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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?

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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.

(Note that this problem has nothing to do with "divergence".)

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  • $\begingroup$ Thank you. This makes much more sense. $\endgroup$ – Leonidas Tsampros Jan 27 '15 at 11:25
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It is not true that any closed curve within the ellipse has maximal work. What you do know if that the value of the curl is positive within the ellipse and negative outside. A curve with maximum work will be a double integral over an area with positive curl. So the maximal curve is the ellipse.

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