Let us live in $\mathbb{R^2}$, $\vec{e_1} = \vec{(1,0)}, \vec{e_2} = \vec{(0,1)}$ are two standard vectors.
Evaluate $\Large{\int_{C_1} \frac{y\vec{e_1}-x\vec{e_2}}{x^2+y^2} d\vec{r}}$ where $C_1$ is a circle of radius $3$ centered at $(0,0)$
I am completely stuck.
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In polar coordinates $(r,\theta)$, we have
$$\begin{align} \color{blue}{\left(\frac{y\vec i -x\vec j}{x^2+y^2}\right)}\cdot \color{red}{d\vec r}&=\color{blue}{\left(\frac{-r\,\hat \theta}{r^2}\right)}\cdot \color{red}{\hat \theta \,r\,d\theta}\\\\ &=-\,d\theta \end{align}$$
Therefore, we can write
$$\begin{align} \oint_{\sqrt{x^2+y^2}=3}\frac{y\vec i -x\vec j}{x^2+y^2}\cdot d\vec r&=\int_0^{2\pi}(-1)\,d\theta\\\\ &=-2\pi \end{align}$$
Just straight parametrization is not that difficult. Assume positive orientation. Parametrize as follows,
$$\vec r(t)=\langle 3 \cos (t), 3\sin (t) \rangle$$
Or if you prefer $x=3\cos (t)$, and $y=3 \sin (t)$.
Then,
$$\vec dr=\langle -3 \sin (t),3\cos (t) \rangle dt$$
We have $x^2+y^2=9$ on our curve so we just need to evaluate,
$$\int_{0}^{2\pi} \frac{1}{9} \langle 3\sin (t), -3\cos (t) \rangle \cdot \langle -3 \sin (t), 3 \cos t \rangle dt$$
$$=\int_{0}^{2\pi} -1 dt$$
$$=-2\pi$$
It is possible to prove the result holds for any closed, simple, curve $C$ enclosing the origin by creating a slit that avoids the origin and connects to a circle centered at the origin which is of radius $a$, then by using Greens Theorem.