- #1
Calculating surface integral using diverg. thm.
- Thread starter Miike012
- Start date
In summary, the conversation discusses a problem with a singularity on the axis and the use of Gauß's Theorem to solve it. The book being referenced may contain an incorrect explanation, but the individual has found a solution by changing the expression to k1/2 instead of [1/r]d(k1)/dr. However, they still do not fully understand why this change works.
- #2
- #3
- 42,139
- 10,204
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
- 16,131
- 4,885
I will check the book thingy, but checking your result is easier: what if k2 = 0 and k1 is not ?
- #5
- 24,488
- 15,033
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
- 42,139
- 10,204
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:
FAQ: Calculating surface integral using diverg. thm.
What is the Divergence Theorem?
The Divergence Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the volume integral of the divergence of that vector field over the enclosed volume.
How is the Divergence Theorem used to calculate surface integrals?
The Divergence Theorem allows us to convert a surface integral into a volume integral, making it much easier to calculate and evaluate. This is because the volume integral over the divergence of a vector field is typically simpler to calculate than a surface integral.
What is the significance of using the Divergence Theorem for surface integrals?
The Divergence Theorem is significant because it provides a powerful tool for solving a wide range of problems involving surface integrals. It allows us to convert a difficult surface integral into a simpler volume integral, making calculations easier and more efficient.
Are there any restrictions on when the Divergence Theorem can be used for calculating surface integrals?
Yes, the Divergence Theorem can only be used for closed surfaces, meaning that the surface must completely enclose a finite volume. Additionally, the vector field must be continuous and differentiable within this volume.
Can the Divergence Theorem be used for any type of vector field?
No, the Divergence Theorem is only applicable to conservative vector fields, meaning that the curl of the vector field must be equal to zero. Non-conservative vector fields may require different methods for calculating surface integrals.