My concern regards solving a class of somewhat ill-defined surface integrals occurring in Mathematical Physics and EMF Theory. I'll be using a simplified, representative example.(adsbygoogle = window.adsbygoogle || []).push({});

Consider the surface S given by

(x/2)^2 + (y/2)^2 + (z/3)^2 = 1

0 <= z.

And the integral ∫F⋅dswhere we have in cylindrical coordinatesF=e_{ρ}/ρ.

That is to say, we are integrating over half an ellipsoid, with the boundary of a radius 2 circle at z = 0 and the height of z = 3. Furthermore we have a singularity at z = 3 -> ρ = 0. The field amounts to, short of a constant, the electric field of an infinitely long, thin wire at the z-axis.

There are various ways of evaluating this, for example by arguments invoking Gauss' law, or through a painstaking direct evaluation using iterated integrals, with a limit around the singularity. However, this is one of those idealized cases where the electric field has no divergence, and so we have a vector potential;F= ∇ ×A.

It is therefore appealing to invoke Stokes' Theorem. However, we clearly cannot use it directly - the singularity means that the surface is not simply connected. But going back to the idea of examining limiting behaviour around the singularity, we can make the following argument that gives us the correct result: Consider the boundary δS_{1}at z = 0, and a small closed loop δS_{2}close to z = 3. In the limit of a small δS_{2}, by taking the difference of the line integrals ofAover the two loops with the appropriate orientations, we should arrive at the original, desired integral over S. We can now observe that the integral over S would, by similar boundaries, be the same as an integral over a cylinder of radius 2 and height 3 with a closed top with a small hole in it. But we have no flux in the z-direction, and we can thus simply open the top, and do our surface integral over a cylinder with open tops and bottoms. This takes us back to the results known from EMFT, and the trivial computation gives us, in this case, the answer of +-6π depending on orientation.

I understand that, despite the apparent problems, this argument is not entirely invalid. My question is, under what parameters is it valid? It appears to me in this case that the argument works because we don't have any flux in the z-direction, so we don't have to consider the line integral in the neighbourhood of the singularity. Are there other pitfalls? Geometrically, it appears to me that you can use the line integral around a hole to "connect" a region even if the integrand cannot be defined in the entire region to be excised, but I'm not entirely sure this works mathematically.

I hope the above is clear enough!

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# Kelvin-Stokes' theorem in the presence of singularities

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