Integrate the differential form over a cardioid $\omega=\dfrac{-ydx+(x-1)dy}{(x-1)^2+y^2}$
Calculate $\int_C\omega$ where $C...r=1+\cos\varphi$ (positively oriented)
I'm still pretty lost when it comes to differential forms but as far as I understand, the form I'm given is actually the change in angle but the origin is shifted to $(1, 0)$. Since the cardioid goes around that point, I think the result should be $2\pi$, right?
If it's true, how can I make it formal?
 A: 
Let $C$ be the positively oriented cardioid given by the polar equation,
  $$r(\theta)=1+\cos{\theta},~~~-\pi\le\theta\le\pi.$$
  Define the one-form $\omega$ as,
  $$\omega:=\dfrac{-y\,\mathrm{d}x+(x-1)\,\mathrm{d}y}{(x-1)^2+y^2}.$$
  Calculate the value of the line integral $\oint_{C}\omega$.

To solve this integral we will use the "shifted" polar coordinates $(\rho,\phi)$, which can be converted into the Cartesian coordinates $(x,y)$ by using the equations:
$$\begin{cases}
\begin{align}
x-1&=\rho\cos{\phi},\\
y&=\rho\sin{\phi}.
\end{align}
\end{cases}$$
For $r\ge0$ and $\phi\in(-\pi,\pi]$, the above pair of equations can be solved for $\rho$ and $\phi$ in terms of $(x,y)$ using the atan2 function:
$$\begin{cases}
\rho=\sqrt{(x-1)^2+y^2},\\
\phi=\operatorname{atan2}{(y,x-1)}=2\arctan{\frac{y}{(x-1)+\sqrt{(x-1)^2+y^2}}}.
\end{cases}$$
Computing the partial derivatives of the last equation we find that:
$$\begin{cases}
\partial_x\phi=\frac{-y}{(x-1)^2+y^2},\\
\partial_y\phi=\frac{x-1}{(x-1)^2+y^2},
\end{cases}$$
and, hence, that the total differential of the angle function $\phi(x,y)$ simply $\omega$:
$$\mathrm{d}\phi=\partial_x\phi\,\mathrm{d}x+\partial_y\phi\,\mathrm{d}y=\frac{-y\,\mathrm{d}x}{(x-1)^2+y^2}+\frac{(x-1)\,\mathrm{d}y}{(x-1)^2+y^2}=\omega.$$
Thus, the line integral we seek is simply:
$$\oint_{C}\omega=\int_{-\pi}^{\pi}\mathrm{d}\phi=\pi-(-\pi)=2\pi$$
