Vector Integration: Solving \int_{S}\int n dS = 0 on Closed Surfaces

Click For Summary

Homework Help Overview

The discussion revolves around the vector integral \(\int_{S}\int n \, dS = 0\) for any closed surface \(S\), focusing on the implications of this equation in the context of vector fields and their properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the conditions under which the integral equals zero, questioning the nature of the vector field \(n\) and its relationship to the surface. Some suggest that the vector field must be conservative or tangential to the surface, while others inquire about specific vector fields that satisfy the condition.

Discussion Status

The discussion is ongoing, with participants providing hints and exploring different interpretations of the problem. Some guidance has been offered regarding the use of vector components and the divergence theorem, but no consensus has been reached on a definitive approach or proof.

Contextual Notes

There is a recurring emphasis on the properties of vector fields, particularly regarding their tangential nature and the implications of being zero everywhere. Participants express uncertainty about the exact question being posed and the expectations for proof within the forum's guidelines.

yusukered07
Messages
18
Reaction score
0

Homework Statement



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

Homework Equations





The Attempt at a Solution


I can't solve this because I don't have any idea in Vector intregrals.
 
Physics news on Phys.org
It should only equal zero if it is conservative since the partial of n with respect to x equals the partial m with respect to y. I am not sure if this what you are looking for since I am not sure what your question is.
 
yusukered07 said:

Homework Statement



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

Homework Equations





The Attempt at a Solution


I can't solve this because I don't have any idea in Vector intregrals.

Homework Statement




Homework Equations




The Attempt at a Solution



Hint: The result is a vector, so look at its components. The x component of a vector V is [itex]V \cdot i[/itex]. So look at, for example:

[tex]i \cdot \int\int_S \hat n\ dS = \int\int_S i\cdot \hat n\ dS[/tex]
 
LCKurtz said:
Hint: The result is a vector, so look at its components. The x component of a vector V is [itex]V \cdot i[/itex]. So look at, for example:

[tex]i \cdot \int\int_S \hat n\ dS = \int\int_S i\cdot \hat n\ dS[/tex]

What is i there?

n is normal line, I think
 
i is the unit vector in the x direction, the usual i,j,k notation. n is the unit outward normal.
 
If the vector field is always tangential to the surface (and therefore perpendicular with the normal of the surface), this relation is trivially true. If the vector field is 0 everywhere, then this relation is also trivially true. What is the question exactly? Find all such vector fields n for which this relation holds?
 
LCKurtz said:
i is the unit vector in the x direction, the usual i,j,k notation. n is the unit outward normal.

So, will you help me to make a proof?
 
Matterwave said:
If the vector field is always tangential to the surface (and therefore perpendicular with the normal of the surface), this relation is trivially true. If the vector field is 0 everywhere, then this relation is also trivially true. What is the question exactly? Find all such vector fields n for which this relation holds?

is there any proof you can show??

that's the question..
 
Matterwave said:
If the vector field is always tangential to the surface (and therefore perpendicular with the normal of the surface), this relation is trivially true. If the vector field is 0 everywhere, then this relation is also trivially true. What is the question exactly? Find all such vector fields n for which this relation holds?

As LCKurtz already pointed out, [itex]\textbf{n}[/itex] is the outward unit normal to whichever closed surface is integrated over. The problem is to show that this integral is zero for any closed surface.
 
  • #10
yusukered07 said:
So, will you help me to make a proof?

We don't make proofs for you here. LCKurtz has given you a very good hint in his first reply, try using it and show us what you get. You should find that the divergence theorem is very useful to you here:wink:
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 26 ·
Replies
26
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
3K