5
$\begingroup$

Could someone point out what is wrong with this equality? Assume that $\mathbf{F}$ is continuous (and hence, its partial derivatives).

$$\begin{align} \oint \mathbf{F}\cdot d\mathbf{s} & =^\text{by Stokes} \iint_S \nabla \times \mathbf{F} \cdot d\mathbf{S} \\ &=^\text{by Div} \iiint_V \nabla\cdot( \nabla \times \mathbf{F} ) \, dV \\ &=\iiint_V 0 \,dV \\ &=0\\ &\implies \oint \mathbf{F}\cdot d\mathbf{s}= 0 \; \forall \mathbf{F} \end{align}$$

Since we assumed $\mathbf{F}$ and its partials are all continuous. But obviously this is wrong if $\mathbf{F}$ is non-conservative. But everything seems to agree. What went wrong?

EDIT. For a refinement of the problem. Let me specifically state that $S$ is a closed surface with a boundary curve that is also closed. So $V$ here is the volume of that surface and since $S$ is closed it has a volume

  • 0
    $S$ is not the boundary of a volume. Edit: Or, as Schmitty says, $\partial S$ is empty.2012-08-15
  • 0
    Isn't $V$ is the volume over the entire enclosed surface?2012-08-15
  • 0
    You need to be more explicit about what curve/surface/volume you are integrating over. Perhaps take a simple example, say a circle, and tell us what you think the domains of the integrals are.2012-08-15
  • 0
    @RahulNarain, take a circle to be the boundary. Then I attach a hemisphere to that boundary to make my surface. My volume integral will integrate the volume of that hemisphere2012-08-15
  • 1
    @jak If you attach a hemisphere without base, you cannot use the divergence theorem. If you attach a hemisphere with base, then your surface is closed, hence the boundary is empty, not the circle you started with.2012-08-15
  • 0
    @Tunococ, why does Schmitty say my surface is closed then?2012-08-15
  • 0
    @jak: Tunococ also said it is closed.2012-08-16
  • 0
    Schmitty starts from not knowing what your original curve is, and says that since you applied the divergence theorem, your intermediate surface must have been closed, so its boundary -- your original curve -- must have been empty. On the other hand, if you start with the curve being a circle, then the surface you obtain will not be closed, so you cannot apply the divergence theorem after that.2012-08-17
  • 0
    If I take my hemisphere to be my surface (which is closed), then I say for my base (which is a circle), I'll attach a circle that is the boundary for the open surface of the open hemisphere or the disk itself. Then doesn't the Div Thrm still hold?2012-08-18

2 Answers 2