The task is to find the maximum of $$\int_{\gamma} (x^2y-2y)dx + (2x-xy^2)dy$$ when $\gamma$ is a smooth, regular, closed and simple curve in $\mathbb{R}^2$, and $\gamma = \partial G$ for a bounded domain $G$. Also $\gamma$ is with positive orientation.
Moreover, I need to find the curve $\gamma$ that gives the maximum.
It's seems obvious to me to use Green's theorem here. I denote $$\omega = (x^2y-2y)dx + (2x-xy^2)dy$$ and then: $$\int_{\gamma} \omega = \int_G (2-y^2-x^2+2)dxdy = \int_G (4-(x^2+y^2))dxdy$$
Now, since $x^2+y^2>0$, the it seems to me that the maximum is $4*vol(G)$, but I am not entirely sure, since it depends on $G$.
Moreover, I was not able to find a curve $\gamma$ that gives me this result.
Help would be appreciated.
You are on the right track. Now note that $(4-(x^2+y^2))\geq 0$ iff $(x,y)\in B(0,2)$ i.e. the disc centered at the origin of radius $2$. Hence, for any bounded domain $G$ with $\gamma=\partial G$,
$$\begin{align}\int_{B(0,2)} (4-(x^2+y^2))dxdy&\geq \int_{B(0,2)\cap G} (4-(x^2+y^2))dxdy\\&\geq \int_G (4-(x^2+y^2))dxdy=\int_{\gamma} \omega.\end{align}$$ Thus the maximum value is attained when $\gamma=\partial B(0,2)$: $$\int_{B(0,2)} (4-(x^2+y^2))dxdy=2\pi\int_0^2(4-r^2)r dr=8\pi.$$