Kelvin-Stokes Theorem

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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.


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:

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


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

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.


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.