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Math Help - An interesting problem for Multivariable Calculus Students

  1. #1
    A Plied Mathematician
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    An interesting problem for Multivariable Calculus Students

    This is aimed a bit more at students, rather than helpers, but I won't go so far as to ask helpers to refrain from answering.

    The divergence theorem states that

    \iiint_{V}\nabla\cdot \mathbf{F}\,dV=\iint_{{\partial}V}\mathbf{F}\cdot d\mathbf{S},

    and the Kelvin-Stokes theorem states that

    \iint_{A}\nabla\times\mathbf{G}\cdot d\mathbf{S}=\oint_{\partial A}\mathbf{G}\cdot d\mathbf{r}.

    If I set

    \mathbf{F}=\nabla\times\mathbf{G},

    then comment on the legitimacy of writing

    \iiint_{V}\nabla\cdot(\nabla\times\mathbf{G})\,dV=  \iint_{\partial V}\nabla\times\mathbf{G}\cdot d\mathbf{S}=\oint_{\partial(\partial V)}\mathbf{G}\cdot d\mathbf{r}.
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  2. #2
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    Not seeing any takers here, so I'll post my thoughts.

    Spoiler:

    There are at least two problems with this expression.

    1. The integrand of the far left expression is always zero; hence, even if the expression were true, it wouldn't tell us much.

    2. There's a problem with what you're integrating over. The divergence theorem really says that for an enclosed volume, the volume integral equals the given surface integral over the enclosing volume. I could be wrong, you topologists out there could probably answer me yea or nay, but if a volume has an enclosing surface, then I don't think it's possible for that enclosing surface to have an enclosing contour, which is what the Kelvin-Stokes theorem requires. That is, the Kelvin-Stokes theorem would, I think, require a surface to be integrated over which has a boundary. But if it has a boundary, I don't think it can enclose a volume.
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  3. #3
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    gah...i would have to review my old books on manifolds, but i believe that under a suitable definition of "integrable region" you have ∂(∂V) = 0, so the equation becomes 0 = 0.
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    Re: An interesting problem for Multivariable Calculus Students

    I asked the exact same question in Calculus Forum.
    personally i do not know the correct answer. But div(curl(G)) is always zero, making the far right side always zero
    whereas the far left side is only zero if G field is conservative. so I guess I need better brain to prove this.....
    I do not know the answer.
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  5. #5
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    Re: An interesting problem for Multivariable Calculus Students

    Hi friends, calculus constitutes a major part of modern mathematics education. And confidently i can say you guys going to rock in calculus if you going to concern with good calculus tutor and if you are not getting better tutor near your residence then you can go for online calculus tutor.
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