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Stokes and Divergence theorem questions

  1. Apr 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Let
    [tex]\vec{F}=xyz\vec{i}+(y^{2}+1)\vec{j}+z^{3}\vec{k}[/tex]
    And let S be the surface of the unit cube in the first octant. Evaluate the surface integral:


    [tex]
    \int\int_{S} \nabla\times \vec{F} \cdot \vec{n} dS
    [/tex]
    using:
    a) The divergence theorem
    b) Stoke's theorem
    c) Direct computation
    2. Relevant equations

    Divergence Theorem:
    [tex]\int\int\int\nabla \bullet \vec{F} dV = \int\int \vec{F} \bullet \vec{n} dS[/tex]

    3. The attempt at a solution
    I think I have b and c worked out but part a I am not sure of. What I don't understand is how I can use the divergence theorem here. If I substitute [tex] F = curl(F)[/tex] into the divergence theorem I will get zero by identity. So, how is that possible to evaluate? If someone could pelase tell me what I'm doing wrong I would greatly appreciate it.
     
    Last edited: Apr 17, 2008
  2. jcsd
  3. Apr 17, 2008 #2

    TMM

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    You can't substitute some curl(A) for some B unless B's divergence is zero. When you take the divergence of F, is it zero?
     
  4. Apr 17, 2008 #3
    The divergence of F is yz + 2y + 3z^2.

    I'm still not too sure how I could apply the divergence theorem here to help me evaluate that integral though.
     
  5. Apr 17, 2008 #4

    Hootenanny

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    I can't see why one would need to take the curl of F? Surely to use the divergence theorem one merely needs to take the divergence of F x n?
     
  6. Apr 17, 2008 #5
    I realized I wrote down the probelm wrong here. The problem is actually Evaluate the surface integral:

    [tex]
    \int\int_{S} \nabla\times \vec{F} \cdot \vec{n} dS
    [/tex]
    That is why i've been trying to substitute the curl of F into the divergence formula. I've been saying that
    [tex] \vec{B} = \nabla\times \vec{F} [/tex]
    and then just using that in the divergence theorem, which is a flawed technique because it would always return zero since I would be taking the divergence of solenoidal fields all the time:

    [tex]\int\int\int\nabla \cdot (\nabla \times \vec{B}) dV = 0[/tex]
     
    Last edited: Apr 17, 2008
  7. Apr 17, 2008 #6

    HallsofIvy

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    It's not clear to me why you call that a "flawed" technique. Yes, the divergence of a solenoidal field is 0 and the integral over the unit cube is obviously 0. Are you saying that you DIDN'T get 0 using the divergence theorem and direct integration?
     
  8. Apr 17, 2008 #7
    I would think that this would be flawed because I could argue that any surface integral of a curl would be zero (by using the divergence theorem like i did above) which conflict with Stokes theorem because this is clearly not always the case. I must not be understanding the relationships between the theorems.
     
  9. Apr 17, 2008 #8

    HallsofIvy

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    Why would Stoke's theorem conflict? The divergence theorem connects the integral over a closed surface to the integral over the region the surface encloses. Stoke's theorem connect the integral over surface to an integral over the boundary of the surface. If the surface is close, it has no boundary and Stoke's theorem necessarily gives 0 as result.
     
  10. Apr 17, 2008 #9
    contnuing my last reply, consider I have:

    [tex]\vec{F}= z \vec{i} +x \vec{j} + y \vec{k}[/tex]
    [tex]\vec{G} = \nabla\times\vec{F} = <1, 1, 1> [/tex]
    So,
    [tex]\nabla\cdot\vec{G} = 0[/tex]
    So, I would get the divergence theorem telling me that:

    [tex]\int\int\nabla\times\vec{F}\cdot \vec{n} dS = 0[/tex]
    But, at the same time,
    [tex] \int\int\nabla\times\vec{F}\cdot \vec{n} dS = \int \vec{F}\cdot d\vec{R} \neq 0[/tex]
    edit: consider my Surface the part of the sphere in the first octant with radius 1 for simplicity
     
    Last edited: Apr 17, 2008
  11. Apr 17, 2008 #10
    Ahh, disregard my last post. I think I now understand why i was having difficulty with this. So then is the only way I can use the divergence method to compute such an integral is if the surface is closed? And in this case, will it always be zero?
     
  12. Apr 17, 2008 #11
    Thank you for all the help!
     
  13. Apr 17, 2008 #12

    Hootenanny

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    Yes as Halls has already said, your version of the divergence theorem states that the integral of a continuous vector field, which has continuous first order partial derivatives, over a closed surface is equal to the integral of the divergence of the same vector field over the volume enclosed by the surface. Hence, the surface must be closed. It would perhaps better to write the RHS explicitly, so as to remind yourself that the surface must be closed,

    [tex]\iiint_V \text{div}\left(\underline{F}\right) dV = {\bigcirc \hspace{-0.85cm} \int \hspace{-0.55cm} \int}_S\underline{F}\cdot\underline{n}dS[/tex]

    It should also be noted that the surface S is oriented by the outward normals n.
    It's a pleasure :smile:
     
    Last edited: Apr 17, 2008
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