surface integral of a vector field over the unit sphere Problem: find the surface integral of the vector field:
$$F = \frac{x - (0,0,-1)}{||x - (0,0,-1)||^3}$$
over the unite sphare Except the point $(0,0,-1)$.
I used polar coordinate for parametrization but then a $\sqrt{2(1 + \sin(\phi))}$ appears in the denomitor which makes it hard to get integral with respect to $\phi$ any hints?
Edit: Here is the polar coordinate approach :
Ok I got the solution using polar coordinate I had a mistake which made it hard to compute:
$$\int\int F .n dS = \int\int \frac{z + 1} { {\sqrt{(x ^ 2 + y ^ 2 + (z + 1) ^ 2)}}^{\frac{3}{2}}} ds $$
$$=\int\int \frac{\cos(\phi) + 1}{8 cos^3(\phi/2)} \sin(\phi)d\phi d \theta = 2 \pi$$
But as hinted in the comments section the divergence is zero so using divergence theorem the flux should also be zero. Is this a contradiction?
 A: $\newcommand{\DS}{\mathbf{dS}}$Here are two calculations. The first uses your approach but avoids converting to spherical coordinates. (The integral obtained by converting to spherical is easily evaluated by converting back to the form below.) The second uses the divergence theorem.

I. As you've shown, at a point $(x, y, z)$ of the unit sphere, the outward unit normal is $n = (x, y, z)$, so the flux density is
\begin{align*}
  F \cdot n
  &= \frac{(x, y, z) \cdot (x, y, z + 1)}{(x^{2} + y^{2} + (z+1)^{2})^{3/2}} \\
  &= \frac{x^{2} + y^{2} + z^{2} + z}{(x^{2} + y^{2} + z^{2} + 2z + 1)^{3/2}} \\
  &= \frac{1 + z}{(2 + 2z)^{3/2}}
  = \frac{1}{2\sqrt{2}\sqrt{1 + z}}.
\end{align*}
Since the scalar surface element of the unit sphere is $dS = 2\pi\, dz$ (compare Archimedes' hat box theorem), the flux is
$$
\iint_{S} F \cdot n\, dS = \frac{\pi}{\sqrt{2}}\int_{-1}^{1} \frac{dz}{\sqrt{1 + z}} = 2\pi.
$$
II. Consider, for each real number $r$ with $0 < r < 1$, the closed surface $S_{r}^{-}$ consisting of the unit sphere $S$ with a small disk removed near $(0, 0, -1)$ and capped off with the "top" part of the sphere of radius $r$ about $(0, 0, -1)$. As $r \to 0^{+}$, the portion of $S$ in $S_{r}^{-}$ "approaches" the entire sphere, while the portion of the small sphere similarly approaches the upper hemisphere of radius $r$, for which the outward flux is $-2\pi$ (either directly, or by the divergence theorem). The flux over $S$ is therefore $2\pi$.
A similar argument can be given using the closed surface $S_{r}^{+}$ consisting of the unit sphere with the same small disk removed near $(0, 0, -1)$ and capped off with the "bottom" part of the sphere of radius $r$ about $(0, 0, -1)$.

This yields a third "more symmetrical" viewpoint: The outward flux through $S_{r}^{-}$ is $0$ by the divergence theorem, and the outward flux through $S_{r}^{+}$ is $4\pi$. Since the outward fluxes of $F$ over the small hemispheres cancel in the limit (because $F$ points inward along the top hemisphere and outward along the bottom), we have
$$
2\iint_{S} F \cdot \DS
= \lim_{r \to 0^{+}} \biggl[\iint_{S_{r}^{+}} F \cdot \DS + \iint_{S_{r}^{-}} F \cdot \DS\biggr]
= 4\pi.
$$
The qualitative point is, the tangent plane to $S$ at $(0, 0, -1)$ splits each ball of radius $r$ about $(0, 0, -1)$ in half, which "splits the source of $F$ in half". The fluid flowing "upward" into $S$ from $(0, 0, -1)$ contributes to the net flux across $S$ and the portion flowing "downward" out of $S$ does not because it's not enclosed by $S$.
