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Thread: Calculating flux without using divergence theorem

  1. #1
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    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.
    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
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  2. #2
    MHF Contributor

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    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 $\displaystyle \pi$.
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  3. #3
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    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

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