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Calculating surface integral using diverg. thm.

  1. Jan 24, 2014 #1
    I believe the book is wrong.. Can some one please check my work.
    PROBLEM IS IN THE 2nd POST ( SORRY I COULDNT ADD BOTH PICS FOR SOME REASON )
     

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  3. Jan 24, 2014 #2
    Problem:
     

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  4. Jan 25, 2014 #3

    haruspex

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    Not an area I know well, but I think your problem is that 1/r gives you a singularity on the axis, so your divergence integral does not vanish there.
     
  5. Jan 25, 2014 #4

    BvU

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    I will check the book thingy, but checking your result is easier: what if k2 = 0 and k1 is not ?
     
  6. Jan 25, 2014 #5

    vanhees71

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    If you could use Gauß's Theorem it would read
    [tex]\int_{V} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{F}=\int_{\partial V} \mathrm{d}^2 \vec{A} \cdot \vec{F}.[/tex]
    However, as already stated by haruspex, there is the problem along the [itex]z[/itex] axis, where the field is singular, and thus there the divergence of the field is not well defined. So you have to do the surface integral directly. I have not checked in detail, whether the book is correct, but the explanation looks correct.
     
  7. Jan 25, 2014 #6
    Instead of making [1/r]d(k1)/dr = 0, I change it to k1/2 where r = 2. Then I get the correct answer.. still don't understand why though.
     
  8. Jan 25, 2014 #7

    haruspex

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    It's because ##\frac 1r \frac \partial {\partial r} r F(r)## is only valid where F(r) is defined, and F(r)=k/r is not defined at r = 0.
     
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