Orientation, cavities and singular-field synthesis

Learning goals and theorem checklist

Distinguish an open surface from a closed boundary, account for every cavity, and diagnose singularities that invalidate an otherwise tempting theorem. Before applying Green, Stokes or divergence, identify the region, its orientation, the required derivative class on a neighborhood, and every excluded point. A small singular set is not automatically harmless.

For a solid with a cavity, “outward” on the inner wall points into that cavity, away from the material. For Stokes, the boundary orientation must be induced by the chosen surface normal; reversing either side alone changes the sign. For a closed surface there is no boundary curve, while an open surface generally has one. Closing an open surface adds a new flux term that must be subtracted afterwards.

Worked example: an open hemisphere

Let $S$ be the upper unit hemisphere, normal outward from the upper half-ball, and $F=(x,y,z+1)$. Add the base disk $D$ in $z=0$ with downward normal. Divergence is three and the half-ball volume is $2\pi/3$, so the closed flux is $2\pi$. On $D$, $F\cdot(0,0,-1)=-1$, giving disk flux $-\pi$. Consequently the requested hemisphere flux is $2\pi-(-\pi)=3\pi$. The constant vertical component does not change divergence but does change each open-surface flux.

Worked example: a radial singularity and a shell

Let $F(x)=x/\|x\|^3$ on $\mathbb R^3\setminus\{0\}$. For $r=\|x\|>0$,

$$\operatorname{div}F=\sum_{j=1}^3(r^{-3}-3x_j^2r^{-5})=0.$$

On the sphere $r=R$ with $R>0$, $F=R^{-2}n$, so its outward flux is $R^{-2}(4\pi R^2)=4\pi$. One cannot apply divergence to the whole ball, since $F$ is undefined at its center. For the shell $\varepsilon\le r\le R$ with $0<\varepsilon<R$, it is $C^1$ on a neighborhood of the closed shell. The inner normal is $-n$, producing inner flux $-4\pi$. Total flux is zero, agreeing with the zero volume divergence. Removing the singularity changes the boundary and restores the theorem exactly.

Worked example: replace a difficult Stokes surface

Let $C$ be the intersection of $x^2+y^2=1$ with $z=2+x$, parametrized by $r(t)=(\cos t,\sin t,2+\cos t)$, $0\le t\le2\pi$. For $F=(-y,x,0)$, choose the planar disk in $z=2+x$, parametrized $R(u,v)=(u,v,2+u)$ on the unit disk. Its upward vector area is $R_u\times R_v=(-1,0,1)$. Curl is $(0,0,2)$, so Stokes gives integral $\iint_D2\,du\,dv=2\pi$. The parametrization of $C$ is the image of the counterclockwise parameter circle, so its direction matches. A reversed curve would give $-2\pi$.

Graded practice

1. Foundation. For $F=(0,0,1)$, find the outward flux through the closed unit sphere and through its upper hemisphere separately.
2. Core. Find the outward flux of $F=(x,y,z)$ through the shell $1\le\|x\|\le2$, including both components.
3. Core. For the planar vortex field $G=(-y/(x^2+y^2),x/(x^2+y^2))$, compute the positively oriented circulation around the annulus $1\le x^2+y^2\le4$. Compare with the circulation around its outer circle alone.
4. Proof. Two surfaces have the same oriented boundary, and a $C^1$ field is defined near both surfaces. Explain why their curl fluxes agree when Stokes applies. Does this imply their fluxes for the field itself agree?

Full solutions

1. Divergence of the constant field is zero, so the closed flux is zero. For the upper hemisphere, closing with the downward disk gives disk flux $-\pi$, hence hemisphere flux $\pi$. The two questions have different answers because one surface is open.
2. The outer sphere contributes $4\pi(2^3)=32\pi$ and the inner wall, oriented toward the center, contributes $-4\pi$. Total is $28\pi$. Divergence three times shell volume $3[4\pi(8-1)/3]$ gives the same result.
3. On any counterclockwise circle centered at zero, circulation is $2\pi$. The annulus boundary has outer counterclockwise circle and inner clockwise circle, so the total is $2\pi-2\pi=0$. The field is smooth near the annulus and has zero scalar curl, consistent with Green. For the outer circle alone the filled disk includes the singularity, so Green cannot be used to claim zero.
4. Each curl flux equals the same boundary circulation by Stokes, so they agree. No analogous claim holds for ordinary field flux without additional hypotheses: the two surfaces can enclose a volume with nonzero divergence. For example the upper hemisphere and upward disk share the same counterclockwise boundary, but $F=(x,y,z)$ has hemisphere flux $2\pi$ and disk flux zero.

Summary and term definitions

A cavity is an excluded interior region whose wall remains part of the solid's boundary. A singularity is a point where the field or required regularity fails. A closing surface completes an open surface into a solid boundary. A theorem may be fully correct while its proposed application is invalid; checking every boundary component and regularity neighborhood is part of the computation.