Calculate surface integral I need some help with the following:
Given $$f(x,y,z)=\left(
\frac{-x}{(x^2+y^2+z^2)^{\frac{3}{2}}}, \frac{-y}{(x^2+y^2+z^2)^{\frac32}}, \frac{-z}{(x^2+y^2+z^2)^{\frac32}}
\right),$$ calculate the flow rate of fuid out of the total surface
$S$, where $$S=\{(x,y,z)\; | \; x^2+y^2+z^2=1 \}.$$
I got $\displaystyle \frac{4\pi}{3}$ but I think I messed up with the normal vector.
Any help would be really appreciated.
Thank you!
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Note that
$\ds{\vec{\fermi}\pars{x,y,z}=-\,{\vec{r} \over r^{3}}=\nabla\pars{1 \over r}}$ where
$\ds{\vec{r} \equiv x\,\hat{x} + y\,\hat{y} + z\,\hat{z}}$.

\begin{align}
\color{#66f}{\large\int_{\rm S}\vec{\fermi}\pars{x,y,z}\cdot\dd\vec{\rm S}}&
=\int_{\rm V}\nabla\cdot\vec{\fermi}\pars{x,y,z}\,\dd^{3}\vec{r}
=\int_{\rm V}\nabla\cdot\nabla\pars{1 \over r}\,\dd^{3}\vec{r}
\\[5mm]&=\int_{\rm V}\nabla^{2}\pars{1 \over r}\,\dd^{3}\vec{r}
=\int_{\rm V}\bracks{-4\pi\,\delta\pars{\vec{r}}}\,\dd^{3}\vec{r}
=\color{#66f}{\large -4\pi}
\end{align}

A $\ds{\tt\mbox{direct calculation}}$ is given by:
$$
\color{#66f}{\large%
\int_{r\ =\ 1}\pars{-\,{\vec{r} \over r^{3}}}\cdot\pars{r^{2}\, {\vec{r} \over r}}
\verts{\dd\vec{\rm S}}}
=-\int_{r\ =\ 1}\verts{\dd\vec{\rm S}}
=\color{#66f}{\large -4\pi}
$$

Otherwise
  $$
\color{#66f}{\large%
\int_{r\ =\ 1}\pars{-\,{\vec{r} \over r^{3}}}\cdot\dd\vec{\rm S}}
=-\int_{r\ =\ 1}{\vec{r}\cdot\dd\vec{\rm S} \over r^{3}}
=-\int_{r\ =\ 1}\dd\Omega_{\vec{r}}=\color{#66f}{\large -4\pi}
$$

