Gauss' Law and area vector Recently I've been doing some physics exercises on electric and magnetic fields and read up somewhere that the vector area of a closed surface is equal to zero. That made me wonder why, when using Gauss' Law, the surface integral over dS (assuming E is constant, like when we use a sphere as the surface around the charge which is spread uniformly along a circle) is ever non-zero (as the law states that it considers net flux through a closed surface). Maybe I'm just mixing up some terms but I really need some clarification on this.
 A: Gauss' Law says $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ whereas the total vector area is $\iint_S \mathbf{dA}=\mathbf{0}$ for some closed surface $S$. The total vector area is taking vector sum of all the differential area vectors that are normal to the surface, whereas $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ is taking the dot product of the electric field and the differential area vector.
Let's use a sphere of radius $R$ centered at the origin as an example. Its clear that $\iint_S \mathbf{dA}=\mathbf{0}$ since, for every point on the sphere, the normal vector of that point cancels out with the point opposite it on the sphere. Therefore, $\iint_S \mathbf{dA}=\mathbf{0}$.
Now suppose there's a charge $Q$ at the origin. The electric field and the normal vector will be parallel at every point on the sphere, so the symmetry of the problem means $\iint_S \mathbf{E} \cdot \mathbf{dA} = EA$ where $E$ and $A$ are the electric field and surface area of the sphere respectively.
A: When you find the vector area of a closed surface you are dealing with vectors and the total sum is zero. However, when you use Gauss' law with a constant E what you are integrating is a scalar, and that gives you the surface (not zero).
