Finding a value R that maximizes the flux a vector field over half a sphere of radius R

Question

Sorry for the bad title, couldn't think of a less convoluted way of writing it.

I have to find $ R\in \mathbb{R}$ so that the flux of $$F(x,y,z) = (xz - x\cos(z), -yz +y\cos(z), -4 - (x^2 + y^2)) $$

Over $S_R = \{ x^2 + y^2 + z^2 = R^2 , z \geq 0 \} $

I started doing the more obvious calculation.

$$ \nabla \cdot F = z - \cos z -z + \cos z + 0 = 0$$

Which must be wrong, since then R wouldn't matter. I started working the other way around, calculating $$\int _0 ^{2\pi} \int_0 ^{\frac{\pi}{2}} F(R\cos \theta \sin \phi, R\sin \theta \sin \phi, R\cos \phi) \cdot T_\theta \times T_\phi d\phi d\theta$$

But at some point I get a malicious $\cos (R \cos\phi)$ that doesn't cancel out like I hoped.

I'm sure there's some basic error I'm making here, but can't seem to find it. This is my first time dealing with maximization problems like this and I'm really at a loss on how to face it.

Any help would be greatly appreciated.

Answer 1

This can be done with the divergence theorem. Since $\nabla\cdot F=0$, the flux of the vector field though the closed surface given by the union of the upper hemisphere and the equator $E=\{(x,y,z):z=0,x^2+y^2\leq R^2\}$ is zero. Hence the flux of $F$ through the upper hemisphere equals the opposite of the flux of $F$ through $E$: $$\Phi(F) = \int_{x^2+y^2\leq R^2}(4+x^2+y^2)\,d\mu = 4\pi R^2+\int_{0}^{2\pi}\int_{0}^{R}\rho^3\,d\rho\,d\theta = 4\pi R^2+\frac{\pi}{2}R^4.$$ Since the RHS is an increasing function of $R$, the flux has no maximum.

Answer 2

No, your first calculation is fine. The point is that you need to close your surface for Stokes' theorem (the version with the divergence in it, i.e. Gauss' theorem) to hop in. What you've realized (or not) is that your integral over the upper side of the sphere is the same as the integral over the flat disk on the $xy$-plane. Can you continue? Here $z=0$, so your vector field ends up being $(-x,y,-4-(x^2+y^2))$.

ADD If I wasn't clear enough, let $S$ be your upper cap positively oriented, and let $D$ be the disk on the $xy$-plane, oriented negatively. Then $\partial M=S\cup D$ is a closed surface which is the boundary of the half ball call it $M$, and Stokes' theorem gives that $$\int_{\partial M}\omega =\int_M d\omega$$ where $\omega$ is (the differential $2$-form associated with) $F$. But $d\omega=0$, so we get that the integrals over the cap and the disk are equal.