find the simple closed curve of $F(x,y) = (y^3-6y)i + (6x-x^3)j$ using Green's Theorem which will have the largest positive value

$$F(x,y) = (y^3-6y)i + (6x-x^3)j$$

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.
• No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $\pm \sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates. – Jordan Green Jan 31 at 3:10
• You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 \pi$ (or, alternatively, any real numbers with a difference of $2 \pi$). – Jordan Green Jan 31 at 3:27