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)
div F =\left[\frac{\partial}{x}\vector{0} + \frac{\partial}{y}\vector{0}+\frac{\partial}{z}\vector{0}\right]
div F = 0 + 0 + 0...
\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...
my problem is how do I show that the \int\int_{S} n dS = 0
for any closed surface [SIZE="3"]S
----> I used the divergence theorem, but I don't think it will help me 'cause it is applicable for
\int\int_{S} F{ \cdot} n dS
not for
\int\int_{S} n dS
Homework Statement
Prove that \int\int_{S} n dS = 0 for any closed surface [SIZE="3"]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).
\int\int_{S} F {\cdot} n dS