Applications of Stokes' Theorem
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.8: “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” (Example 2, Remark on surface independence) |
Quick Reference
The key application strategy:
- If $\oint_C \mathbf{F}\cdot d\mathbf{r}$ is hard, swap to $\iint_S\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ over a convenient $S$.
- If $\iint_S\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ is hard over one surface, swap to a simpler surface with the same boundary.
- If $\operatorname{curl}\mathbf{F} = \mathbf{0}$, the line integral around any closed curve (in a simply connected domain) is zero.
Motivation
Stokes’ Theorem becomes a strategic tool once you see the surface as a choice rather than a given. The boundary curve $C$ determines the answer -- the surface is just the vehicle you use to compute it. Picking a flat disk instead of a complicated curved surface can turn a hard double integral into an easy one. The habit to build: ask what the simplest surface is before computing.
Key Concept
Two surfaces $S_1$ and $S_2$ that share the same oriented boundary $C$ always satisfy $$\iint_{S_1}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}_1 = \iint_{S_2}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}_2,$$ provided $\mathbf{F}$ has continuous partials in the region between them. This follows from the Divergence Theorem applied to the closed surface $S_1 \cup (-S_2)$ and the identity $\operatorname{div}(\operatorname{curl}\mathbf{F}) = 0$.
The practical upshot: when a problem specifies a complicated surface, check whether the boundary curve admits a much simpler surface -- often a flat disk.
Worked Example
Compute $\oint_C \mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F} = \langle -y^3, x^3, -z^3\rangle$ and $C$ is the circle $x^2+y^2 = 1$, $z = 0$, counterclockwise.
A direct parametrization would work but requires computing cubes along a circle. Stokes’ Theorem is faster.
Curl: $$\operatorname{curl}\mathbf{F} = \left\langle\frac{\partial(-z^3)}{\partial y}-\frac{\partial x^3}{\partial z},\;\frac{\partial(-y^3)}{\partial z}-\frac{\partial(-z^3)}{\partial x},\;\frac{\partial x^3}{\partial x}-\frac{\partial(-y^3)}{\partial y}\right\rangle = \langle 0, 0, 3x^2+3y^2\rangle.$$
Surface: Choose the flat disk $S$: $x^2+y^2\leq 1$, $z = 0$, with upward normal $\langle 0,0,1\rangle$.
Integral: $$\iint_S\operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = \iint_D 3(x^2+y^2)\,dA = \int_0^{2\pi}\int_0^1 3r^2\cdot r\,dr\,d\theta = 2\pi\cdot 3\int_0^1 r^3\,dr = 2\pi\cdot 3\cdot\frac{1}{4} = \frac{3\pi}{2}.$$
$$\oint_C \mathbf{F}\cdot d\mathbf{r} = \frac{3\pi}{2}.$$
The direct computation would have required integrating $(-\sin^3 t)(-\sin t) + (\cos^3 t)(\cos t)$ over $[0,2\pi]$ -- the Stokes approach is faster.
Stokes’ Theorem allows replacing any surface with any other surface. Replacing surfaces is valid only when the two surfaces have the same oriented boundary and $\operatorname{curl}\mathbf{F}$ is defined and continuous throughout the region between them. If the curl is undefined somewhere inside (say, at the origin for a field like $\mathbf{F} = \langle y/r^2, -x/r^2, 0\rangle$), the two integrals may differ. Always check that $\mathbf{F}$ (and hence $\operatorname{curl}\mathbf{F}$) is smooth in the region bounded by $S_1$ and $S_2$.
Common Misconceptions
any two surfaces sharing the same boundary curve give the same flux integral for any field $\mathbf{F}$.
This is the concept-image-conflicts-definition error. Two surfaces with the same oriented boundary give equal values of $\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ only when $\operatorname{curl}\mathbf{F}$ is defined and continuous throughout the region between them. If $\mathbf{F}$ has a singularity in that region -- for example, $\mathbf{F}$ is undefined at the origin and the two surfaces enclose it differently -- the two integrals can differ. Surface-swapping is valid only when the field is smooth in the enclosed region.
Leveled Practice
Problem 1. Use Stokes’ Theorem to compute $\oint_C\mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F} = \langle z^2, y^2, x\rangle$ and $C$ is the boundary of the square $0\leq x\leq 1$, $0\leq y\leq 1$ in the plane $z = 3$, counterclockwise when viewed from above.
Show answer
$\operatorname{curl}\mathbf{F} = \langle 0 - 0, 2z - 1, 0-0\rangle = \langle 0, 2z-1, 0\rangle$.
Surface $S$: square $z = 3$, $0\leq x,y\leq 1$, upward normal $\langle 0,0,1\rangle$.
$\operatorname{curl}\mathbf{F}\cdot\langle 0,0,1\rangle = 0$.
$\oint_C \mathbf{F}\cdot d\mathbf{r} = 0$.