Stoke's theorem and normal vectors

In summary, Stokes' theorem involves the curl of a vector field dotted with the normal vector of a surface, which can be written in two different ways depending on whether the surface is a graph of a function. The unit normal vector is typically used, but it is possible to use a non-unit normal vector as long as it is accounted for in the calculation. This is not specific to Stokes' theorem and applies to any formula involving the term \hat{n} dS.
  • #1
BrianC12
18
0
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!
 
Mathematics news on Phys.org
  • #2
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)-
[tex]\int\int \vec{F}\times d\vec{S}[/tex]
and
[tex]\int\int \vec{F}\times \vec{n} dS[/tex]

where [itex]\vec{n}[/itex] is, indeed, the unit normal vector. [itex]d\vec{S}[/itex] is just [itex]\vec{n}dS[/itex]. Where did you get the idea that the length of the normal vector was arbitrary?
 
Last edited by a moderator:
  • #3
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>
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.)
 
  • #7
Is the surface y-2z=0?
 
Last edited:
  • #8
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 = <z^2, 2xy, 4y^2>, 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.
 
  • #9
Yes, as I said the surface is y-2z=0. Then the vector [itex]d\bar{S}=(0,-1/2,1)dxdy[/itex]. They write [itex]\bar{n}[/itex] for some of the formulas, are you using Stewart. He then stops writing [itex]\bar{n}[/itex] and worries about other ways of finding [itex]d\bar{S}[/itex]. 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.
 
  • #10
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.
 
  • #11
BrianC12 said:
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
[tex]
dS = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + 1} dy dx
[/tex]
Also, the unit normal is
[tex]
\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}}
[/tex]
This means that
[tex]
\hat{n} dS = \left\langle -\frac{ \partial f}{\partial x}, -\frac{ \partial f}{\partial y}, 1 \right\rangle dy dx
[/tex]
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 [itex]\hat{n} dS[/itex] 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).
 
Last edited:
  • #12
slider142 said:
If you are using this special normal vector for a surface S defined by z = f(x, y), then
[tex]
dS = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + 1} dy dx
[/tex]
Also, the unit normal is
[tex]
\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}}
[/tex]
This means that
[tex]
\hat{n} dS = \left\langle -\frac{ \partial f}{\partial x}, -\frac{ \partial f}{\partial y}, 1 \right\rangle dy dx
[/tex]
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 [itex]\hat{n} dS[/itex] 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.
 
  • #13
Alright thanks for clearing that up! Btw LCKurtz, I like your sig haha, that's so true.
 

1. What is Stoke's theorem?

Stoke's theorem is a fundamental theorem in multivariable calculus that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field over the boundary of the surface.

2. How is Stoke's theorem related to the divergence theorem?

Stoke's theorem is a higher-dimensional analogue of the divergence theorem, which relates a surface integral to a triple integral. Both theorems are a consequence of the generalized Stokes' theorem, which unifies the two concepts.

3. What is a normal vector?

A normal vector is a vector that is perpendicular to a surface at a specific point. It is often used in vector calculus to define the direction and orientation of a surface.

4. How is a normal vector used in Stoke's theorem?

In Stoke's theorem, the normal vector is used to determine the direction of the line integral over the boundary of a surface. It is also used to calculate the dot product with the vector field in the surface integral.

5. What are some real-life applications of Stoke's theorem?

Stoke's theorem is used in various fields such as physics, engineering, and fluid mechanics to calculate the flow of a vector field through a surface. It is also used in computer graphics for 3D modeling and animation.

Similar threads

Replies
72
Views
4K
  • General Math
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
26
Views
1K
  • Science and Math Textbooks
Replies
10
Views
2K
Replies
5
Views
3K
  • General Math
Replies
6
Views
749
  • Calculus and Beyond Homework Help
Replies
4
Views
751
  • Calculus and Beyond Homework Help
Replies
1
Views
546
  • General Math
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Back
Top