# Vector flux through a segment of a sphere

Given the vector field $\vec A(\vec r) = \vec r$, I have to calculate the vector flux through a sphere whose center is located in the origin. I want to apply Gauß-Theorem and use spherical coordinates.

The vector field in spherical coordinates is $\vec A (\vec r) = \begin{pmatrix} r \\ 0 \\ 0 \end{pmatrix}$. Now I have to calculate $\int \vec A d\vec F$, and here i am very unsure what $\vec F$ is because I never did it in spherical coordinates.

My attempt is $\vec F = \vec e_r = \begin{pmatrix} \cos(\phi)\sin(\theta) \\ \sin(\phi)\sin(\theta) \\ \cos(\theta)\end{pmatrix}d\phi d\theta$ .

Is this correct and if no, what did I miss?

$d\vec{F}$ is the vector pointing out of the infinitesimal area element of your sphere. Therefore: $d\vec{F}=dS\cdot\vec{n}$, where $dS=r^2\sin\theta d\theta d\phi$ and $\vec{n}=\vec{r}$ (by symmetry of the problem).
• why $r^2 \sin(\theta)$ ? I know this as "volume element", is there a deeper meaning of that or is it just coincidence? Jan 18, 2015 at 19:40
• The surface element is $r^2\sin\theta d\theta d\theta$. Imagine a small square on the surface of your sphere. The "vertical" side is equal to an arc of angle $d\theta$, i.e. the length (angle times radius) is $r\cdot d\theta$. The "horizontal" side of the square is again an arc, this time of angle $d\phi$, but with the radius of a small circle, given by $r\cdot\sin\theta$. The surface area of the square is then $r^2\sin\theta d\theta d\phi$. If you want the volume element, you move from a square to a cube, i.e. you also integrate with respect to $r$: $dV=r^2\sin\theta d\theta d\phi dr$. Jan 18, 2015 at 19:51