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Verify Stokes' Theorem

  1. Jun 7, 2008 #1
    1. The problem statement, all variables and given/known data
    A = sin([tex]\phi[/tex]/z)* a([tex]\phi[/tex])

    I'm having problem verifying Stokes Theorem. I have to verify the theorem over the upper half of the sphere with radius b and the sphere is centered at the origin. The problem also says z > = 0

    Could someone help me with this.
     
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  3. Jun 7, 2008 #2

    Dick

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    No one can help you until you try to make some effort to solve it on your own. How might you try to do this?
     
  4. Jun 8, 2008 #3
    Well i'm new to EM theory and i'm just soo stuck on this whole vector calculus thing. I know stokes theorem state's that Integral of [tex]\oint(A.dl)[/tex] = [tex]\int(\nabla x A).dS[/tex]

    So I have to evaluate the 2 integrals separately.
    The part that i'm stuck on is how to evaluate [tex]\oint(A.dl)[/tex] integral when no equation is given or no boundaries are given...
     
  5. Jun 8, 2008 #4
    Also in this problem if I were to evaluate the RHS integral, since there is no r or theta, would the del x A be 0?
     
  6. Jun 8, 2008 #5

    Dick

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    You are going to have trouble computing curl of A or the line integral because it doesn't look to me like the A you've stated is even a vector. Can you check you've transcribed the problem correctly?
     
  7. Jun 8, 2008 #6
    Yes I have double checked it. It's actually called F, but I don't think that should matter. It is a vector for sure, because the problem states, you are given a vector field.
    Any suggestions?
     
  8. Jun 8, 2008 #7

    Dick

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    It may actually called F, but A still doesn't look like a vector. What are the x, y and z components? Can you maybe scan this in or something. It's not making sense.
     
  9. Jun 8, 2008 #8
    I just sent you the link to this problem.
     
  10. Jun 8, 2008 #9

    Dick

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  11. Jun 8, 2008 #10
    I'm thinking the phi is the angular co-ordinate in the spherical co-ordinate system since we are working with a sphere. But i'm as confused as the next person, so I would go with a unit vector to make things easier and see if we can get a solution this way. This is pretty much the whole problem I didn't leave anything out.
     
  12. Jun 9, 2008 #11

    Dick

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    Well, ok. Let's just try it that way. That makes A.dl pretty easy. The boundary of the sphere is the unit circle in the xy plane. What does that make the integral of A.dl? But it still looks to me like A is discontinuous over the surface. Which make me nervous. Are you sure you can't look up what a_phi really is?
     
  13. Jun 9, 2008 #12
    Unless you want to convert a_phi into cartesian co-ordinates or try some funky business there... So we have one equation where x^2 + y^2 = 1. When we are going to do the integral of A.dl, dl = dxdy right? I really have no idea what a_phi could be to be honest...
     
  14. Jun 9, 2008 #13
    Any new thoughts / suggestions?
     
  15. Jun 9, 2008 #14

    Dick

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    Well, actually I misspoke. The boundary in the xy plane isn't the unit circle. It's the circle of radius b, sorry. It's a line integral, not dxdy. And is phi the polar coordinate, i.e phi=pi/2 on the positive z axis and -pi/2 on the minus? But the problem still looks funny. I don't think you can be expected to do a problem where you don't know what one of the terms means. Can you ask?
     
  16. Jun 9, 2008 #15
    i do believe phi goes from pi/2 to -pi/2. So the equation would become x^2 + y^2 = b^2. I don't think the problem wants us to really solve the whole thing, just do both integrals and show that they are both equal to each other.
     
  17. Jun 9, 2008 #16

    Dick

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    But what is a_phi?
     
  18. Jun 9, 2008 #17
    sorry sir, don't know that... so we have the equation of the circle which is in cartesian (x^2 + y^2) and we can covert the x's and y's to spherical co-ordinates and then do the integral ?
     
  19. Jun 9, 2008 #18

    Dick

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    You can't 'do an integral' if you don't know what the integrand is.
     
  20. Jun 9, 2008 #19
    Well i guess then we can't solve the problem ? Thanks for your help bud!
     
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