# Kelvin-Stokes Theorem

(Redirected from Classical Stokes' Theorem)

## Theorem

Let $S$ be some orientable smooth surface with boundary in $\R^3$.

Let $\mathbf F:\R^3 \to \R^3$ be a vector-valued function with Euclidean coordinate expression:

$F = f_1 \mathbf i + f_2 \mathbf j + f_3 \mathbf k$

where $f_i: \R^3 \to \R$.

Then:

$\displaystyle \oint_{\partial S} f_1 \ \mathrm d x + f_2 \ \mathrm d y + f_3 \ \mathrm d z = \iint_S \left({\nabla \times \mathbf F}\right) \cdot \mathbf n \ \mathrm d A$

where $\mathbf n$ is the unit normal to $S$ and $\mathrm d A$ is the area element on the surface.

## Proof

Let $\mathbf r:\R^2 \to \R^3, \mathbf r \left({s, t}\right)$ be a smooth parametrization of $S$ from some region $R$ in the $st$-plane, so that:

$\mathbf r \left({R}\right) = S$

and:

$\mathbf r \left({\partial R}\right) = \partial S$

First, we convert the left hand side into a line integral:

 $\displaystyle \oint_{\partial S} f_1 \ \mathrm d x + f_2 \ \mathrm d y + f_3 \ \mathrm d z$ $=$ $\displaystyle \oint_{\partial S} \mathbf F \cdot \ \mathrm d \mathbf r$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \oint_{\partial R} \mathbf F \cdot \frac{\partial \mathbf r} {\partial s} \ \mathrm d s + \mathbf F \cdot \frac{\partial \mathbf r} {\partial t} \ \mathrm d t$ $\quad$ $\quad$

so that if we define:

$\mathbf G = \left({G_1, G_2}\right) = \left({\mathbf F \cdot \dfrac{\partial \mathbf r}{\partial s}, \mathbf F \cdot \dfrac{\partial \mathbf r} {\partial t} }\right)$

then:

$\displaystyle \int_{\partial S} \mathbf F \cdot \ \mathrm d \mathbf r = \int_{\partial R} \mathbf G \cdot \ \mathrm d \mathbf s$

where $\mathbf s$ is the position vector in the $s t$-plane.

We turn now to the right-hand expression and write it in terms of $s$ and $t$:

 $\displaystyle \iint_S \left({\nabla \times \mathbf F}\right) \cdot \mathbf n \ \mathrm d A$ $=$ $\displaystyle \iint_R \nabla \times \mathbf F \cdot \left({\frac{\partial \mathbf r}{\partial s} \times \frac{\partial \mathbf r}{\partial t} }\right) \ \mathrm d s \ \mathrm d t$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \iint_R \left({\frac{\partial G_2}{\partial s} - \frac{\partial G_1}{\partial t} }\right) \ \mathrm d s \ \mathrm d t$ $\quad$ $\quad$

By Green's Theorem, this can be written as:

$\displaystyle \int_{\partial R} \mathbf G \cdot \ \mathrm d \mathbf s$

Hence both sides of the theorem equation are equal.

$\blacksquare$

## Also known as

Also known as the Classical Stokes' Theorem.

## Source of Name

This entry was named for Lord Kelvin and George Stokes.