Surface Integrals

  • #1

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]
 

Answers and Replies

  • #2
Dick
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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
Dick
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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
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
Dick
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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
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
Dick
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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
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
Dick
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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
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
Dick
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[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
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
Dick
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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
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
ideasrule
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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
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
Dick
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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
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
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
Dick
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Is this for me????

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

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