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Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

For a sufficiently smooth vector field on an oriented piecewise-smooth surface $S$ with boundary $\partial S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is the scalar surface-area element, and $d\mathbf{r}$ follows the positively oriented boundary. The boundary orientation is fixed by the right-hand rule: the fingers curl along $\partial S$ when the thumb points along $\mathbf{n}$.

For a smooth oriented surface $S$ with piecewise smooth boundary $C=\partial S$ and a continuously differentiable vector field $\mathbf{G}$, the Stokes theorem statesThe boundary orientation is fixed by the right-hand rule: curling the fingers along the positive boundary direction makes the thumb point along the surface normal. For an outward upper cap, its boundary ellipse is traversed counterclockwise when viewed from the positive $z$ direction. The figure shows two possible caps with the same oriented boundary, anticipating the last part.

The modified upper surface has exactly the same ellipse at $z=0$ and the same boundary orientation. The field retains the global polynomial vector potential from part (a). By the Stokes theorem, both surface fluxes equal the same boundary line integral; the cap's height and interior shape do not enter. HenceThis is an instance of the fact that solenoidal flux depends on the oriented boundary, with the domain condition automatically satisfied by the smooth polynomial field on all of ${\mathbb{R}}^3$. When $a,b$ are chosen positive, the formula simplifies to $\pi a^{3}b/4$; the absolute-value form respects the original allowance of arbitrary nonzero signs.

Consider a connected assembly of $n$ closed discs formed by identifying $e$ pairs of boundary arcs, each arc used once and all arcs in a given tile having disjoint endpoints. Its gluing graph has one vertex per tile and one edge per paired arc. A spanning tree joins the initial discs into a single closed disc; each of the remaining $e-n+1$ pairings attaches a band. This is a thickening of the gluing graph, so it is homotopy equivalent to that graph. Thus its Euler characteristic is $n-e$ and its fundamental group is the free group $F_{e-n+1}$. If the gluing graph is a tree, the assembled surface is itself a closed disc.

Suppose each pairing preserves the positive boundary parameter of the reference discs. A global orientation must then assign opposite orientation signs to adjacent tiles, because their induced boundary orientations on an interior seam must oppose one another. Such signs exist exactly when the gluing graph is bipartite. An odd gluing cycle therefore proves that the assembly is a nonorientable surface. A Dirichlet boundary condition or a Neumann boundary condition on an unpaired arc has no effect on this topological calculation.