Proving \int\int_{S} n dS = 0 for Closed Surface S

In summary: I mean, the F is an arbitrary vector field. So, it can be expressed in terms of its components as F=(Fx,Fy,Fz). Then, div(F)=d/dx(Fx)+d/dy(Fy)+d/dz(Fz), which is equal to zero since all the components are zero. This, in turn, leads to the conclusion that \int\int_{S} F\cdot n dS=0 for any closed surface S, as required. In summary, the conversation discusses how to prove that the integral of ndS is equal to zero for any closed surface S using the divergence theorem. The participants consider different vector fields and their components before realizing that the divergence theorem can be applied to prove
  • #1
fibonacci101
15
0

Homework Statement


Prove that [tex]\int\int_{S} n dS = 0 [/tex] for any closed surface S.

Homework Equations


The Attempt at a Solution



I used divergence theorem. But i thought it is applicable only if there is another vector multiplied to that outward unit vector (n).

[tex]\int\int_{S} F {\cdot} n dS[/tex]
 
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  • #2
The integral of ndS is a vector equation. Split it into components. n=(nx,ny,nz). What's a vector field F that might have the property that F.n=nx, for example?
 
  • #3
my problem is how do I show that the [tex]\int\int_{S} n dS = 0[/tex]

for any closed surface S

----> I used the divergence theorem, but I don't think it will help me 'cause it is applicable for

[tex]\int\int_{S} F{ \cdot} n dS [/tex]

not for

[tex]\int\int_{S} n dS [/tex]
 
  • #4
You basically just reposted the same thing. Look, if you show integral of nx*dS, ny*dS and nz*dS are zero, you are done right? What's an appropriate choice of F for each?
 
  • #5
Dick said:
You basically just reposted the same thing. Look, if you show integral of nx*dS, ny*dS and nz*dS are zero, you are done right? What's an appropriate choice of F for each?

Is there a value of n(outward unit vector)?
 
  • #6
fibonacci101 said:
Is there a value of n(outward unit vector)?

No, n is the outward unit vector. Whatever that is. You don't have much control over that. You can choose F. Suppose you choose F=(1,0,0). What do you conclude?
 
  • #7
Dick said:
No, n is the outward unit vector. Whatever that is. You don't have much control over that. You can choose F. Suppose you choose F=(1,0,0). What do you conclude?

Yeah, I get what you are saying... So If I choose an F a vector the solution will lead to Divergence Theorem? Am I right?
 
  • #8
fibonacci101 said:
Yeah, I get what you are saying... So If I choose an F a vector the solution will lead to Divergence Theorem? Am I right?

You can use the divergence theorem to conclude something, yes. But what do you conclude?
 
  • #9
Dick said:
You can use the divergence theorem to conclude something, yes. But what do you conclude?

I will conclude that [tex]\int\int_{S} F \cdot n dS = 0 [/tex]
 
  • #10
fibonacci101 said:
I will conclude that [tex]\int\int_{S} F \cdot n dS = 0 [/tex]

Why do you conclude that and what is F.n and how does that help you conclude integral ndS is zero? You really aren't giving me much to go on except repeating the divergence theorem over and over again. You have to apply the divergence theorem in a specific way to solve a problem.
 
  • #11
Dick said:
Why do you conclude that and what is F.n and how does that help you conclude integral ndS is zero? You really aren't giving me much to go on except repeating the divergence theorem over and over again. You have to apply the divergence theorem in a specific way to solve a problem.


[tex]\vec{n}=(\vec{i}\cdot\vec{n})\vec{i}+(\vec{j}\cdot \vec{n})\vec{j}+(\vec{k}\cdot\vec{n})\vec{k}[/tex]

Then,

[tex]\int\int_{S}\vec{n}dS=\int\int_{S}(\vec{i}\cdot\vec{n})dS\vec{i}+\int\int_{S}(\vec{j}\cdot\vec{n})dS\vec{j}+\int\int _{S}(\vec{k}\cdot\vec{n})dS\vec{k}[/tex]

I stacked to this step...
 
  • #12
fibonacci101 said:
[tex]\vec{n}=(\vec{i}\cdot\vec{n})\vec{i}+(\vec{j}\cdot \vec{n})\vec{j}+(\vec{k}\cdot\vec{n})\vec{k}[/tex]

Then,

[tex]\int\int_{S}\vec{n}dS=\int\int_{S}(\vec{i}\cdot\vec{n})dS\vec{i}+\int\int_{S}(\vec{j}\cdot\vec{n})dS\vec{j}+\int\int _{S}(\vec{k}\cdot\vec{n})dS\vec{k}[/tex]

I stacked to this step...

Now that's GOOD! You've got it. Ok, so what does the divergence theorem tell you about the first term where F=i? You may have known this all along. But I just couldn't figure out if you did from your posts.
 
Last edited:
  • #13
Dick said:
Now that's GOOD! You've got it. Ok, so what does the divergence theorem tell you about the first term where F=i?

Divergence Theorem said that

[tex]\int\int_{S} \vec{F} \cdot \vec{n} dS = \int\int\int_{V} \nabla \cdot \vec{F} dV [/tex]
 
  • #14
fibonacci101 said:
Divergence Theorem said that

[tex]\int\int_{S} \vec{F} \cdot \vec{n} dS = \int\int\int_{V} \nabla \cdot \vec{F} dV [/tex]

You did it again. You just quoted the divergence theorem. What DO YOU CONCLUDE from the divergence theorem? What's F and what's div(F)?
 
  • #15
Dick said:
You did it again. You just quoted the divergence theorem. What DO YOU CONCLUDE from the divergence theorem? What's F and what's div(F)?

[tex]div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{i} + \vector {j} +\vector{k}\right) [/tex]

[tex]div F =\left[\frac{\partial}{x}\vector{0} + \frac{\partial}{y}\vector{0}+\frac{\partial}{z}\vector{0}\right][/tex]

[tex]div F = 0 + 0 + 0[/tex]

[tex]div F = 0 [/tex]

Is this right? I hope so...
 
  • #16
Yes, kind of. [tex]
div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{i} + \vector {j} +\vector{k}\right)
[/tex] should be [tex]
div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{i}+\frac{\partial}{z}\vector{i}\right]\left(\vector{i} + \vector {j} +\vector{k}\right)
[/tex] since F=i, and your next line is not correct since i is not equal to 0. However, it is true that div F=0. What do you conclude from that?
 
  • #17
ideasrule said:
Yes. So what do you conclude from div F=0?

Yeah.. That is 0...

Thanks for helping and guiding me,Sir.

Hope to guide me in my further studies..

Thanks again... I LOVE YOU! LOL
 
  • #18
The argument looks a little strange. What is the F you are talking about? If F=i then Fx=1, Fy=0 and Fz=0. So sure, div(F)=d/dx(Fx)+d/dy(Fy)+d/dz(Fz)=0. All of the terms are zero.
 
  • #19
Dick said:
The argument looks a little strange. What is the F you are talking about? If F=i then Fx=1, Fy=0 and Fz=0. So sure, div(F)=d/dx(Fx)+d/dy(Fy)+d/dz(Fz)=0. All of the terms are zero.

Is this for me?
 
  • #20
ideasrule said:
Yes, kind of. [tex]
div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{i} + \vector {j} +\vector{k}\right)
[/tex] should be [tex]
div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{i}+\frac{\partial}{z}\vector{i}\right]\left(\vector{i} + \vector {j} +\vector{k}\right)
[/tex] since F=i, and your next line is not correct since i is not equal to 0. However, it is true that div F=0. What do you conclude from that?

By the definition of divergence...see??
 
  • #21
fibonacci101 said:
Is this for me?

Yes. Divergence is a scalar, not a vector. Your derivation just looks weird.
 
  • #22
Dick said:
Yes. Divergence is a scalar, not a vector. Your derivation just looks weird.

[tex]

div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{ni} + \vector {nj} +\vector{nk}\right) = 0

[/tex]

right?
 
  • #23
fibonacci101 said:
[tex]

div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{ni} + \vector {nj} +\vector{nk}\right) = 0

[/tex]

right?

If F=i then Fx=1, Fy=0 and Fz=0. So sure, div(F)=d/dx(Fx)+d/dy(Fy)+d/dz(Fz)=0. I know I'm repeating myself here but I don't see how what you are doing is related to div(F).
 
  • #24
Dick said:
If F=i then Fx=1, Fy=0 and Fz=0. So sure, div(F)=d/dx(Fx)+d/dy(Fy)+d/dz(Fz)=0. I know I'm repeating myself here but I don't see how what you are doing is related to div(F).

So, How is it to be equal to zero?

i don't know what to do. I thought I've already got the right solution. but it is quite wrong...
 
  • #25
It is zero. What's the definition of divergence? How can you say div(i), div(j) and div(k)=0. This is really pretty simple. I don't know why you are making this look so difficult.
 
  • #26
Dick said:
It is zero. What's the definition of divergence? How can you say div(i), div(j) and div(k)=0. This is really pretty simple. I don't know why you are making this look so difficult.

Okay!

This is my last try and I hope I will be right ..
Given a closed Surface S, the vector Area of this is zero. so that is i.t.
 
  • #27
You solved it in post 11. div(i)=0, div(j)=0 and div(k)=0. Put those in for F. Not all at once, one at a time. You were all done.
 
Last edited:

1. What does the given integral represent in the context of a closed surface?

The integral \int\int_{S} n dS represents the flux of a vector field through a closed surface S. It is a measure of how much of the vector field passes through the surface in a given direction.

2. Why is it important to prove that the integral is equal to zero for a closed surface?

This integral is important because it helps us understand the behavior of a vector field and its interaction with a closed surface. It also has many applications in physics, such as in the study of electric and magnetic fields. Proving that the integral is equal to zero for a closed surface allows us to make accurate predictions and calculations in various scientific fields.

3. What is the significance of a closed surface in this context?

A closed surface is one that has no boundary or openings, and it completely encloses a region. In the context of the integral \int\int_{S} n dS = 0, the closed surface S represents a physical boundary or a theoretical boundary between two regions. This integral is used to analyze the flow of a vector field through this boundary.

4. What are some common methods used to prove that the integral is equal to zero for a closed surface?

There are various methods used to prove that the integral \int\int_{S} n dS = 0 for a closed surface S. Some of the commonly used methods include the divergence theorem, Stokes' theorem, and Green's theorem. These theorems provide a way to convert the surface integral into a line or volume integral, which can then be easily evaluated to show that it is equal to zero.

5. Can this integral be non-zero for a closed surface?

Yes, it is possible for the integral \int\int_{S} n dS to be non-zero for a closed surface S. This can occur if the vector field is not continuous or if the surface has a hole or opening. In such cases, the integral represents the net flow of the vector field through the surface.

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