24.1 Statement and proof of Stokes' theorem

Vector Calculus Fundamentals

Vector Fields and Differential Forms

A vector field $\mathbf{F}(x, y, z) = (F_1, F_2, F_3)$ assigns a vector to each point in a region of space. The curl of $\mathbf{F}$, written $\nabla \times \mathbf{F}$, measures the infinitesimal rotation of the field at each point:

$\nabla \times \mathbf{F} = \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z},\; \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x},\; \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right)$

You can also remember this as the determinant of the symbolic matrix with $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ in the first row, partial operators in the second, and $F_1, F_2, F_3$ in the third. Curl shows up whenever you need to describe rotational behavior: fluid vortices, magnetic fields curling around currents, etc.

Stokes' theorem has a particularly clean formulation in the language of differential forms:

• A 1-form $\omega = P\,dx + Q\,dy + R\,dz$ is the object you integrate over curves.
• A 2-form $\eta = P\,dy \wedge dz + Q\,dz \wedge dx + R\,dx \wedge dy$ is the object you integrate over surfaces. The wedge product $\wedge$ is antisymmetric: $dx \wedge dy = -dy \wedge dx$.

The exterior derivative $d$ takes a $k$-form to a $(k+1)$-form and unifies gradient, curl, and divergence into a single operation. For a 1-form $\omega = P\,dx + Q\,dy + R\,dz$:

$d\omega = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)dy \wedge dz + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)dz \wedge dx + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dx \wedge dy$

Notice that $d\omega$ is exactly the 2-form whose coefficient functions are the components of $\nabla \times \mathbf{F}$. This is why Stokes' theorem, phrased in forms as $\int_S d\omega = \int_{\partial S} \omega$, directly encodes the curl version.

Integration Concepts

Surface and Line Integrals

The surface integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$ measures the total flux of $\mathbf{F}$ through an oriented surface $S$. Here $d\mathbf{S} = \mathbf{n}\,dS$, where $\mathbf{n}$ is the unit outward normal and $dS$ is the scalar area element. If $S$ is parametrized by $\mathbf{r}(u,v)$, then $d\mathbf{S} = (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv$, and the direction of the cross product determines the orientation.

The line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ evaluates the component of $\mathbf{F}$ tangent to a curve $C$, accumulated along the path. With $d\mathbf{r} = (dx, dy, dz)$, this integral computes quantities like work done by a force or circulation of a velocity field.

Orientable Surfaces and Boundary Curves

Stokes' theorem requires the surface to be orientable, meaning you can choose a continuously varying normal vector across the entire surface without contradiction. Spheres, tori, and planes are all orientable. A Möbius strip is the classic non-orientable counterexample: if you try to push a normal vector continuously around the strip, it flips when you return to the starting point.

The boundary curve $\partial S$ is the edge of the surface $S$, and its orientation must be compatible with the surface orientation via the right-hand rule: if your right thumb points in the direction of the chosen surface normal $\mathbf{n}$, your fingers curl in the direction of traversal along $\partial S$. Getting this orientation agreement right is essential; reversing it flips the sign of the line integral.

Stokes' Theorem

Statement

Stokes' theorem states that for a smooth vector field $\mathbf{F}$ defined on an open region containing an oriented, piecewise-smooth surface $S$ with piecewise-smooth boundary curve $\partial S$:

$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}$

In the language of differential forms, this is $\int_S d\omega = \int_{\partial S} \omega$, which is the generalized Stokes' theorem applied to a 1-form on a 2-dimensional domain.

The theorem says: the total curl flux through a surface equals the total circulation around its boundary. This is a higher-dimensional analog of the Fundamental Theorem of Calculus, where integrating a derivative over a region reduces to evaluating something on the boundary.

Proof Outline

The full proof handles general oriented surfaces by decomposing them into patches, but the core argument is clearest for a surface $S$ that is the graph of a function $z = g(x,y)$ over a region $D$ in the $xy$-plane. Here's the structure:

1. Reduce to components. Write $\mathbf{F} = (F_1, F_2, F_3)$. Because both sides of Stokes' theorem are linear in $\mathbf{F}$, it suffices to prove the theorem separately for $\mathbf{F} = (F_1, 0, 0)$, then $(0, F_2, 0)$, then $(0, 0, F_3)$. The full result follows by adding.

2. Work out one component. Take $\mathbf{F} = (F_1, 0, 0)$. Its curl is: $\nabla \times \mathbf{F} = \left(0,\; \frac{\partial F_1}{\partial z},\; -\frac{\partial F_1}{\partial y}\right)$

3. Compute the surface integral. Parametrize $S$ by $(x, y, g(x,y))$ over $D$. The outward normal (with upward orientation) gives $d\mathbf{S} = (-g_x, -g_y, 1)\,dx\,dy$. Dotting with the curl: $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_D \left(\frac{\partial F_1}{\partial z}(-g_y) - \frac{\partial F_1}{\partial y}\right) dx\,dy$

By the chain rule, since $F_1(x, y, g(x,y))$ depends on $y$ both directly and through $z = g(x,y)$: $\frac{\partial}{\partial y}\bigl[F_1(x, y, g(x,y))\bigr] = \frac{\partial F_1}{\partial y} + \frac{\partial F_1}{\partial z}\,g_y$ So the surface integral equals $\iint_D -\frac{\partial}{\partial y}\bigl[F_1(x, y, g(x,y))\bigr]\,dx\,dy$.

1. Compute the line integral. On the boundary curve $\partial S$, the $\mathbf{F} \cdot d\mathbf{r}$ integral reduces to $\oint_{\partial S} F_1\,dx$. Since $\partial S$ lies above $\partial D$ (the boundary of the planar region), and $z = g(x,y)$ on the surface, this becomes $\oint_{\partial D} F_1(x, y, g(x,y))\,dx$.

2. Apply Green's Theorem in the plane. Green's theorem gives: $\oint_{\partial D} F_1(x,y,g(x,y))\,dx = \iint_D -\frac{\partial}{\partial y}\bigl[F_1(x,y,g(x,y))\bigr]\,dx\,dy$ This matches the surface integral from step 3.

3. Repeat for the other components $(0, F_2, 0)$ and $(0, 0, F_3)$ using analogous arguments, then sum.

For a general oriented surface that isn't a single graph, you decompose $S$ into patches, each of which is a graph over one of the coordinate planes. The interior boundary contributions from adjacent patches cancel (they're traversed in opposite directions), leaving only the integral over the outer boundary $\partial S$.

Applying Stokes' Theorem

When you encounter a problem, follow these steps:

1. Identify the vector field $\mathbf{F}$, the surface $S$, and the boundary curve $\partial S$.
2. Check orientations. Make sure the surface normal and boundary traversal direction are compatible via the right-hand rule.
3. Decide which side to compute. Stokes' theorem is useful precisely because one side is often much easier than the other. A complicated surface integral of curl might reduce to a simple line integral around a circle, or vice versa.
4. Compute the curl $\nabla \times \mathbf{F}$ if you're evaluating the surface side.
5. Parametrize whichever side you chose and evaluate the integral.