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Compute circulation of vector around the contour

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Compute the circulation of the vector a = yi+x2j - zk around the contour L: {x2 +y2 = 4; z = 3}, a) directly and b) via the Stokes Theorem.
    Plot the contour and show its orientation.

    2. Relevant equations

    Stokes theorem is [itex]\oint[/itex]F.dr = [itex]\int[/itex]∇ X F . n dS

    3. The attempt at a solution

    For (a) which is solving it directly, i used the left side of the equation [itex]\oint[/itex]F.dr and I obtained 0 as my answer. Having problem with part b. What should my ∇f be? Is it 2x i + 2y j + 0 k?
     
  2. jcsd
  3. May 11, 2013 #2

    HallsofIvy

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    No, it isn't. Don't you know how [itex]nabla\times \vec{F}[/itex] is defined? If [itex]\vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex] then
    [tex]\nabla\times \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}-\left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}[/tex]
     
  4. May 11, 2013 #3
    Yes I know that. But to obtain a unit normal vector n, i need to first find [itex]\frac{∇f}{|∇f|}[/itex] , isn't it?

    Anyway, I managed to find a mistake I made for the left hand side, and obtained my answer as -4pi. Mind guiding me through for the right hand side?
     
  5. May 17, 2013 #4
    How did you get -4pi? Also, did you manage to finish part b?
     
  6. May 18, 2013 #5

    I like Serena

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    No.

    The unit normal vector ##\mathbf n## is normal to the surface inside the contour.
    The obvious choice for that surface is in the plane z=3, bounded by a circle.
    The unit normal vector to that plane is ##\mathbf n = \mathbf k##.

    So first you have to determine ##\nabla \times \mathbf a##, and then you have to take the dot product with ##\mathbf n = \mathbf k##.
     
  7. May 21, 2013 #6
    How do you know it's "k" and not negative "k"
     
  8. May 22, 2013 #7

    I like Serena

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    You can choose the direction.
    The consequence is that for the contour integral you have to follow the contour according to the right hand rule.
     
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