Homework Help: Integral of a closed surface

1. Jan 6, 2005

meteorologist1

I've been stuck on the following problem: If S is a closed surface that bounds the volume V, prove that: integral over this surface dS = 0.

I've been reading several textbooks that discuss flux, Stokes' Theorem, Divergence Theorem, but I can't seem to relate them to the problem I'm doing. The examples in the text all have a vector F and present the integral: integral over a surface of F dS, which I understand it as the flux. Is my case a flux problem? There is no vector F given in my problem.

Should I divide the closed surface into two halves and argue that pairs of normal vectors, one from each half cancel and therefore the integral over this surface dS = 0? What about Stokes' Theorem -- transforming it into a line integral?

Thanks.

2. Jan 6, 2005

dextercioby

$$\oint\oint_{\partial \Omega} dS=\oint\oint_{\partial\Omega} \vec{n}\cdot \vec{n} dS=\int\int\int_{\Omega} (\nabla\cdot \vec{n}) dV=\int\int\int_{\Omega} 0 dV=0$$

Okay??I made use of the fact that
$$\vec{n}\cdot\vec{n}=n^{2}\cos 0=n^{2}=1$$
,as unitvectors of the exterior normal to the surface.
Because this unit vector is constant (the director cosines are constants),its flux is zero,because its divergence is zero.

Daniel.

3. Jan 6, 2005

meteorologist1

Thanks. Actually there's a slight problem: I forgot to tell you that the dS is a vector in my problem. Yours is a scalar. I'm not sure what the difference here is between integrating a scalar and integrating a vector. I don't think the first equality holds anymore for vector dS. Sorry for the confusion. Thanks for your help.

(Show that: $$\oint_{S} d\vec{S}=0$$)

Last edited: Jan 6, 2005
4. Jan 7, 2005

HallsofIvy

Then perhaps it would be a good idea to tell us what the problem really is! You can integrate dS alone (getting surface area) but you can't integrate $\vec{dS}$ alone over a surface: you integrate the its dot product with some vector function.

In dextercioby's $\oint\oint_{\partial\Omega} \vec{n}\cdot \vec{n} dS$,
$\vec{n}dS$ IS the vector $\vec{dS}$.

5. Jan 7, 2005

arildno

meteorologist:
It seems to me that you want the VECTOR result:
$$\int_{S}d\vec{S}=\int_{S}\vec{n}dS=\vec{0}$$
This is achieved as follows:
$$\vec{n}=(\vec{i}\cdot\vec{n})\vec{i}+(\vec{j}\cdot\vec{n})\vec{j}+(\vec{k}\cdot\vec{n})\vec{k}$$
So that we have:
$$\int_{S}\vec{n}dS=\int_{S}(\vec{i}\cdot\vec{n})dS\vec{i}+\int_{S}(\vec{j}\cdot\vec{n})dS\vec{j}+\int_{S}(\vec{k}\cdot\vec{n})dS\vec{k}$$
since the unit vectors $$\vec{i},\vec{j},\vec{k}$$ are constants you may take out of the integral.

Use the normal form of the divergence theorem to get your result.

6. Jan 7, 2005

meteorologist1

Thanks. I was looking for the vector result, but it was also helpful to know how the scalar result is proved. I wasn't paying attention to dS as a vector or a scalar when I was first posting it, so sorry once again for the confusion.

7. Jan 7, 2005

arildno

Glad to be of assistance; welcome to PF!

8. Mar 17, 2010

fibonacci101

how do you prove this???

Prove that $$\int\int_{S} n dS = 0$$

for any closed surface S.