Help stokes theorem - integral problem

In summary, Sarah was trying to find N, the unit normal field, but was unsure of how to do so. She was given some lead-in work and was then asked how to find N. She explained that the "fundamental vector product" is normal to the surface and the "differential of area" is equal to the fundamental vector product time du dv. She was also asked how to find the direction of the normal. She explained that by convention you traverse the boundary in such a way that if you were walking along the boundary with your head in the direction of the normal vector, you would have your left arm inside the region.
  • #1
sarahisme
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Hello all,

http://img244.imageshack.us/img244/218/picture8ce5.png

I am completely new to this stokes theorem bussiness..what i have got so far is the nabla x F part, but i am unsure of how to find N (the unit normal field i think its called).

any suggestions people?

i get that nabla x F or curl(F) = i + j + k

cheers

-sarah :)
 
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  • #2
Well, a way to find N is to first find any normal to the surface and then normalize it. Some lead-in work: Put the surface S into vector form,

[tex]S:\, \, \, \vec{r}(u,v)=\left< u\cos v,u\sin v, v\right> \, \, 0\leq u\leq 1,\, 0\leq v\leq\frac{\pi}{2}[/tex]​

Also, recall that the vectors [tex]\frac{d\vec{r}}{du}\mbox{ and }\frac{d\vec{r}}{dv}[/tex] are tangent to S, and that the cross product of two vectors is normal to both vectors, so a normal vector to S is

[tex] \frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv} [/tex]​

but we want a unit normal to S, so we normalize the above vector (divide it by its magnitude) to get

[tex]\vec{N}(u,v)= \frac{\frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}}{\left| \frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}\right|} [/tex]​

PS: I gave the same reply on www.mathlinks.ro
 
  • #3
how do i work out what dS is?, i think that is my main problem...

how does this look for the integral we need to evaluate?

http://img151.imageshack.us/img151/5053/picture9fn3.png

that is, do we have to use the normal or the unit normal when doing this stuff?
 
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  • #4
Some textbooks refer to the "fundamental vector product": If a surface is given by the parametric equations x= f(u,v), y= g(u,v), z= h(u,v), then the fundamental vector product is
[tex]\left<\frac{\partial f}{\partial u}, \frac{\partial g}{\partial u},\frac{\partial h}{\partial u}\right> \times \left<\frac{\partial f}{\partial v}, \frac{\partial g}{\partial v},\frac{\partial h}{\partial v}\right>[/itex]

The "fundamental vector prodct" for that surface is normal to the surface and the "differential of area" is equal to the fundamental vector product time du dv. the "scalar differential of area", for the surface is the length of the fundamental vector product times du dv. You certainly would not use the unit normal since it is the length of the normal vector that gives the information about the area.

In your case, since you are integrating the vector [itex]\nabla \times F[/itex], you need to use the vector differential of area.
 
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  • #5
hmmm i can't quite decide whether to go with my answer of pi/4 or try to redo the question and get -pi/4 as the answer (which is what a friend got...)
 
  • #6
Now I am confused! What do you mean "go with my answer of pi/4"? How did you get that answer to begin with? Why would you redo the question if you know you will get the same answer?
 
  • #7
HallsofIvy said:
Now I am confused! What do you mean "go with my answer of pi/4"? How did you get that answer to begin with? Why would you redo the question if you know you will get the same answer?

its ok, i think its to do with which direction you choose the normal to be pointing...
 
  • #8
Ah, I missed the "-". It was on the previous line from [itex]\frac{\pi}{4}[/itex] and I though it was a hyphen! Yes, swapping the direction of the normal will multiply the answer by -1. Notice that you will still have Stokes' theorem true since, by convention, you traverse the boundary in such a way that if you were walking along the boundary with your head in the direction of the normal vector, you would have your left arm inside the region.

[itex]d\vec r[/itex] here, however, because of the way the parameters u and v are given, should be the outer normal so that, since [itex]\vec F[/itex] is also pointing away from the origin, the answer is positive. You might want to ask your teacher for more detail on how to select the direction of the normal in a problem like this.
 
  • #9
HallsofIvy said:
Ah, I missed the "-". It was on the previous line from [itex]\frac{\pi}{4}[/itex] and I though it was a hyphen! Yes, swapping the direction of the normal will multiply the answer by -1. Notice that you will still have Stokes' theorem true since, by convention, you traverse the boundary in such a way that if you were walking along the boundary with your head in the direction of the normal vector, you would have your left arm inside the region.

[itex]d\vec r[/itex] here, however, because of the way the parameters u and v are given, should be the outer normal so that, since [itex]\vec F[/itex] is also pointing away from the origin, the answer is positive. You might want to ask your teacher for more detail on how to select the direction of the normal in a problem like this.

ah, yep i see now i think , that makes a bit more sense. :) thanks for all the help everyone! :D
 

FAQ: Help stokes theorem - integral problem

What is Stokes' theorem and how is it used?

Stokes' theorem is a fundamental theorem in vector calculus that relates a line integral over a closed curve to a surface integral over the region enclosed by that curve. It is used to calculate integrals in three-dimensional space by converting them into more easily solvable integrals in two-dimensional space.

What is the formula for Stokes' theorem?

The formula for Stokes' theorem is ∫C F · dr = ∫S (∇ x F) · dS, where ∫C F · dr denotes the line integral of the vector field F along the closed curve C, and ∫S (∇ x F) · dS denotes the surface integral of the curl of F over the surface S enclosed by C.

What are the requirements for Stokes' theorem to be applicable?

Stokes' theorem is applicable when the vector field F is continuously differentiable in the region enclosed by the closed curve C. This means that all of the partial derivatives of F exist and are continuous.

Can Stokes' theorem be used to solve any type of integral problem?

No, Stokes' theorem can only be used to solve certain types of integrals. It is specifically applicable to integrals involving vector fields in three-dimensional space.

Can Stokes' theorem be extended to higher dimensions?

Yes, Stokes' theorem can be extended to higher dimensions through the use of differential forms and the generalized Stokes' theorem. This allows for the calculation of integrals in spaces with more than three dimensions.

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