- #1
Calculating surface integral using diverg. thm.
- Thread starter Miike012
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Homework Help Overview
The discussion revolves around calculating a surface integral using the divergence theorem, with a focus on potential singularities in the field being analyzed. Participants are examining the implications of these singularities on the validity of the divergence integral.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants are questioning the assumptions related to singularities in the field, particularly at the axis where the divergence integral may not vanish. Some are exploring alternative approaches to handle the singularity, while others are checking the correctness of the book's explanation.
Discussion Status
The discussion is ongoing, with participants providing insights into the nature of the singularity and its impact on the divergence theorem. Some guidance has been offered regarding the need to evaluate the surface integral directly due to the singularity, but there is no explicit consensus on the correctness of the book or the approaches being discussed.
Contextual Notes
There are constraints related to the singularity at the axis, which complicates the application of the divergence theorem. Participants are also considering specific cases, such as when certain parameters are set to zero, to explore the behavior of the integral.
- #2
- #3
- 42,890
- 10,523
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.
- #4
BvU
Science Advisor
Homework Helper
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I will check the book thingy, but checking your result is easier: what if k2 = 0 and k1 is not ?
- #5
- 24,488
- 15,059
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.
- #6
Miike012
- 1,009
- 0
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.
- #7
- 42,890
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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:
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