Stokes's Theorem

From Green's to Stokes

Green's theorem has a circulation form that we've already seen. For a 2D vector field $\mathbf{u} = P\,\mathbf{i} + Q\,\mathbf{j}$ and a flat region $R$ bounded by a closed curve $C$:

The left side measures the circulation - how much the field pushes you along $C$. The right side measures the total rotational tendency inside $R$.

Now imagine $C$ is a curve in 3D space, not necessarily lying in a plane. It bounds some surface $S$ - think of a soap film stretched across a wire loop. The surface could be flat, curved, or wildly warped. Stokes's theorem says the same relationship holds, with the full 3D curl replacing the scalar $\text{curl}_z$:

Green's theorem is just Stokes with $S$ taken as a flat region in the $xy$-plane and $\mathbf{n} = \mathbf{k}$.

Stokes's Theorem

Let $S$ be a piecewise smooth oriented surface whose boundary $C$ is a piecewise smooth simple closed curve, directed in accordance with the orientation of $S$ (right-hand rule: curl fingers in the direction of $C$, thumb points along $\mathbf{n}$). Let $\mathbf{u} = L\,\mathbf{i} + M\,\mathbf{j} + N\,\mathbf{k}$ have continuous differentiable components. Then:

In component form:

The orientation convention matters: the direction of $C$ must match the choice of normal $\mathbf{n}$ on $S$ via the right-hand rule. Reversing either $C$ or $\mathbf{n}$ flips the sign.

Surface Independence

Here's one of the most beautiful consequences. Suppose two different surfaces $S_1$ and $S_2$ share the same boundary curve $C$ and are oriented to induce the same direction on $C$. Stokes says both give the same surface integral of curl:

Why? Because both equal the same line integral around $C$. We can also prove it directly: $S_1$ and $S_2$ together form a closed surface $S = S_1 \cup (-S_2)$ bounding some solid $R$. By the Divergence Theorem:

$$\iint_S \text{curl}\,\mathbf{u}\cdot\mathbf{n}\,d\sigma = \iiint_R \text{div}(\text{curl}\,\mathbf{u})\,dV = 0$$

since $\text{div}(\text{curl}\,\mathbf{u}) = 0$ is one of our fundamental identities from §3.6. So the integral over $S_1$ must equal the integral over $S_2$.

This is powerful: to compute $\iint_S (\text{curl}\,\mathbf{u}\cdot\mathbf{n})\,d\sigma$ over some complicated surface, you can replace $S$ with any simpler surface sharing the same boundary.

Proof Sketch

As with the Divergence Theorem, we prove one component at a time. We'll show

$$\oint_C L\,dx = \iint_S \left(-\frac{\partial L}{\partial y}\,dx\,dy + \frac{\partial L}{\partial z}\,dz\,dx\right)$$

Assume $S$ has the form $z = f(x,y)$ over a region $R_{xy}$, with boundary curve $C$ projecting to $C_{xy}$ in the $xy$-plane.

Left side: On $C$, we have $z = f(x,y)$, so $L(x,y,z) = L[x, y, f(x,y)]$. Thus

$$\oint_C L\,dx = \oint_{C_{xy}} L[x, y, f(x,y)]\,dx$$

Apply Green's theorem in the $xy$-plane:

$$= -\iint_{R_{xy}} \frac{\partial}{\partial y}L[x, y, f(x,y)]\,dx\,dy = -\iint_{R_{xy}} \left(\frac{\partial L}{\partial y} + \frac{\partial L}{\partial z}\frac{\partial f}{\partial y}\right) dx\,dy$$

Right side: Using the surface integral formula for $z = f(x,y)$:

$$\iint_S \left(-\frac{\partial L}{\partial y}\,dx\,dy + \frac{\partial L}{\partial z}\,dz\,dx\right) = \iint_{R_{xy}} \left(-\frac{\partial L}{\partial y} - \frac{\partial L}{\partial z}\frac{\partial f}{\partial y}\right) dx\,dy$$

Both sides match. The same argument works for $M\,dy$ and $N\,dz$; adding them gives the full theorem. General surfaces are handled by decomposition and a limit process.

Curl as Circulation per Unit Area

Just as the Divergence Theorem gave us a new interpretation of divergence (flux per unit volume), Stokes gives us a coordinate-free meaning for curl.

Take a small circular disk $S_r$ of radius $r$ centered at $(x_1, y_1, z_1)$ with normal $\mathbf{n}$, bounded by a circle $C_r$. Stokes says

$$\oint_{C_r} u_T\,ds = \iint_{S_r} (\text{curl}\,\mathbf{u}\cdot\mathbf{n})\,d\sigma$$

By the Mean Value Theorem, the right side equals $\text{curl}_n\,\mathbf{u}(x^*, y^*, z^*) \cdot A_r$ where $A_r = \pi r^2$. Dividing by $A_r$ and letting $r \to 0$:

In words: the component of curl in direction $\mathbf{n}$ equals the circulation per unit area around a small loop with that normal. This is coordinate-free - it tells us curl has intrinsic geometric meaning, just like divergence does.

Choosing $\mathbf{n} = \mathbf{i}, \mathbf{j}, \mathbf{k}$ recovers the three components of $\text{curl}\,\mathbf{u}$. And remember the paddle wheel from §3.5? This formula makes it rigorous: the paddle wheel spins fastest when its axis aligns with $\text{curl}\,\mathbf{u}$.

The Big Picture

We now have three great integral theorems, each connecting an integral over a region to an integral over its boundary:

Theorem	Interior	Boundary	Operator
Green's	Double integral over $R$	Line integral over $C$	$\text{curl}_z$ or $\text{div}$
Divergence	Triple integral over $R$	Surface integral over $S$	$\text{div}$
Stokes	Surface integral over $S$	Line integral over $C$	$\text{curl}$

The pattern is always the same: integral of a derivative over a region = integral of the function over the boundary. This is the generalized Stokes theorem, and it's one of the deepest ideas in all of mathematics.