# Question about singularities and path integrals

I'm given a vector field that has an obvious singularity at a point $(a,b)$. In order to learn more about the singularity I place a circle around it with the singularity at it's center. The line integral for the field across the circle gives me $18\pi \,r$. What conclusion can I get from the solution to the line integral?

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Field = Vector field? – Dylan Moreland Jan 8 '12 at 19:13
Yes a vector field. – JDD Jan 8 '12 at 19:16

Do you mean "along the circle"? If so, the line integral tells you that the singularity is of the form $(-9y/r,9x/r$). The two-dimensional "curl" of that vector field is
$$\def\part#1{\frac{\partial}{\partial #1}} \part x\frac{9x}r+\part y\frac{9y}r=\frac9r\;.$$
$$\int_0^r\mathrm dr' r'\int_0^{2\pi}\mathrm d\phi\frac9{r'}=18\pi r\;.$$