Your application of Green’s Theorem is justified. You can think of $r$ and $\theta$ as the labels of axes in a different Cartesian plane. You have to be a little careful about $\mathcal C$ and $\mathcal R$ with this point of view, though—they need to be replaced by their preimages under the polar-to-Cartesian map.
As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:
$$\oint_{\partial D} (P\, dx+Q\, dy) = \iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$
Now in our case, $$\frac{\partial Q}{\partial x}=2,\frac{\partial P}{\partial y}=3$$
Hence, We get, by the formula:
$$\iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)=\iint_D 2-3dx\,dy=-\iint_D 1dx\,dy$$
This is simply $-1$ times the area of the enlosed region, which we know is simply $\sqrt{2}\pi$. So:
$$\oint_{\partial D} (P\, dx+Q\, dy) =-\sqrt{2}\pi$$
Regarding the parametrization, it should be $x=\sqrt{2}\cos(t)$ and $y=\sin(t)$ and it should give you the same answer.
Best Answer
HINT:
$$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=1 \tag 1$$
Now, what is the area enclosed by $C$?
SPOILER ALERT: Scroll over the highlighted area to reveal the solution