Here's a question I've been stuck on:
Find a piecewise smooth, simple, closed, oriented curve $C$ which maximizes
$$
\int\limits_{C} \vec{F} \,\mathrm{d}\vec{x},\quad \vec{F}(x,y,z)=\big(-y(z+1),x(z+1),0\big)
$$ among all curves $C$ restricted to lie on the three-dimensional unit sphere.
I began with computing
$$\nabla \times F=\big(-x,-y,2(z+1)\big)$$ which doesn't really simplify the question a lot. Next, I restricted myself to deal with situations where $z$ is held constant. In this situation, by the Stokes' Theorem, and by using Polar Coordinates, I can write out, $$\int\limits_{C} \vec{F}\, \mathrm{d}\vec{x}=\iint\limits_S \big((\nabla \times \vec{F})\cdot \vec{n}\big) r\,\mathrm{d}r\,\mathrm{d}\theta \quad\vec{n}=(0,0,1), r\in(0,a],\theta\in(0,2\pi), z^2=1-a^2 $$ Upon solving this, I got that the integrand is maximized when, $$a=1 \implies z=0\implies x^2+y^2=1$$
I'm trying to generalize this approach to any curve which lies on the unit sphere. However, the primary issue I'm encountering is a lack of a neat expression for the unit normal (for any arbitrary surface satisfying the constraints) which appears in the Stokes' Theorem. I'm not really sure how to proceed any further.
Update: 23rd August 2020
So, based on some comments and help from other forums, I got the following idea. The integrand, after the application of Stokes' Theorem, reduces to:
$$\iint\limits_S 3z^2+2z-1 dA$$
If you consider the function in the integral on the domain $$z\in[-1,1]$$, you'll see that the function is non-negative when $$\frac{1}{3}\leq z<1$$, and non-positive otherwise. Therefore, in order to maximize the integral, we need to consider the surface on the sphere enclosed between the planes $$z=\frac{1}{3}, z=1$$
I used the following parameterization:
$$(x,y,z)=(\cos(\theta)\sin(\phi), \sin(\theta)\sin(\phi), \cos(\phi)), 0\leq \theta \leq 2\pi, 0 \leq \phi \leq \frac{\pi}{2}-\arctan{\frac{1}{2\sqrt{2}}}$$
Then,
$$\iint\limits_S 3z^2+2z-1 dA=\iint\limits_S (3\cos^2(\phi)+2\cos(\phi)-1)\sin(\phi)d\phi d\theta = \frac{64\pi}{27}$$
I believe this is the maximum…
Best Answer
This is an excellent solution. I have one sneaky improvement on your final calculation. It is "well known" that radial projection from the unit sphere to the circumscribing right circular cylinder (of radius $1$, as well) is area-preserving (away from the north and south poles), so let's just do the integral on the corresponding piece of the surface of the cylinder. Then you get $$2\pi\int_{1/3}^1 (3z^2+2z-1)\,dz = 2\pi\cdot\frac{32}{27},$$ confirming your result.