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So this part I'm struggling with on Stokes' Theorem:

$$\iint_S ~(\text{curl}~\vec{F} \cdot \hat{n})~ dS$$

I don't really understand why we would want to dot it with the unit normal vector at that point. This is going to tell us how much of the curl is in the normal direction but why would we want this surely we only care about how much the curl is actually on the surface as opposed to normal to the surface it seems to me like this is actually the opposite of what we want. Very counter intuitive to me. I'm guessing I am misinterpreting something here so if someone would explain that would be fantastic.

Also in the next line it says

$$\iint_S ~(\text{curl}~\vec{F} \cdot \hat{n})~ dS=\iint_S ~\text{curl}~\vec{F} \cdot d\vec{S}$$

so is this just a notation to say that $$\hat{n}\cdot~ dS= d\vec{S}$$

why is this so it it just purely for convenience or is there some reason to write it like this, I struggle to see why some differential of the surface would be a vector? To me it's just a little chunk of the surface area.

I really need this clarified and cleared up thanks.

FreeMind
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Peter H
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  • the vector surface element is very common and very useful. Once you build familiarity with more multivariable calculus it'll become the more intuitive choice. As to why exactly it's right I hope somebody else can elucidate, I've never had intuitive understanding for Stokes – ShakesBeer Jul 23 '15 at 12:32
  • @PeterH "surely we only care about how much the curl is actually on the surface as opposed to normal to the surface" what makes you say that? – Ben Grossmann Jul 23 '15 at 12:36
  • One thing to keep an eye on is what quantities are vectors and what are scalars (i.e. individual numbers). Rewriting $(v\cdot \hat{n}) d S$ as $v \cdot d \vec{S}$ is a little mysterious and deserves an explanation, but it is not for the reason you suspected. – hardmath Jul 23 '15 at 12:37
  • @PeterH I really like like the Kahn academy explanation of Stokes' theorem; maybe you should take a look. – Ben Grossmann Jul 23 '15 at 12:39
  • And yes; $d \vec S$ is defined as $\hat n \cdot dS$ – Ben Grossmann Jul 23 '15 at 12:43
  • @Omnomnomnom I would say that because the surface is defined by the curve that we are doing the line integral over so surely we would be concerned with what is happening on the surface, not what is happening perpendicular to the surface?

    Also I just watched that Khan video, indeed he says "we care about the curl on the surface" and then proceeds to multiply by the unit normal vector. How does this tell you what is happening on the surface when you multiply by it's normal???

     – Peter H Jul 23 '15 at 13:03
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    Here's a question: for the vector field $F = (-y,x,0)$, curl occurs in the $z = 0$ plane. Compute the curl. In which direction does the curl vector "point"? – Ben Grossmann Jul 23 '15 at 13:28
  • The curl is $(0,0,2)$ if I have done it correctly so the curl points in the $z$ direction. So then the curl at any point $(x,y,z)$ depends only on the $z$ component? – Peter H Jul 23 '15 at 13:54
  • There are many different things we might wish to calculate. If you want to measure a thing like "the degree to which curl at points on the surface aims parallel to the surface" then you could certainly calculate $\iint_S\left\Vert\text{curl}~\vec{F} \times \hat{n}\right\Vert~\mathrm dS$ instead. But that wouldn't equal the single integral in the Kelvin-Stokes theorem. Are you asking 1. Why do we need to take the dot product to equal the single integral? 2. What does a dot product integral tell us? or 3. With [a particular goal], does the dot product calculate it? (Just realized FreeMind bumped – Mark S. May 21 '21 at 12:36
  • It makes more sense if you use bivectors. – mr_e_man Oct 25 '22 at 21:14

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