Sum Of The Normal Vectors Of a Gaussian Surface The thought occurred to me while I was studying Gauss's Law(Electrostatics).
Let there be a Gaussian surface and let $\hat{\mathbf{n}}$ denote the normal vector associated with each of the  individual infinitesimal areas $ds$,
Then the vector sum of all the normal vectors of the gaussian surface must be zero, or
$\sum\hat{\mathbf{n}} = 0$
The reasoning is as follows -
$1)$ All gaussian surfaces are closed, i.e they don't have any breakages or gaps.
$2)$ All gaussian surfaces are continuous, i.e the surface is differentiable(i think that's what it's called.)
Since the surface is continuous, for a given normal vector $\hat{\mathbf{n}}$ on the surface, I shall always be able to find a normal vector $-\hat{\mathbf{n}}$ and their vector sum will be zero. This implies that the vector sum of all the normal vectors must be zero. Therefore the above mentioned equation must be true.
Also, since Gauss law is true for all closed surfaces(Gaussian surface), there must be some property which is common for all closed surfaces. I think this is that property.
Is this actually true? Please reply.
 A: This is essentially true. The only problem is that the sum 
$$
\sum \mathbf{n} $$ 
makes no sense, since you have a continuum of normal vectors to sum. What is true is that the surface integral vanishes: 
$$
\iint_{\rm surface} \mathbf n dS = \mathbf 0.$$
To prove this you can use the divergence theorem: 
$$
\left(\iint_{\rm surface} \mathbf n dS \right)\cdot \mathbf i = \iint \mathbf n\cdot \mathbf i\, dS \overset{\rm div.\, theorem}{=} \iiint_{\rm volume} \nabla \cdot \mathbf i\, dV = 0,$$
because $\nabla \cdot \mathbf i =0$. Repeat this argument with $\mathbf j$ and $\mathbf k$ in place of $\mathbf i$ to conclude that
$$
\iint \mathbf n \, dS = \mathbf 0.$$
A: The sum of normal unit vectors is not zero. But the sum of all area element vectors is always zero for a closed surface.
By divergence theorem
$$\int_V \nabla\cdot \mathbf{c}d\tau=\oint_{\partial V}\mathbf{c}\cdot d\mathbf{a}$$
If $\mathbf{c}$ is a constant vector, then
$$\mathbf{c}\cdot\oint_{\partial V}d\mathbf{a}=0$$
Since $\mathbf{c}$ is arbitrary constant vector, we have
$$\oint_{\partial V}d\mathbf{a}=0$$
