Kelvin-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:
 * $\mathbf F = f_1 \mathbf i + f_2 \mathbf j + f_3 \mathbf k$

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

Then:
 * $\ds \oint_{\partial S} f_1 \rd x + f_2 \rd y + f_3 \rd z = \iint_S \paren {\nabla \times \mathbf F} \cdot \mathbf n \rd A$

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

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

and:
 * $\map {\mathbf r} {\partial R} = \partial S$

First, we convert the into a line integral:

so that if we define:
 * $\mathbf G = \paren {G_1, G_2} = \paren {\mathbf F \cdot \dfrac {\partial \mathbf r} {\partial s}, \mathbf F \cdot \dfrac {\partial \mathbf r} {\partial t} }$

then:


 * $\ds \int_{\partial S} \mathbf F \cdot \rd \mathbf r = \int_{\partial R} \mathbf G \cdot \rd \mathbf s$

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

Note that:

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

Let us investigate the integrand:

That is:
 * $\ds \iint_S \paren {\nabla \times \mathbf F} \cdot \mathbf n \rd A = \iint_R \paren {\frac {\partial G_2} {\partial s} - \frac {\partial G_1} {\partial t} } \rd s \rd t$

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


 * $\ds \int_{\partial R} \mathbf G \cdot \rd \mathbf s$

Hence both sides of the theorem equation are equal.

Also known as
Also known as the Classical Stokes' Theorem.