# 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

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!

• Without specifying the orientation of the curve, the problem isn't well-defined. If we take curves which are traversed counterclockwise when viewed from the origin, the maximum is infinity. Jul 8, 2019 at 11:27

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.
• Yeah - that's what I meant, of course. And about what you've said - I haven't thought about that, that's smart! However, I still don't understand why $\Sigma$ must lie in the plane. I mean - I can find a surface which $\gamma\in\Pi$ would be its boundary, yet the surface won't be planar Jun 15, 2019 at 21:49
• Maybe it has something to do with the fact that $\vec{\nabla}\times\vec{F}$ is not dependent on $z$? Jun 15, 2019 at 22:04
• You're right. I was assuming the region should be contained in the plane. Our formula for the unit normal, of course, presumes that. But the point is that Stokes's Theorem gives the same result for any surface $\Sigma$ having the given boundary curve $\gamma$. So allowing the surface to be more complicated gains you absolutely nothing. Jun 15, 2019 at 22:32
• Thank you. However, I don't understand why do I need to normalize the vector. I learned that: $\iint_S\vec{F}\cdot d\vec{s}=\iint_S\vec{F}\cdot\hat{n}\ ds=\iint_D(\vec{F}\cdot\hat{n}\cdot|n|)\ dA$, when the LHS is vector surface integral, the MHS is scalar surface integral, and the RHS is double integral. Jun 21, 2019 at 9:25