Find a simple, closed and piecewise-smooth curve $\gamma\in\Pi$ such that $\oint_{\gamma}\vec{F}\cdot d\vec{r}$ gets its maximum value

Question

Recently, I stumbled upon this question, which seemed very interesting to me. I must say that this is not the first time I encounter this specific kind of questions - but I can ever handle them. I think I've done a big part of the solution - but I don't know how to continue. Secondly, I have made an assumption during my solution, which might not be true - so I would be glad to hear, in addition, whether it should be done, or not, and of course how to prove it (if so).

The task: Given the vector field:

$$\vec{F}(x,y,z)=(xy^2,3z-xy^2,4y-x^2y)$$

We are interested in finding a parametrization of a simple, closed and piecewise-smooth curve $\gamma$, such that $\gamma$ is lying on the plane $\Pi:x+y+z=1$, and:

$$W(\gamma)\equiv\oint_{\gamma}\vec{F}\cdot d\vec{r}$$

gets its maximum value.

My partial solution: Since $\vec{F}\in C^1$ and $\gamma$ is simple, closed, and piecewise-smooth, we can use the Kelvin-Stokes Theorem, which states that:

$$\oint_{\partial\Sigma}\vec{F}\cdot d\vec{r}=\iint_{\Sigma}\left(\vec{\nabla}\times\vec{F}\right)\cdot d\vec{s}$$

As long as $\gamma\equiv\partial\Sigma$ is positively oriented.

Thus, after computing the curl of $\vec{F}$, we can say that we are looking for a surface $\Sigma$ such that:

$$I\equiv\iint_{\Sigma}(1-x^2,2xy,-y^2-2xy)\cdot d\vec{s}$$

gets its maximum value.

Assumption (which might not be correct, and need a solid proof): Assume the surface $\Sigma$ lies on the plane $\Pi$; Thus given a general parametrization of $\Sigma$, we can say that:

$$\Sigma(u,v)=(x(u,v),y(u,v),1-x(u,v)-y(u,v))\implies d\vec{s}=(\frac{\partial y}{\partial v}\frac{\partial x}{\partial u}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v})\cdot(1,1,1)$$

Now we can calculate $\left(\vec{\nabla}\times\vec{F}\right)\cdot \hat{n}ds$ and receive:

$$\left(\vec{\nabla}\times\vec{F}\right)\cdot \widehat{(1,1,1)}ds=\frac{1}{\sqrt{3}}(1-x^2-y^2)ds$$

Which seems much more easy to work with - but I still couldn't figure out how.

I would be very glad for your help. Thank you very much!

Answer 1

This is very close. You mean $(\nabla\times\vec F)\cdot d\vec s$, right? Also, remember you need to dot with the unit normal, which scales this by $1/\sqrt3$. Now finish: You want to choose $\Sigma$ (lying in the plane, of course) so as to maximize $\displaystyle{\frac1{\sqrt3}\iint_\Sigma (1-x^2-y^2)\,dA}$. Shouldn't $\Sigma$ thus be the portion of the plane lying inside the cylinder $x^2+y^2=1$?

EDIT: Note that choosing any other surface $\Sigma'$ with the same boundary curve gives the same flux of the curl. So it is no limitation to look for regions in the plane $x+y+z=1$ to start with.