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?

  • $\begingroup$ Field = Vector field? $\endgroup$ Jan 8, 2012 at 19:13
  • $\begingroup$ Yes a vector field. $\endgroup$
    – JDD
    Jan 8, 2012 at 19:16

1 Answer 1


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

$$ \def\part#1{\frac{\partial}{\partial #1}} \part x\frac{9x}r+\part y\frac{9y}r=\frac9r\;. $$

By Green's theorem, the integral of that curl over the interior of the circle is equal to the line integral along the circle, and indeed

$$\int_0^r\mathrm dr' r'\int_0^{2\pi}\mathrm d\phi\frac9{r'}=18\pi r\;.$$

(This is the solenoidal component of the field; the field may also have an irrotational component that doesn't contribute to the line integral.)


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