Here is a diagram depicting the question:
$C_2$ is I think parameterized by $r(t)=\cos t\vec i + \sin t\vec j$, where $3\pi /4 \leq t \leq 7\pi /4 $
The vector field in question is $\vec G=3y\vec i-3x\vec j$, and I want to find $\int_{C_2} \vec G \cdot d\vec r $.
$\vec G \cdot r(t) = 3\cos t\sin t - 3\cos t\sin t =0$, so we have:
$$\int_{C_2} \vec G \cdot d\vec r = \int_{C_2} 0 = 0 $$
But apparently the line integral evaluates to $-6\pi$. Where did I go wrong? Sorry if this is a stupid question, I'm learning this for the first time and without supervision
You need to evaluate $\vec G \cdot d\vec r(t)$, not $\vec G \cdot \vec r(t)$.
$$\vec G \cdot d\vec r(t) =3(\sin t i - \cos t j)\cdot(-\sin t i + \cos t j) d\theta$$
$$=-3(\sin^2t+\cos^2 t)d\theta = -3d\theta \ne0$$