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Show Divergence theorem works

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Show divergence theorem works
    For the vector field [tex] E = \hat{r}10e^{-r}-\hat{z}3z[/tex]

    2. Relevant equations

    [tex]\int_{v}\nabla \cdot E dv = \oint_{s} E \cdot ds [/tex]


    3. The attempt at a solution

    [tex] \nabla \cdot E = 1/r \frac{d}{dr}(rAr)+1/r\frac{dA\phi}{d\phi}+\frac{dAz}{dz} [/tex]
    Ar=10e^(-r)
    Aphi=0
    Az=-3z

    [tex] \nabla \cdot E = \frac{1}{r}(10e^{-r}-10re^{-r})+3 [/tex]

    [tex] \int_{0}^{2\pi}\int_{0}^{4}\int_{0}^{2} (r)(\frac{1}{r}(10e^{-r}-10re^{-r})+3) dr dz d\phi = -82.77 [/tex]

    [tex] \oint_{s} E \cdot ds= \int_{0}^{2} \int_{0}^{4} (\hat{r}10e^{-r}-\hat{z}3z)\cdot(16 \pi \hat{r}+4\pi\hat{z}) = 2341.7 [/tex]

    wow that took awhile to type
     
  2. jcsd
  3. Oct 24, 2007 #2
    The surface form you chose to integrate over wasn't my first choice to use. The volume integral looks correct (barring calculator errors), but could you explain your thought process for the surface geometry?
     
  4. Oct 24, 2007 #3

    HallsofIvy

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    You haven't given the full problem. What are s and v? What is the surface to be integrated over and its boundary?

     
  5. Oct 24, 2007 #4
    oh shoot i forgot to say it is bounded by the cylinder z=0 z=4 r=2

    my thought process from the surface integral was
    i found ds in r direction or the surface area normal to r 16 pi
    and surface area normal to z and doted it with E
     
  6. Oct 24, 2007 #5

    Dick

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    The surface integral has two parts. There's the cylinder surface and then the two disks at each end. Looking at the 'normal' part of your integral, I think you know this, but you can't combine them both into a single integral. Split them up. It also looks like you are including some sort of area factor into the normal and then also integrating.
     
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