The discussion centers on proving that the surface integral \(\int\int_{S} n dS = 0\) for any closed surface \(S\). Participants emphasize the application of the Divergence Theorem, which states that \(\int\int_{S} \vec{F} \cdot \vec{n} dS = \int\int\int_{V} \nabla \cdot \vec{F} dV\). By selecting appropriate vector fields \(F\), such as \(F = (1, 0, 0)\), and demonstrating that the divergence \(\nabla \cdot \vec{F} = 0\), it is concluded that the integral of the outward unit normal vector \(n\) over the closed surface must equal zero. This conclusion is reached by breaking down the integral into its component parts and applying the Divergence Theorem correctly.
PREREQUISITESStudents and professionals in mathematics, physics, and engineering who are working with vector calculus, particularly those focusing on surface integrals and the Divergence Theorem.
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?
fibonacci101 said:Is there a value of n(outward unit vector)?
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?
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?
Dick said:You can use the divergence theorem to conclude something, yes. But what do you conclude?
fibonacci101 said:I will conclude that \int\int_{S} F \cdot n dS = 0
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.
fibonacci101 said:\vec{n}=(\vec{i}\cdot\vec{n})\vec{i}+(\vec{j}\cdot \vec{n})\vec{j}+(\vec{k}\cdot\vec{n})\vec{k}
Then,
\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}
I stacked to this step...
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?
fibonacci101 said:Divergence Theorem said that
\int\int_{S} \vec{F} \cdot \vec{n} dS = \int\int\int_{V} \nabla \cdot \vec{F} dV
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)?
ideasrule said:Yes. So what do you conclude from div F=0?
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.
ideasrule said:Yes, kind of. <br /> 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) <br /> should be <br /> 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) <br /> 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?
fibonacci101 said:Is this for me?
Dick said:Yes. Divergence is a scalar, not a vector. Your derivation just looks weird.
fibonacci101 said:<br /> <br /> 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<br /> <br />
right?
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).
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.