Calculate surface integral

Question

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!

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $$ \color{#66f}{\large% \int_{r\ =\ 1}\pars{-\,{\vec{r} \over r^{3}}}\cdot\dd\vec{\rm S}} =-\int_{r\ =\ 1}{\vec{r}\cdot\dd\vec{\rm S} \over r^{3}} =-\int_{r\ =\ 1}\dd\Omega_{\vec{r}}=\color{#66f}{\large -4\pi} $$