The electric field of a point charge $q$ at the origin is given by \begin{equation} \EE = \frac{q}{4\pi\epsilon_0} \frac{\rhat}{r^2} = \frac{q}{4\pi\epsilon_0} \frac{x\,\xhat+y\,\yhat+z\,\zhat}{(x^2+y^2+z^2)^{3/2}} \end{equation} where $\rhat$ is the unit vector in the radial direction in spherical coordinates. Note that the first expression clearly indicates both the spherical symmetry of $\EE$ and its $\frac{1}{r^2}$ fall-off behavior, while the second expression does neither.

It is easy to determine $d\SS$ on the sphere by inspection; we nevertheless go through the details of the differential approach for this case. We use “physicists' conventions” for spherical coordinates, so that $\theta$ is the angle from the North Pole, and $\phi$ the angle in the $xy$-plane. We use the obvious families of curves, namely the lines of latitude and longitude. Starting either from the general formula for $d\rr$ in spherical coordinates, namely \begin{equation} d\rr = dr\,\rhat + r\,d\theta\,\that + r\sin\theta\,d\phi\,\phat \end{equation} or directly using the geometry behind that formula, one quickly arrives at \begin{eqnarray} d\rr_1 &=& r\,d\theta\,\that \\ d\rr_2 &=& r\sin\theta\,d\phi\,\phat \\ d\SS &=& d\rr_1 \times d\rr_2 = r^2\sin\theta\,d\theta\,d\phi\,\rhat \end{eqnarray} so that \begin{equation} \Sint \EE\cdot d\SS = \int_0^{2\pi} \int_0^\pi \frac{q}{4\pi\epsilon_0} \frac{\rhat}{r^2} \cdot r^2 \sin\theta\,d\theta\,d\phi \,\rhat = \frac{q}{\epsilon_0} \label{Gauss0} \end{equation} You may recognize this result as Gauss' Law, which says that (\ref{Gauss0}) gives the relationship between the total charge $q$ inside any closed surface $S$ and the flux of the electric field through the surface.