# Stokes theorem and the simple closed curve on which work is maximum

I have a problem that states:

Given the vector field $$\vec{F} = y^3\hat{i} + \left(4x - 2x^3 \right)\hat{j}$$ find the simple closed curve (with $\frac{d\vec{r}}{dt}\gt0$) on which the work performed by $\vec{F}$ is maximum.

The solution is stated as the following using Stokes theorem: \begin{equation*} \oint_c \textbf{F}\centerdot d\textbf{r}= \int\int_S (\nabla \times\textbf{F})\centerdot d \textbf{S}\end{equation*}

Then in order for the work to be maximum $\nabla \times\textbf{F} \gt 0$. Calculating the above divergence gives us $4 - 6x^2 - 3y^2>0$ which leads to an inequality like $$\frac{x^2}{a^2} + \frac{y^2}{b^2} \lt 1$$

So according to this textbook every simple closed curve within the limits of the above ellipsis would give us the maximum work.

Could someone explain why this happens or if the above statement is correct?

Green's theorem for the plane states that $$\int_{\partial B}(Pdx+Qdy)=\int_B(Q_x-P_y)\>{\rm d}(x,y)\ .\tag{1}$$ In your case $$Q_x-P_y=4-6x^2-3y^2\ .$$ It follows that the right hand side of $(1)$ is maximal when $B$ is the elliptic domain in ${\mathbb R}^2$ where $4-6x^2-3y^2\geq0$, or $${x^2\over 4/6}+{y^2\over4/3}\leq1\ .$$ This optimal $B$ has boundary $$\partial B:\quad t\mapsto\left(\sqrt{2\over3}\cos t,\>{2\over\sqrt{3}}\sin t\right)\qquad(0\leq t\leq 2\pi)\ ,$$ which is the curve you are looking for.