I have to evaluate a line integral using Green's theorem:
$$\overrightarrow{F}(x,y)=\langle y-\cos{y}, x\sin{y} \rangle$$
The curve $C$ is the circle $(x-3)^2+(y+4)^2=4$, oriented clockwise.
It would be much easier to solve the double integral using polar coordinates, except the circle has been translated and writing the translation in polar coordinates makes the function "ugly". At which point you may as well just evaluate it using Cartesian coordinates.
My idea was to move the circle and the vector field together such that the circle will be centered at the origin and the vector field follows. What I get is:
$$\overrightarrow{F}(x,y)=\langle (y-4)-\cos{(y-4)}, (x+3)\sin(y-4) \rangle$$
and $C$ becomes $x^2+y^2=4$. Now you can convert to polar coordinates, with $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$.
Is this valid? Or is there some way to simplify the coordinates that I am missing?
Your approach is correct, but here we don't really need it.
Indeed, by Green's theorem, the line integral is equal to $$\begin{align}-\iint_{D}({F_2}_x-{F_1}_y)\,dx dy&=-\iint_{D}(\sin(y)-(1+\sin(y))\,dx dy\\ &=\iint_{D}1\,dx dy=|D|=4\pi \end{align}$$ where $D$ is the disc $\{(x,y):(x-3)^2+(y+4)^2\leq 4\}$ and $|D|=\pi \cdot 2^2=4\pi$ is its area (which is invariant under translations and rotations).