For any $R>0$ let the surface $S={(x,y,z):x^2+y^2+(z-1)^2 = R^2; z≥1}$ oriented with the normal whose third coordinate is positive. And let the field $F(x,y,z) = (zx-x cos (z), -zy+ y cos(z), 4-x^2-y^2)$ Determine $R$ so that the flux of the $F$ field through $S$ is maximum.
It occurs to me to use the Stokes theorem since S is an orientable surface in $ℝ^3$ and $F$ is a continuous vector field in $ℝ^3$. Then Stokes' theorem ensures that $Rot F(S) = \frac{k Rot F(S)}{||Rot F(s)||}$ where $k$ is a constant. But I do not know how to continue with this, how to develop or if it really serves this approach with which I started. So from now I appreciate the help with this exercise .
I don't think Stokes' theorem is of much use here. I would instead rely on the divergence theorem.
Consider the surface $S'$ which is the union of $S$ and the disk of radius $R$ centered at $(0,0,1)$ on the plane $z=1$. Then by the divergence theorem, the flux of $\mathbf F$ over this closed surface is $$\iint_{S'}\mathbf F\cdot\mathbf n\,\mathrm dS=\iiint_\Sigma\nabla\cdot\mathbf F\,\mathrm dV=0$$ where $\Sigma$ is the region with $S'$ as its boundary, so that the flux of $\mathbf F$ over $S$ is the negative of the flux of $\mathbf F$ over the disk: $$\iint_S\mathbf F\cdot\mathbf n\,\mathrm dS=-\iint_{S'\setminus S}\mathbf F\cdot n\,\mathrm dS$$ The latter integral is easier to compute. Parameterize $S'\setminus S$ by $$\mathbf s(u,v)=(u\cos v,u\sin v,1)$$ where $0\le u\le R$ and $0\le v\le2\pi$. Then the integral becomes $$-\int_{u=0}^{u=R}\int_{v=0}^{v=2\pi}(u\cos v(1-\cos1),u\sin v(\cos1-1),4-u^2)\cdot(0,0,-u)\,\mathrm dv\,\mathrm du$$ where $\mathbf s_v\times\mathbf s_u=(0,0,-u)$ is the vector normal to the disk pointing in the negative $z$-direction (the "outward" direction from the total surface $S'$). $$=-2\pi\int_{u=0}^{u=R}(u^3-4u)\,\mathrm du=\dfrac{\pi R^2(8-R^2)}2$$ You'll find that the maximum flux occurs for $R=2$.