swapping surfaces under Stokes' theorem

Since Stokes' Theorem's curl-flux only depends on the boundary curve, any two oriented surfaces sharing that same boundary give the identical integral.

If $S_1$ and $S_2$ are two oriented surfaces sharing the same boundary curve $C$ (with matching orientation), then $\iint_{S_1}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_{S_2}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ — both surface integrals equal the same circulation, hence each other, regardless of how differently $S_1$ and $S_2$ bulge in between. This traces back to $\operatorname{div}(\operatorname{curl}\mathbf{F})\equiv 0$ for any $\mathbf{F}$ with continuous second partials.

Examples

For $\mathbf{F}(x,y,z)=(-y,\ x,\ z)$ and $S$ the paraboloid $z=4-x^2-y^2$ for $z\ge0$ : swapping in the flat disk $x^2+y^2\le4$ with the same rim gives $\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = 8\pi$ without integrating over the curved paraboloid.