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Stokes's Theorem and the Right Hand Rule

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data

    For [itex]\mathbf{F} = y \mathbf{i} - x \mathbf{j} + 4z \mathbf{k}[/itex] evaluate the surface integral of [itex](\nabla \times \mathbf{F}) · k[/itex] over the 2D surface of the disc [itex]x^{2} + y^{2} \leq 4, z = \frac{H}{2}[/itex]

    2. Relevant equations


    3. The attempt at a solution

    I am unsure of my answer to this question for a few reasons.

    1. Does it make sense for my answer to be negative? I thought we usually choose the unit vector such that the answer comes out to be positive, but it was chosen for me here.

    2. Is the 'z' information only relevant to performing the closed loop integral?

    My working:

    [itex]\nabla \times \mathbf{F} = -2 \mathbf{k}[/itex]

    [itex](\nabla \times \mathbf{F}) · k = -2[/itex]

    [itex]\int_{s}(\nabla \times \mathbf{F}) · k = \int^{2\pi}_{0} \int^{4}_{0} -2 \rho d\rho d\phi[/itex]

    [itex]\int_{s}(\nabla \times \mathbf{F}) · k = (\int^{4}_{0} -2 \rho d\rho)(\int^{2\pi}_{0} d\phi)[/itex]

    [itex]\int_{s}(\nabla \times \mathbf{F}) · k = [-\rho^{2}]^{4}_{0} [\phi]^{2\pi}_{0} = -32\pi[/itex]

    Also, should it be [itex](-\rho)^2[/itex] or [itex]-(\rho^2)[/itex]? The former would solve my problem of having a negative answer, but i'm not sure...

    Thanks!
     
  2. jcsd
  3. Apr 6, 2015 #2

    HallsofIvy

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    There certainly is no reason why an integral should not be negative but perhaps I don't understand what problem you are trying to solve. Where did that "[itex]\rho[/itex]" come from in the integral? There is no "[itex]\rho[/itex]" in your statement of the problem.
     
  4. Apr 6, 2015 #3

    Orodruin

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    There is nothing wrong with getting a negative flux integral. It just means the net flux is in the opposite direction from the chosen normal direction.

    Exactly what do you mean by "the z information"? Given the vector field, it would not matter which z the loop was at, but this may not be true for a different vector field.

    First of all, you should check your integration boundaries in the radial direction ... You should also be able to tell from your original expression whether or not the integral should be negative and from the value of curl(F) what it should be.
     
  5. Apr 6, 2015 #4
    Ok, I thought so but wanted to be sure.


    I mean the height at which the disk is located above the x-y plane.

    It's a disk of radius 4. Am I wrong in thinking that this means I need to integrate between 0 and 4 in the radial direction?

    By my 'original expression' are you referring to the vector field or (curl F).k?
     
  6. Apr 6, 2015 #5

    Orodruin

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    You are wrong in thinking it is a disk of radius 4 ...

    I am referring to the integrand of your integral.
     
  7. Apr 6, 2015 #6
    Ah, yes it's a disk of radius 2.

    I can see from my integrand that my answer will be negative - I was just checking that in instances like this, it's not necessary to change my unit vector to get a positive answer.
     
  8. Apr 6, 2015 #7
    [itex]\rho[/itex] is the radius, from cylindrical polar coordinates.

    [itex]x^2 + y^2 = \rho^2[/itex]

    [itex]dA = \rho d\rho d\phi[/itex]
     
  9. Apr 6, 2015 #8

    Orodruin

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    Then, as I mentioned, there is no a priori reason you should get a positive answer to a flux integral. A negative answer just means the net flux is in the opposite direction.
     
  10. Apr 6, 2015 #9
    Thank you for your help :)
     
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