# Calculating flux without using divergence theorem

• Jun 13th 2011, 07:40 AM
imagemania
Calculating flux without using divergence theorem
Ok i have done majority of the working for this, but get a bit stuck on a conventional basis. Though i may of gone wrong.
Quote:

Evaluate the flux $\displaystyle \iint_{s} \vec{F}.d\vec{S}$
of the vector field
$\displaystyle \vec{F} = r\vec{r}$.
Where $\displaystyle \vec{r}$
is the radius vector and r is its modulus, through the surface of a sphere of radius R.
My attempt:
$\displaystyle \iint_{s}\vec{F}.d\vec{S} = \iint r\vec{r}.\vec{n}ds = \iint r\vec{r}.\vec{n} \frac{da}{\vec{n}.\vec{k}}$
where
$\displaystyle \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$
Surface must be :
$\displaystyle x^{2} + y^{2} + z^{2} = R^{2}$

Im not suppose to use divergence theorem for this (as it asks to prove both ways)

Hence:
$\displaystyle \vec{N} = \nabla(g(x,y,z)) = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}$
$\displaystyle \vec{n} = \frac{\vec{N}}{|\vec{N}|}$
sub in sphereical polar coordinates etc:

If you dot it with the integration above i get:
$\displaystyle \iint r^{2} ds = \iint r^{2} \frac{dA}{\vec{n}.\vec{k}}$
This is where i am stuck, there is no reason why i picked the k direction to be dotted with n, by doing so i get $\displaystyle cos(\theta)$ but it does matter in terms of the integral, dA will have to be in terms of dr but the other can be$\displaystyle d\phi$or $\displaystyle d\theta$ both of which have different limits (2pi, 0 and pi, 0) in which case the result will change :(

Any help is appreciated :D
• Jun 13th 2011, 02:03 PM
HallsofIvy
Re: Calculating flux without using divergence theorem
What you are doing, apparently, is projecting the surface of the sphere onto the xy-plane but once you have projected to the xy-plane, there is no longer a $\displaystyle \phi$ variable. You are using polar coordinates with r from 0 to R and $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$. Be careful to note that the both the "top" and "bottom" hemispheres project to the same disk in the xy-plane.

Personally, I would use the fact that the "vector differential of surface area" for a sphere of radius R is $\displaystyle R^2(cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ sin(\phi)cos(\phi)\vec{k})d\theta d\phi$ and integrate with $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$, $\displaystyle \phi$ from 0 to $\displaystyle \pi$.
• Jun 15th 2011, 08:18 PM
ojones
Re: Calculating flux without using divergence theorem
Calculating the flux in this case is straightforward given that $\displaystyle S$ is a sphere. We have that $\displaystyle \vec{n}=\vec{r}$ and

$\displaystyle \iint_S\vec{F}\cdot d\vec{S }=\iint_Sr \vec{ r}\cdot \vec{n} dS = R\iint_S dS = R\cdot 4\pi R^2=4\pi R^3$.