at 16:26
  • $\begingroup$ Yes -- but sometimes you may not realize you're doing it. Parametrizing a surface just means expressing it as some vector equations or set of scalar equations in $2$ variables. If there is an immediate way to write down the surface in terms of easy variables like $x,y$ or $r, \theta$, you may not realize that the $\hat {\mathbf N}\,dx\,dy$ that you immediately wrote down is really just $\left(\frac {\partial \mathbf r}{\partial x} \times \frac {\partial \mathbf r}{\partial y}\right)\,dx\,dy$ or that your $\mathbf f(x,y)$ is already in terms of your parameters (variables). $\endgroup$ – user137731 Apr 29 '15 at 17:21

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