Stokes' Theorem
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.7: “Stokes’ Theorem” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-8-stokes-theorem |
| Textbook used in class | Stewart, Calculus, Section 16.8: “Stokes’ Theorem” (Examples 1, 2) |
Quick Reference
Stokes’ Theorem: Let $S$ be an oriented, piecewise smooth surface bounded by a simple closed curve $C = \partial S$. If $\mathbf{F}$ has continuous partial derivatives on an open region containing $S$, then $$\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}.$$
Orientation: $C$ must be traversed in the direction that the right-hand rule gives from the normal to $S$ -- curl the fingers of the right hand in the direction of $C$, and the thumb points in the direction of the surface normal.
Motivation
Green’s Theorem relates a line integral around a closed curve in the $xy$-plane to a double integral over the enclosed region. Stokes’ Theorem lifts this to three dimensions: the line integral around the boundary curve $C$ of a surface $S$ equals the flux of the curl through $S$.
This is enormously useful in both directions. If the line integral is hard (the curve wiggles through space), choose a convenient flat disk as $S$ and integrate the curl instead. If the surface integral is hard, choose a simpler surface with the same boundary.
Key Concept
The curl of $\mathbf{F}$ measures the local rotation of the field. Stokes’ Theorem says the total circulation around the boundary $C$ equals the total curl summed over the interior. Tiny little circulation loops in the interior cancel at shared interior edges; only the boundary contribution survives. This is the same cancellation argument behind Green’s Theorem, now lifted to surfaces in space.
When $S$ is a flat region in the $xy$-plane with normal $\mathbf{k}$, the formula $\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot\mathbf{k}\,dA$ reduces to Green’s Theorem with $\operatorname{curl}\mathbf{F}\cdot\mathbf{k} = Q_x - P_y$.
Worked Example
Use Stokes’ Theorem to compute $\oint_C \mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F} = \langle y^2, x, z^2\rangle$ and $C$ is the triangle with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ traversed counterclockwise when viewed from above. (Based on Stewart 16.8, Example 1.)
Step 1: Choose a surface. Take $S$ to be the triangular region in the plane $x+y+z=1$, $x,y,z\geq 0$, with upward-pointing normal (consistent with counterclockwise orientation of $C$).
Step 2: Compute the curl. $$\operatorname{curl}\mathbf{F} = \left\langle\frac{\partial z^2}{\partial y}-\frac{\partial x}{\partial z},\;\frac{\partial y^2}{\partial z}-\frac{\partial z^2}{\partial x},\;\frac{\partial x}{\partial x}-\frac{\partial y^2}{\partial y}\right\rangle = \langle 0, 0, 1-2y\rangle.$$
Step 3: Set up the surface integral. The plane $z = 1-x-y$ has $g_x = -1$, $g_y = -1$. Upward element: $\langle 1, 1, 1\rangle\,dA$.
$$\iint_S\operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = \iint_D \langle 0, 0, 1-2y\rangle\cdot\langle 1,1,1\rangle\,dA = \iint_D (1-2y)\,dA,$$
where $D$ is the triangle $x\geq 0$, $y\geq 0$, $x+y\leq 1$.
Step 4: Evaluate. $$\int_0^1\int_0^{1-x}(1-2y)\,dy\,dx.$$
Inner integral: $\left[y - y^2\right]_0^{1-x} = (1-x) - (1-x)^2 = (1-x)\bigl[1-(1-x)\bigr] = (1-x)\cdot x = x - x^2$.
Outer integral: $\int_0^1 (x-x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \frac{1}{6}$.
$$\oint_C \mathbf{F}\cdot d\mathbf{r} = \frac{1}{6}.$$
any surface bounded by $C$ gives a different answer. The value of $\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ is the same for every oriented surface that shares the same boundary curve $C$ and the same orientation convention. This is a consequence of the fact that $\operatorname{div}(\operatorname{curl}\mathbf{F}) = 0$. You can choose the most convenient surface -- a flat disk, a paraboloid, whatever makes the curl integral simplest -- and the answer will not change.
Common Misconceptions
the orientation of the surface $S$ and the orientation of its boundary curve $C$ can be chosen independently.
This is the concept-image-conflicts-definition error. Stokes’ theorem requires the orientations of $S$ and $\partial S$ to be compatible: the right-hand rule connects them. If the thumb points in the direction of the surface normal, the fingers curl in the direction that $C$ must be traversed. Choosing the outward normal for $S$ but traversing $C$ clockwise (when viewed from outside) violates this convention and produces a sign error. The theorem gives the correct answer only when orientation consistency is verified before computing.
Leveled Practice
Problem 1. Let $\mathbf{F} = \langle yz, xz, xy\rangle$ and $C$ the circle $x^2+y^2 = 1$, $z = 0$, counterclockwise. Use Stokes’ Theorem.
Show answer
$\operatorname{curl}\mathbf{F} = \langle x - x, y - y, z - z\rangle = \langle 0,0,0\rangle$.
$\iint_S \mathbf{0}\cdot d\mathbf{S} = 0$, so $\oint_C \mathbf{F}\cdot d\mathbf{r} = 0$.
(Note: $\mathbf{F} = \langle yz, xz, xy\rangle$ is conservative because it equals $\nabla(xyz)$. The line integral of a conservative field around a closed curve is always zero.)