Interesting Line Integral

Let $C$ be a positively oriented piecewise smooth simple closed curve enclosing the origin, and choose a sufficiently small radius $a$ so that the circle $C’$ given by $x^2 + y^2 = a^2$ is contained in $C$ . Parametrize $C’$ by $\vec{r}(t) = \langle a \cos t, a \sin t \rangle$ , $0 \leq t \leq 2\pi$ . Let $C’' = C \cup (-C’)$ and let $D$ be the region between $C$ and $C’$ . By Green’s Theorem,

$\displaystyle{\oint_{C’'} \vec{F} \cdot \, \vec{dr}} = \displaystyle{\oint_{C’'} P(x,y)dx + Q(x,y)dy} = \displaystyle{\iint_D \left( \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right)\,dA}$

where $P(x,y) = \dfrac{-y}{x^2+y^2} \;\; \mbox{and} \;\; Q(x,y) = \dfrac{x}{x^2+y^2}$ Since $\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = 0$ ,

$\displaystyle{\oint_{C’'} Pdx + Qdy} = \displaystyle{\oint_{C \cup (-C’)} Pdx + Qdy} = \displaystyle{\oint_C Pdx + Qdy} - \displaystyle{\oint_{C’} Pdx + Qdy} = 0$ so that $\displaystyle{\oint_C Pdx + Qdy} = \displaystyle{\oint_{C’} Pdx + Qdy}$

$\displaystyle{\oint_{C’} Pdx + Qdy} = \displaystyle{\int_0^{2\pi} \dfrac{-a \sin t}{a^2 \cos^2(t) + a^2 \sin^2(t)}(-a \sin t) dt + \dfrac{a \cos t}{a^2 \cos^2(t) + a^2 \sin^2(t)}(a \cos t)\, dt} = \displaystyle{\int_0^{2\pi} \,dt = 2\pi}$

I used Amperes circuital Law. Notice , this field is anagalous to magnetic field due to an infinite wire passing through origin along z axis . Comparing with the magnetic field and applying amperes law directly gives the answer.

Consider a smooth simple closed curve $C$ in $\mathbb{R}^2$ that does not run through the origin. Let's parameterize this curve, $\vec{s}(t)$ , with $0\leq t \leq 1$ . We can write this parameterization in terms of polar coordinates, $x=r(t)\cos(\theta(t)), y=r(t)\sin(\theta(t))$ , where the phase angle $\theta$ is measured continuously along the curve, of course. A straightforward if slightly tedious computation shows that $\int_{C}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\theta(1)-\theta(0)$ , the change of $\theta$ along the way. Since $\vec{s}(1)=\vec{s}(0)$ for a closed curve, these two polar angles will differ by a multiple of $2\pi$ ,showing that $\oint_C \vec{F}\cdot d\vec{s}$ is an integer multiple of $2\pi$ . If the curve is simple and running in the counterclockwise direction around the origin, then the line integral must be $\boxed{2\pi}$ .