Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
$S$ 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? –  sidht 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. –  Rahul 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 –  sidht Aug 15 '12 at 22:21
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. –  Tunococ Aug 15 '12 at 22:53
show 4 more comments

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
    
This closed surface (consisting of both the hemisphere surface and the disc base) does not have a boundary curve. Isn't it a circle? –  sidht Dec 23 '12 at 3:49
    
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? –  sidht Dec 25 '12 at 1:05
    
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 1:11
show 8 more comments

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.

share|improve this answer
    
But like Tunococ said, my surface is actually open –  sidht Aug 16 '12 at 0:01
    
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 –  sidht Aug 23 '12 at 21:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.