# Math Help - An interesting problem for Multivariable Calculus Students

1. ## 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

$\iiint_{V}\nabla\cdot(\nabla\times\mathbf{G})\,dV= \iint_{\partial V}\nabla\times\mathbf{G}\cdot d\mathbf{S}=\oint_{\partial(\partial V)}\mathbf{G}\cdot d\mathbf{r}.$

2. 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.

3. 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.

4. ## 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.

5. ## Re: An interesting problem for Multivariable Calculus Students

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