Line Integral $\int_{C} \frac{x dy - y dx}{x^{2}+y^{2}}$ over circle with centre not at origin How can I calculate the Line Integral $\int_{C} \overrightarrow{B}\cdot d\overrightarrow{r}$, where $\overrightarrow{B}$ is given by:
$\overrightarrow{B}=\frac{(-y,x)}{x^2+y^2}$
and $C$ is a circle in the xy-plane, with radius $a$ and centre at $(x_0,y_o,0)$?
My friend suggested that the answer would depend on either the radius, the centre of the circle or both, but I am having a hard time solving this problem. I really need help with clearing things up here...
 A: The vector field $B$ is defined everywhere except at the origin. We can calculate its curl:
$$ \begin{aligned}
\operatorname{curl} B
  &= \frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right) - \frac{\partial}{\partial y} \left(\frac{-y}{x^2 + y^2} \right) \\
  &= \frac{2x^2 - (x^2 + y^2)}{(x^2 + y^2)^4} + \frac{2y^2 - (x^2 + y^2)}{(x^2 + y^2)^4} \\
  &= 0.
\end{aligned} $$
Therefore by Green's theorem, whenever $C$ is a circle oriented counterclockwise not enclosing the origin and $D$ is the closed disc bounded by $C$, we have
$$ \int_C B \cdot dr = \iint_D (\operatorname{curl} B)  dA = 0.$$
This equation makes sense because $D$ does not contain the origin. If we were to try to apply it in the case where $C$ encloses the origin, then $D$ contains the origin and so in the second integral we are trying to integrate over a place where the function is undefined, which doesn't make sense.
In the case that $C$ does enclose the origin, we can take a different approach. First let's try integrating $B \cdot dr$ over a circle $C_R$ of radius $R$ centred on the origin. Let $f(\theta) = (R \cos \theta, R \sin \theta)$.
$$ \begin{aligned}
\int_{C_R} B \cdot dr
  &= \int_0^{2 \pi} B(f(\theta)) \cdot f'(\theta) d \theta \\
  &= \int_0^{2 \pi} \frac{(-R \sin \theta, R \cos \theta)}{R^2} \cdot (- R \sin \theta, R \cos \theta) d \theta \\
  &= \int_0^{2 \pi} 1 \,\,d \theta\\
  &= 2 \pi.
\end{aligned}$$
Notice that the result does not depend on the radius $R$. More is true in fact: it does not even depend on the position (or shape) of the circle, only on the fact that the circle encloses the origin. This can be seen as follows: if $C$ is a closed path enclosing the origin, then we can take $R$ small enough so that $C_R$ is contained entirely inside $C$. We know the path integral over $C_R$, then applying Green's theorem as before to the region enclosed by $C$ minus the region enclosed by $C_R$ (thereby removing the problematic origin point) will give that $\int_C B \cdot dr = \int_{C_R} B \cdot dr = 2\pi$.
Therefore the answer is either $0$ or $2 \pi$, depending on whether or not the curve $C$ encloses the origin.
