Singularities when applying Stokes' theorem

In summary: In fact, if you know that the singularity is contained within a certain closed curve, you can just ignore it and the divergence theorem will still hold.
  • #1
Homework Statement
I am arguing with myself if the way i solved this problem is right
Relevant Equations
It is more or less a generic problem of stokes theorem:
##\int_{\gamma} F dr##, where ##F = (-y/(x²+y²) + z,x/(x²+y²),ln(2+z^10))## and gamma is the intersection of ##z=y^2, x^2 + y^2 = 9## oriented in such way that its projection in xy is traveled clockwise.

So, i decided to apply stokes theorem to the cone surface, considering its boundary to be the one with z=0 and the other being the intersection said above. Call the answer A

##A + \int_{\alpha} (-y/(x^2+y^2), x/(x^2+y^2), ln(2)) * dC = \int \int grad F * ds##

Alpha is the circle at z=0, x²+y²=9, and the right side of the equation will give zero if you calculate it. Doing all the integrals it end as:

##A + 2\pi = 0 => A = -2\pi##

In fact the answer is right. But that is not the problem, i want to know about the singularities!
I mean i just ignored it, following the reasoning that
1: the grad will be calculate on the cylinder surface, which does not intersect the origin
2: the line integral of the circle at z=0 will also not intersect the origin

So, taking in account that the integral line and integral surface "does not know" about the singularity at the origin, i just ignored it. Is it right? (If it were necessary to calculate div F over the volume, in this case, i would need to pay attention at the singularity since the volume spread over there)
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  • #2
You are right to think twice about singularities when talking about Stokes theorem. I can say that the divergence theorem has difficulties when the input vector field has singularity. The classic example is ##\frac{1}{|\vec{x}|}## at the origin. If the divergence theorem has these problems, then I'm inclined to say that Stokes theorem has them too, since they're both versions of the generalized Stokes theorem for the exterior derivative. I don't have a counterexample on hand, though.

Hopefully, someone who's studied their cohomology more recently me than me can expand on this.
  • #3
You talk about the gradient of F when you actually mean the curl of F, right?
And yes as long as the closed curve in which you calculate the line integral, and the surface through which you calculate the curl surface integral do not contain the singularity, you don't need to worry about it.
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1. What is a singularity when applying Stokes' theorem?

A singularity in the context of Stokes' theorem is a point in a vector field where the field is not well-defined or behaves in a strange manner. This can happen when there is a discontinuity or a divergence in the field at that point.

2. How does Stokes' theorem handle singularities?

Stokes' theorem is a mathematical tool used to evaluate the surface integral of a vector field over a closed surface. It does not directly address singularities, but rather provides a way to calculate the integral over a surface that may contain singularities. The theorem states that the integral over a closed surface is equal to the line integral of the vector field along the boundary of the surface.

3. Can singularities affect the validity of Stokes' theorem?

No, singularities do not affect the validity of Stokes' theorem. The theorem holds true as long as the surface and the vector field meet certain conditions, regardless of the presence of singularities.

4. Are there any limitations to using Stokes' theorem with singularities?

One limitation of using Stokes' theorem with singularities is that the surface must be smooth and well-behaved. This means that the surface cannot have any sharp edges or corners, as these would cause the integral to be undefined. Additionally, the vector field must be continuous and differentiable over the surface.

5. How can singularities be handled when applying Stokes' theorem in practice?

In practice, singularities can be handled by choosing a surface that avoids the singularity or by using a different mathematical tool, such as the divergence theorem or Green's theorem, to evaluate the integral. In some cases, the singularity may also be removable through a change of variables or by smoothing out the field near the singularity.

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