I am working on the following exercise:
Let $0 < \phi < \pi$ and consider the curve
$$C_{\phi} := \bigg\{ (2+ \cos(t), 1+ \sin(t)) \ \bigg\lvert \ \lvert t \rvert \le \phi \bigg\}. $$
Calculate the curve integral $$ \frac{1}{2} \int_{C_\phi} y \ dx - x \ dy. $$
I do not know what $y \ dx $ and $x \ dy $ should mean in this context. Could you please explain?
If you parametrize the curve with functions $x=f(t)$, $y=g(t)$ over an interval $a \leq t \leq b$, then $$ \frac{1}{2} \int_{C_\phi} (y\,dx - x\,dy) = \frac{1}{2}\int_a^b \left(g(t)f'(t) -f(t)g'(t)\right)\,dt $$ Formally, you take the functions defining the parametrization and substitute them into the integrand for $x$ and $y$. Then you take the differentials of those functions and substitute them for $dx$ and $dy$.
Your curve $C_\phi$ is already described by a parametrization $x = 2+ \cos t,y=1 + \sin t$, $-\phi \leq t \leq \phi$. Therefore $dx = - \sin t \,dt$ and $dy = \cos t\,dt$ and $$ \frac{1}{2} \int_{C_\phi} (y\,dx - x\,dy) = \frac{1}{2}\int_{-\phi}^\phi \left((1+\sin t)(-\sin t) -(2+\cos t)(\cos t)\right)\,dt $$ Can you take it from here?