Why Does the Integral Over a Closed Surface Equal Zero?

[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.

 

[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.

 

[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: $$\oint_{S} d\vec{S}=0$$)

 

[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 $\vec{dS}$.

 

[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.

 

[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.

 

[arildno]

Glad to be of assistance; welcome to PF!

 

[fibonacci101]

how do you prove this?

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

for any closed surface S.