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Help! stokes theorem - integral problem

  1. Aug 24, 2006 #1
    Hello all,

    http://img244.imageshack.us/img244/218/picture8ce5.png [Broken]

    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


    -sarah :)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Aug 24, 2006 #2


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    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
  4. Aug 24, 2006 #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 [Broken]

    that is, do we have to use the normal or the unit normal when doing this stuff?
    Last edited by a moderator: May 2, 2017
  5. Aug 25, 2006 #4


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    Some text books 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.
    Last edited by a moderator: Aug 26, 2006
  6. Aug 26, 2006 #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...)
  7. Aug 26, 2006 #6


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    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?
  8. Aug 27, 2006 #7
    its ok, i think its to do with which direction you choose the normal to be pointing...
  9. Aug 27, 2006 #8


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    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.
  10. Aug 28, 2006 #9
    ah, yep i see now i think , that makes a bit more sense. :) thanks for all the help everyone! :D
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