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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

Lemon
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  • $S$ is not the boundary of a volume. Edit: Or, as Schmitty says, $\partial S$ is empty. – Neal Aug 15 '12 at 21:28
  • Isn't $V$ is the volume over the entire enclosed surface? – Lemon Aug 15 '12 at 21:30
  • 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. –  Aug 15 '12 at 22:02
  • @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 hemisphere – Lemon Aug 15 '12 at 22:21
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    @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. – Tunococ Aug 15 '12 at 22:53
  • @Tunococ, why does Schmitty say my surface is closed then? – Lemon Aug 15 '12 at 22:59
  • @jak: Tunococ also said it is closed. – timur Aug 16 '12 at 03:54
  • 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. –  Aug 17 '12 at 01:15
  • 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? – Lemon Aug 18 '12 at 01:37
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    Does this answer your question? Stokes and Gauss' Divergence theorem on a closed smooth surface in $\Bbb R^3$ – tryst with freedom May 12 '21 at 12:56

2 Answers2

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Actually nothing is wrong with that. You start with a vector field integrated over a closed curve. Your first equality which does use Stokes's Theorem goes to an integral over a surface S for which your original curve must be the boundary. Your next equality uses the divergence theorem and goes to an integral over a volume for which your surface S must be the boundary implying S is a closed surface. Since your assumptions indicate that S is a closed surface S doesn't have a boundary- or rather, the boundary of S is the empty set. So the integral you started with is over the empty set----> hence it's zero.

Schmitty
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  • But like Tunococ said, my surface is actually open – Lemon Aug 16 '12 at 00:01
  • As Tunococ said for an open surface the divergence theorem doesn't apply which gives you the problem you were looking for. In the case of a closed surface the argument is valid. Whatever F is if you integrate it over the empty set you're going to get zero. – Schmitty Aug 17 '12 at 16:47
  • How could you integrate over the empty set (the boundary)? if it is closed? I am confused – Lemon Aug 23 '12 at 21:58
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This is one of my favorite phenomena in multivariable calculus. I remember noticing this when I was first learning the subject, and spent many hours wondering how this could be.

The explanation for this phenomenon lies in the following geometric principle:**

The boundary of a boundary is empty.

This geometric fact is in some sense "dual" to the fact that $\text{div}(\text{curl}\,\mathbf{F}) = 0$ for all $\mathbf{F}$.

In particular, if you have a volume $V$ that bounds a surface $S$, then the surface $S$ cannot have a boundary curve. Said another way, the boundary curve $C = \partial S$ is the empty set, so integrating anything over it is zero.

Example: In the comments, you consider a solid hemisphere $V$. The boundary of $V$ will then be the surface of the hemisphere and also the disc base. This closed surface (consisting of both the hemisphere surface and the disc base) does not have a boundary curve.

On the other hand, the surface which is just the hemisphere (without the base) does have a boundary curve: namely, the circle. However, this surface cannot be said to enclose any volume.

Note 1: Typically when one talks about a "closed" surface, one specifically means a surface which does not have a boundary curve. This is an unfortunate piece of terminology since the term "closed" can also refer to being a closed subset of $\mathbb{R}^3$, and these two definitions are not equivalent.

For instance, the hemisphere together with its boundary curve (but not including the disc base) is a closed as a subset of $\mathbb{R}^3$, but is generally not called a "closed surface." However, the hemisphere together with the disc base is a closed surface (and is also closed as a subset of $\mathbb{R}^3$).

** Note 2: This principle is somewhat vague as stated. In order to make it precise, one needs to rigorously define the notion of "boundary." This can be done in a couple of ways; some definitions will satisfy this principle, while others won't. For now, let's not get into these details.

Jesse Madnick
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  • This closed surface (consisting of both the hemisphere surface and the disc base) does not have a boundary curve. Isn't it a circle? – Lemon Dec 23 '12 at 03:49
  • No. I understand your confusion, though a simple explanation escapes me... All I can say is: topologically speaking, the hemisphere together with the disc base is like a sphere that has been squashed and then pinched. Certainly you wouldn't say that a sphere has a boundary curve, right? Same thing here. – Jesse Madnick Dec 23 '12 at 17:34
  • As another example, a (circular) cylinder together with both caps does not have a boundary curve. However, a cylinder with no caps has a boundary consisting of 2 (disjoint) circles. Still different is a cylinder together with one cap, which has boundary consisting of 1 circle (namely the circle that isn't capped off). – Jesse Madnick Dec 23 '12 at 17:36
  • Isn't the cap bounded by a circle anyways? – Lemon Dec 25 '12 at 01:05
  • Yes, the boundary of the cap is a circle. But the boundary of a cylinder together with one cap consists of 1 circle -- namely, the circle that is the edge of the uncapped side. Is all of this clear? – Jesse Madnick Dec 25 '12 at 01:11
  • I think I do. So in the original problem, after I'd applied the Div Theorem. I got 0 not just because div(curlF) = 0, but also the volume is 0? – Lemon Dec 27 '12 at 21:18
  • I'm not clear as to the original problem. What exactly is the surface you are given? (Or what are the conditions on the surface?) – Jesse Madnick Dec 27 '12 at 21:35
  • None. I started out with a circle on the xy-plane, then I attached a surface (in this case a hemisphere). – Lemon Dec 28 '12 at 01:21
  • OK, so the surface you started with is the hemisphere (without the cap). You cannot apply the Divergence Theorem to this surface because this surface is not closed (because the cap is not included). On the other hand, if you start instead with the hemisphere together with the cap, then you can apply the Div Thm $\iiint \text{div}, \mathbf{F},dV = \iint \mathbf{F}\cdot d\mathbf{S}$. After that, you could apply Stokes Theorem if you wanted (assuming $\mathbf{F} = \text{curl} \mathbf{G}$) but your boundary curve would not be the circle in this case: it would just be the empty set. – Jesse Madnick Dec 28 '12 at 03:30
  • No, let's do this analytically perhaps. I start out with a sphere $x^2 + y^2 + z^2 = 1$ for $z \leq 0$, so in other words $z = +\sqrt{1-x^2 - y^2}$. The boundary is the unit circle on the xy-plane. Then I incorrectly applied the divergence theorem to the hemisphere because the hemisphere has no base to begin with. Is that right? – Lemon Dec 28 '12 at 21:06
  • Yes, that's exactly right (except I think you mean $z \geq 0$) :-) – Jesse Madnick Dec 29 '12 at 06:39
  • Yes I did. But was my original explanation of my mistake also correct? That I got $0$ not just because div(curlF) = 0, but also the volume is $0$? Btw, I checked in some advanced calc book and they also say that the concept of the "boundary" seems to be different in the context of Stoke's Theorem. It's by Folland – Lemon Jan 01 '13 at 05:10
  • Intuitively, yes, I suppose you could say that. I think it's more accurate, however, to say that the Divergence Theorem simply doesn't apply to the hemisphere (without cap), meaning that you shouldn't have gotten $0$ at all: the calculation should have stopped at Stokes' Theorem. Said another way, the hemisphere doesn't have a well-defined volume. And yes, I mentioned in my post that there are essentially 2 different definitions of "boundary" -- one for vector calculus (and differential geometry) and one for point-set topology. – Jesse Madnick Jan 01 '13 at 07:39