External unit normal field of cylinder Hello, I consider
    $$
A:=\left\{(x,y,z)\in\mathbb{R}^3:x^2+y^2\leq 1,0\leq z\leq 1\right\}
$$
    and
    $$
v(x,y,z):=(x^3,x^2y,zx^2).
$$
    and the task is to calculate the surface of $A$ by using the integral theorem of Gauß.
  


I guess it is meant to use
$$
\int\limits_{\partial A}\langle v,w\rangle\, dS=\int\limits_A\mbox{div } v\, d^3x,
$$
where $w$ is the external unit normal field of $\partial A$.

Am I right that the external normal field of $\partial A$ is given by $w(x)=e_3$, when $x$ is in the "cover" of the cylinder $A$, $w(x)=-e_3$, when $x$ is in the "ground" of the cylinder "A" and $w(x)=(x,y,0)$ is $x$ is the rest of the cylinder $A$?
$\displaystyle{\vec{r}_{\perp} \equiv x\,\hat{x}\ +\ y\,\hat{y}}$.


\begin{eqnarray*}
\int\left\vert{\rm d}\vec{S}\right\vert
&amp; = &amp;
\int\hat{r}_{\perp}\cdot{\rm d}\vec{S}
=
\int\nabla\cdot\left(\vec{r}_{\perp} \over r_{\perp}\right)\,{\rm d}V
=
\int\left\lbrack
\vec{r}_{\perp}\cdot\nabla\left(1 \over r_{\perp}\right)
+
{1 \over r_{\perp}}\,\nabla\cdot\vec{r}_{\perp}
\right\rbrack\,{\rm d}V
\\
&amp; = &amp;
\int\left\lbrack
\vec{r}_{\perp}\cdot\left({-1 \over r_{\perp}^{2}}\,{\vec{r}_{\perp} \over r_{\perp}}\right)
+
{1 \over r_{\perp}}\,2
\right\rbrack\,{\rm d}V
=
\int{1 \over r_{\perp}}\,{\rm d}V
=
\int_{0}^{r}{\rm d}r_{\perp}\int_{0}^{2\pi}\,{\rm d}\theta\int_{0}^{h}{\rm d}z
\\
&amp; = &amp;
2\pi rh
\end{eqnarray*}
$\displaystyle{\vec{r}_{\perp} \equiv x\,\hat{x}\ +\ y\,\hat{y}}$.


\begin{eqnarray*}
\int\left\vert{\rm d}\vec{S}\right\vert
&amp; = &amp;
\int\hat{r}_{\perp}\cdot{\rm d}\vec{S}
=
\int\nabla\cdot\left(\vec{r}_{\perp} \over r_{\perp}\right)\,{\rm d}V
=
\int\left\lbrack
\vec{r}_{\perp}\cdot\nabla\left(1 \over r_{\perp}\right)
+
{1 \over r_{\perp}}\,\nabla\cdot\vec{r}_{\perp}
\right\rbrack\,{\rm d}V
\\
&amp; = &amp;
\int\left\lbrack
\vec{r}_{\perp}\cdot\left({-1 \over r_{\perp}^{2}}\,{\vec{r}_{\perp} \over r_{\perp}}\right)
+
{1 \over r_{\perp}}\,2
\right\rbrack\,{\rm d}V
=
\int{1 \over r_{\perp}}\,{\rm d}V
=
\int_{0}^{r}{\rm d}r_{\perp}\int_{0}^{2\pi}\,{\rm d}\theta\int_{0}^{h}{\rm d}z
\\
&amp; = &amp;
2\pi rh
\end{eqnarray*}