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Math Help - 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 \iint_{s} \vec{F}.d\vec{S}
    of the vector field
    \vec{F} = r\vec{r}.
    Where \vec{r}
    is the radius vector and r is its modulus, through the surface of a sphere of radius R.
    My attempt:
    \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
    \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}
    Surface must be :
    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:
    \vec{N} = \nabla(g(x,y,z)) = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}
    \vec{n} = \frac{\vec{N}}{|\vec{N}|}
    sub in sphereical polar coordinates etc:

    If you dot it with the integration above i get:
    \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 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  d\phi or 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 \phi variable. You are using polar coordinates with r from 0 to R and \theta from 0 to 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 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 \theta from 0 to 2\pi, \phi from 0 to \pi.
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  3. #3
    Senior Member
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    Re: Calculating flux without using divergence theorem

    Calculating the flux in this case is straightforward given that S is a sphere. We have that \vec{n}=\vec{r} and

    \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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