## stoke's theorem and normal vectors

Stokes' theorem uses the curl of a vector field dotted by the normal vector of a surface...but what happens when you use a larger or smaller normal vector? You get a different answer but the theorem doesn't specify to use the unit normal vector. So which normal is the correct one to use? Thanks for any help!
 Recognitions: Gold Member Science Advisor Staff Emeritus There are two different ways of writing the "Kelvin-Stoke's theorem" (just "Stoke's theorem" is used for a more general version that includes that)- $$\int\int \vec{F}\times d\vec{S}$$ and $$\int\int \vec{F}\times \vec{n} dS$$ where $\vec{n}$ is, indeed, the unit normal vector. $d\vec{S}$ is just $\vec{n}dS$. Where did you get the idea that the length of the normal vector was arbitrary?
 I kept getting a homework problem wrong and I couldn't figure out why until I looked at the normal vector that the book answer used, which turned out to not be the unit normal. I checked cramster and they used the same vector as the book. They both used <0, -1/2, 1>

## stoke's theorem and normal vectors

It's not typical to use a non-unit vector there, but it could be they took a constant factor from elsewhere in the integral and rolled it into that vector. Usually you would do that if the differential dS ends up with some constant factors when expressed in terms of dx, dy, and dz.
 mmm nope there's no extra factor that they took out based on the work. Is it possible that the book and cramster are just wrong? And either way thanks, just wanted to clarify.
 Can you show us the work? It may be part of the dS vector (you could probably just give us the paramtrization of the surface) (When i think the book or the professor is wrong, 90% of the time i find in the end that they were right.)
 Is the surface y-2z=0?
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 A particle moves along line segments from the origin to the points (1,0,0),(1,2,1),(0,2,1) and back to the origin under the influence of the force field F = , find the work done. Cramster and the textbook used the normal vector <0, -1/2, 1> which, if I'm not mistaken, isn't a normal vector and got an answer of 3. And does that go against the textbook problem rule..? Because I'm just using it to verify whether the book is wrong, I already know how to do it.
 Yes, as I said the surface is y-2z=0. Then the vector $d\bar{S}=(0,-1/2,1)dxdy$. They write $\bar{n}$ for some of the formulas, are you using Stewart. He then stops writing $\bar{n}$ and worries about other ways of finding $d\bar{S}$. In particular, he gives a convenient formula for when the surface is a graph, like here we have from y-2z=0, z(x,y)=y/2. If you're not using Stewart, and you can't find how to clarify this, let us know we'll type out what happened to the magnitude.
 Well I see that they used <-fx, -fy, 1>, but yea I'm still not seeing what happened to the magnitude. And yes I'm using Stewarts.

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 Quote by BrianC12 Well I see that they used <-fx, -fy, 1>, but yea I'm still not seeing what happened to the magnitude. And yes I'm using Stewarts.
If you are using this special normal vector for a surface S defined by z = f(x, y), then
$$dS = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + 1} dy dx$$
Also, the unit normal is
$$\hat{n} = \frac{\left\langle -\frac{ \partial f}{\partial x}, -\frac{ \partial f}{\partial y}, 1 \right\rangle }{\sqrt{\left(\frac{ \partial f}{\partial x}\right)^2 + \left(\frac{ \partial f}{\partial y}\right)^2 + 1}}$$
This means that
$$\hat{n} dS = \left\langle -\frac{ \partial f}{\partial x}, -\frac{ \partial f}{\partial y}, 1 \right\rangle dy dx$$
As you can see from the derivation, this is only true for this particular type of normal vector. This has nothing in particular to do with the Stokes theorem and can be used in any formula that includes the term $\hat{n} dS$ where you can write the surface S as the graph of a function of two of the Cartesian coordinates. Ie., z = f(x, y), y = f(x, z) or x = f(y, z).

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 Quote by slider142 If you are using this special normal vector for a surface S defined by z = f(x, y), then $$dS = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + 1} dy dx$$ Also, the unit normal is $$\hat{n} = \frac{\left\langle -\frac{ \partial f}{\partial x}, -\frac{ \partial f}{\partial y}, 1 \right\rangle }{\sqrt{\left(\frac{ \partial f}{\partial x}\right)^2 + \left(\frac{ \partial f}{\partial y}\right)^2 + 1}}$$ This means that $$\hat{n} dS = \left\langle -\frac{ \partial f}{\partial x}, -\frac{ \partial f}{\partial y}, 1 \right\rangle dy dx$$ As you can see from the derivation, this is only true for this particular type of normal vector. This has nothing in particular to do with the Stokes theorem and can be used in any formula that includes the term $\hat{n} dS$ where you can write the surface S as the graph of a function of two of the Cartesian coordinates. Ie., z = f(x, y), y = f(x, z) or x = f(y, z).
This works for any parameterization of a surface. Say your surface is parameterized as ##\vec R =\vec R(u,v)##. Then$$\hat n =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}$$ with the sign chosen to agree with the orientation of the surface. And since ##dS =|\vec R_u\times \vec R_v|dudv##, then$$d\vec S =\hat n dS =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}|\vec R_u\times \vec R_v|dudv=\pm{\vec R_u\times\vec R_v}dudv$$and you never have to include the magnitude calculation in integrals containing this.
 Alright thanks for clearing that up! Btw LCKurtz, I like your sig haha, that's so true.