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?
Asked
Active
Viewed 1,041 times
6
-
Field = Vector field? – Dylan Moreland Jan 08 '12 at 19:13
-
Yes a vector field. – JDD Jan 08 '12 at 19:16
1 Answers
2
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.)
joriki
- 238,052