Kelvin-Stokes Theorem

Theorem

Let $S$ be some orientable smooth surface with boundary in $\R^3$, and 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:

$\displaystyle \oint_{\partial S} f_1 dx + f_2 dy + f_3 dz = \iint_S \left({ \nabla \times \mathbf F}\right) \cdot \mathbf ndA$

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

Proof

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

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

$\displaystyle \oint_{\partial S} f_1dx+f_2dy+f_3dz = \oint_{\partial S} \mathbf F \cdot d\mathbf r = \oint_{\partial R} \mathbf F\cdot \frac{\partial \mathbf r}{\partial s} ds + \mathbf F \cdot \frac{\partial \mathbf r}{\partial t} dt$

so that if we define:

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

then:

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

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \iint_S \left({ \nabla \times \mathbf F}\right) \cdot \mathbf n dA\)	\(=\)	\(\displaystyle \)	\(\displaystyle \iint_R \nabla \times \mathbf F \cdot \left({ \frac{\partial \mathbf r}{\partial s} \times \frac{\partial \mathbf r}{\partial t} }\right) dsdt\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \iint_R \left({ \frac{\partial G_2}{\partial s} - \frac{\partial G_1}{\partial t} }\right) dsdt\)	\(\displaystyle \)	\(\displaystyle \)

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

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

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 George Stokes and William Thomson, 1st Baron Kelvin.