Problem
Determine all circles $\mathcal C$ on $\mathbb R^2$ such that $$\int_{\mathcal C}-y^2dx+3xdy=6\pi$$
My attempt at a solution
If I call $P(x,y)=-y^2$ and $Q(x,y)=3x$, then I can apply Green's theorem to the region $D$ enclosed by the circle of radius $R$ centered at $(a,b)$. I'll use a change of variables to describe $D$ with the transformation $T(\theta,r)=(r\cos(\theta)+a,r\sin(\theta)+b)$. We have the identity $$\int_{\mathcal C} Pdx+Qdy=\iint_D (\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dxdy$$
The double integral equals to $$\int_{0}^{2\pi}\int_0^R((3+2(r\sin(\theta)+b)r)drd\theta$$$$=\int_0^{2\pi}((\dfrac{3+2b}{2})R^2+\dfrac{2R^3}{3}\sin(\theta))d\theta$$$$=(3+2b)R^2\pi$$
So, by the chain of equalities, we have $$\int_{\mathcal C}-y^2dx+3xdy=6\pi$$ if and only if $$=(3+2b)R^2\pi=6\pi$$
Notice that the condition isn't satisfied for any circle of arbitrary radius $R$ centered at $(a,b)$ with $b \leq \dfrac{-3}{2}$.
Now, if the circle is centered at an arbitrary point $(a,b)$ with $b>\frac{-3}{2}$ then the radius must satisfy $R=\sqrt{\dfrac{6}{3+2b}}$.
I would appreciate if someone could take a look at my solution and let me know if there are any mistakes. Anybody who wants to add an alternative answer or comments that complement the computational aspect of the solution is welcome to share it.