Miike012
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Calculating surface integral using diverg. thm.
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Miike012
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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.
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I will check the book thingy, but checking your result is easier: what if k2 = 0 and k1 is not ?
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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.
[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.
Miike012
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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.
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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.Miike012 said: