If ${\text{?}\mathrm{}}_{\mathrm{s}}\stackrel{?}{\mathrm{E}}\cdot \stackrel{?}{\mathrm{d}\mathrm{S}}=0$  over a surface, then:

Gauss Law is only concerned with the total enclosed charge that finally tells us the total flux. The charges outside may change field patterns. They not affect the total flux. It's actually incorrect to assume the field due to the external charges should also affect the flux through the Gaussian sur

Gauss Law does not directly give the electric field in all cases. It can only be used in calculations for symmetrical surfaces: spherical, cylindrical, or planar. 

The integral form of Gauss Law is considered as an indirect form and only in theory. It will still create a mathematical problem. The Gaussian surface passing through a discrete charge means it lies on the surface. Half of the electric flux is outside and half in. Not on the boundary. And we know Ga

All the charge given to a conducting sphere resides on outer surface.

Using gauss law in differential form

$\overrightarrow{\nabla }.\overrightarrow{E}=\frac{\rho }{{\epsilon }_0}$

$\overrightarrow{\nabla }.\left({\epsilon }_0\overrightarrow{E}\right)=\rho$

$\Rightarrow \overrightarrow{\nabla }.\overrightarrow{D}=\rho$

$\Rightarrow \left(\frac{\partial }{\partial x}\widehat{i}+\frac{\partial }{\partial y}\widehat{j}+\frac{\partial }{\partial z}\widehat{k}\right).\overrightarrow{D}=\rho$

$\begin{array}{l}\Rightarrow \rho =-e^{-x}siny+e^{-x}siny+2=2\\ \Rightarrow Q=2\left(Volume\right)=4\times 10^{-9}c=4nc\end{array}$