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I have a problem understanding Stokes' Theorem and Gauss' Divergence Theorem. Suppose the following:

Let $F$ be a vector field in $\Bbb R^3$. Let $S$ be an oriented closed smooth Surface enclosing a volume $V$ and let $C$ be a positively-oriented closed curve surrounding $S$

Stokes' Theorem says: $$ \int_C F·dr=\iint_S (\nabla \times F) · dS $$

Then, by the Divergence Theorem: $$ \iint_S (\nabla \times F)·dS = \iiint_V \nabla·(\nabla \times F) dV $$

But $ \nabla·(\nabla \times F)=0$. So everything is $0$

What is it that I am not seeing?

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    $\begingroup$ If $S$ encloses $V$, then there is no curve $C$ surrounding $S$, since $S$ is a closed surface (or a "sum" of several such). $\endgroup$ Commented Feb 23, 2014 at 21:28
  • $\begingroup$ Thanks, I know understand how Stockes' theorem works, I was quite confused! $\endgroup$
    – Francisco
    Commented Feb 24, 2014 at 13:05

1 Answer 1

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While Daniel Fischer's answer probably is more direct, I find this way of showing that the circulation over a closed surface is zero (using Stokes theorem) more intuitive:

A closed surface split in two open surfaces

Here I split the closed surface $S$ into two surfaces $S_1$ and $S_2$ with a shared boundary, the curve $C_1$. If we apply Stokes theorem to the two surfaces separately we get:

$$\int{\int_{S_1} (\nabla \times \bar F) d \bar S} = \int_{C_1} {\bar F \cdot d \bar r}$$

and

$$\int{\int_{S_2} (\nabla \times \bar F) d \bar S} = - \int_{C_1} {\bar F \cdot d \bar r}$$

notice the negative sign before the right hand integral in the second equation. This is because the curve $C_1$ runs in the opposite direction to the normals of $S_2$ compared to $S_1$.

Combing the two surface integrals to get the integral over the entire surface, $S = S_1 \cup S_2$:

$$ \begin{eqnarray} \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int{\int_{S_1} (\nabla \times \bar F) d \bar S} + \int{\int_{S_2} (\nabla \times \bar F) d \bar S} \\ \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int_{C_1} {\bar F \cdot d \bar r} - \int_{C_1} {\bar F \cdot d \bar r} \\ &=& 0 \end{eqnarray} $$

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  • $\begingroup$ Beautiful proof $\endgroup$
    – Babu
    Commented May 12, 2021 at 12:46

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