Stokes' theorem has a surprising consequence: the flux of a curl through a surface depends only on the surface's boundary curve. Any two surfaces with the same edge, oriented to match, give the same answer. So a hard surface integral of $\nabla \times \mathbf{F}$ can be replaced by an easy one, usually over a flat disk, or by the line integral around the edge.
Key idea
If $S_1$ and $S_2$ are oriented surfaces with the same boundary curve $C$, and both orientations induce the same direction on $C$, then
$$\iint_{S_1} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_{S_2} (\nabla \times \mathbf{F}) \cdot d\mathbf{S}.$$
For a closed surface (no boundary curve), $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = 0$.
Why it works
Both surface integrals equal the same line integral. For a closed surface, cut it along a curve into two caps. With the outward normal on both, the right-hand rule sends the cutting curve in opposite directions for the two caps, so their circulations cancel and the total is ${}0$. This matches $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ and the divergence theorem.
Only for curls. For a general field $\mathbf{F}$, the flux through two surfaces with the same edge can differ; the difference is the total divergence of $\mathbf{F}$ in the solid between them.
Only for curls. For a general field $\mathbf{F}$, the flux through two surfaces with the same edge can differ; the difference is the total divergence of $\mathbf{F}$ in the solid between them.
Worked examples
Worked example: Swap a paraboloid for a disk
Find $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ for $\mathbf{F} = \langle -y^3, x^3, e^{z} \rangle$, where $S$ is $z = 4 - x^2 - y^2$, $z \ge 0$, oriented upward.
- $\nabla \times \mathbf{F} = \langle 0 - 0, 0 - 0, 3x^2 + 3y^2 \rangle$.
- The boundary is the circle $x^2 + y^2 = 4$, $z = 0$. Use the flat disk $D$ with upward normal $\mathbf{k}$ instead.
- $$\iint_D 3(x^2 + y^2)\,dA = \int_0^{2\pi} \int_0^2 3r^2 \cdot r\,dr\,d\theta = 2\pi \cdot 12 = 24\pi.$$
Answer: ${}24\pi$
Worked example: A closed surface
Find the outward flux of $\nabla \times \mathbf{F}$ through the sphere $x^2 + y^2 + z^2 = 9$, for $\mathbf{F} = \langle xy^2, \sin(xz), e^{xyz} \rangle$.
- The sphere is closed: it has no boundary curve.
- So the flux of any curl through it is ${}0$, no computation needed.
Answer: ${}0$
Worked example: An open can
$S$ is the side and top of the cylinder $x^2 + y^2 = 1$, ${}0 \le z \le 2$, with outward normal, and $\mathbf{F} = \langle -y, x, z^2 \rangle$. Find $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.
- The boundary of $S$ is the bottom circle. The outward normal on the side, with the right-hand rule, sends it counterclockwise seen from above.
- The bottom disk with upward normal has the same oriented boundary. There $\nabla \times \mathbf{F} = \langle 0, 0, 2 \rangle$, so the flux is ${}2 \cdot \pi = 2\pi$.
Answer: ${}2\pi$