$F(x,y) = (y^3-6y)i + (6x-x^3)j$
a. Using Green's Theorem, find the simple closed curve C for which the integral $ ∳F \cdot dr $ (with positive orientation) will have the largest positive value.
b. Compute this largest possible value.
I'm quite certain that this is just $ \iint Nx-My $ $dA$ but I do not know how to find the bounds in this scenario for both integrals. Though I'm also sure that this problem can also be done using just $ ∳ F \cdot dr $ as there is an equation given for F.
You are correct that if we integrate around a closed curve $C$ that bounds a region $\Omega$, then $$ \oint F \cdot dr = \iint_{\Omega} (N_x - M_y) dA. $$ Here, we have $$ (N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). \tag{1} $$ In order to maximize $\iint_{\Omega} (N_x - M_y) dA$, we want $\Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) \geq 0$ if and only if $x^2 + y^2 \leq 4$, i.e. $(N_x - M_y)(x,y) \geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.