Green's theorem says the circulation around a flat region equals the total curl inside. Stokes' theorem says the same thing for a curved surface in space: the circulation around its edge equals the flux of the curl through the surface. Cut the surface into tiny patches; each patch's circulation is about its curl times its area, and the shared edges cancel.
Key idea
Stokes' theorem. Let $S$ be an oriented, piecewise smooth surface whose boundary $C$ is a simple closed curve, oriented by the right-hand rule (thumb along $\mathbf{n}$, fingers curl along $C$; equivalently, walking along $C$ with your head along $\mathbf{n}$, the surface is on your left). If $\mathbf{F}$ has continuous partial derivatives on an open region containing $S$, then
$$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}.$$
Why it works
Green's theorem is the flat case: if $S$ is a region $D$ in the $xy$-plane with $\mathbf{n} = \mathbf{k}$, then $(\nabla \times \mathbf{F}) \cdot \mathbf{k} = Q_x - P_y$ and $d\mathbf{S} = \mathbf{k}\,dA$, so Stokes reads $\oint_C P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA$.
Why curl appears: a tiny flat patch with unit normal $\mathbf{n}$ and area $\Delta S$ is, in its own coordinates, a small region where Green's theorem applies. Its circulation is about (the component of the curl along $\mathbf{n}$) $\times\,\Delta S = (\nabla \times \mathbf{F}) \cdot \mathbf{n}\,\Delta S$. Adding over all patches, the inside edges cancel in pairs (each is traced twice in opposite directions), leaving the circulation around $C$.
Why curl appears: a tiny flat patch with unit normal $\mathbf{n}$ and area $\Delta S$ is, in its own coordinates, a small region where Green's theorem applies. Its circulation is about (the component of the curl along $\mathbf{n}$) $\times\,\Delta S = (\nabla \times \mathbf{F}) \cdot \mathbf{n}\,\Delta S$. Adding over all patches, the inside edges cancel in pairs (each is traced twice in opposite directions), leaving the circulation around $C$.
Worked examples
Worked example: A hemisphere
Let $S$ be the upper hemisphere $x^2 + y^2 + z^2 = 4$, $z \ge 0$, oriented upward, and $\mathbf{F} = \langle -y, x, z \rangle$. Find $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.
- The boundary is the circle $x^2 + y^2 = 4$, $z = 0$; upward normal means counterclockwise seen from above.
- With $\mathbf{r}(t) = \langle 2\cos t, 2\sin t, 0 \rangle$: $\mathbf{F} \cdot \mathbf{r}' = 4\sin^2 t + 4\cos^2 t + 0 = 4$.
- $\oint_C \mathbf{F} \cdot d\mathbf{r} = 8\pi$. (Check: $\nabla \times \mathbf{F} = \langle 0, 0, 2 \rangle$, and through the flat disk of area ${}4\pi$ its flux is ${}8\pi$.)
Answer: ${}8\pi$
Worked example: A triangle
Let $S$ be the triangle with vertices $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 1)$, oriented away from the origin, and $\mathbf{F} = \langle z, x, y \rangle$. Find $\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}$.
- $$\nabla \times \mathbf{F} = \langle 1 - 0, 1 - 0, 1 - 0 \rangle = \langle 1, 1, 1 \rangle.$$
- $S$ lies in the plane $x + y + z = 1$ with unit normal $\frac{1}{\sqrt{3}}\langle 1, 1, 1 \rangle$, so $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = \sqrt{3}$.
- The triangle has area $\frac{\sqrt{3}}{2}$ (its sides have length $\sqrt{2}$). Circulation $= \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}$.
Answer: $\frac{3}{2}$
Worked example: Reading the orientation
A surface lies above the $xy$-plane with boundary the unit circle in that plane, and is oriented downward. Which way must the circle be traced?
- Thumb pointing down: the fingers curl clockwise as seen from above.
- So the circle is traced clockwise (seen from above), and with that direction Stokes' theorem holds.
Answer: clockwise as seen from above
Common mistake
Orienting the boundary against the normal. If $\mathbf{n}$ points up, $C$ must go counterclockwise as seen from above; if $\mathbf{n}$ points down, clockwise. Pairing an upward normal with a clockwise curve gives the right number with the wrong sign: $-8\pi$ instead of ${}8\pi$ in the first example. Use the right-hand rule every time.