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

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

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

Also known as the Kelvin-Stokes theorem.

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

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

so that if we define $$\mathbf{G}=(G_1, G_2)= \left({ \mathbf{F}\cdot \frac{\partial \mathbf{r}}{\partial s}, \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial t} }\right) \ $$, then

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

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 $$t \ $$:

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

$$= \iint_R \left({ \frac{\partial G_2}{\partial s} - \frac{\partial G_1}{\partial t} }\right) dsdt \ $$

By Green's Theorem, this can be written as

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

Hence both sides of the theorem equation are equal.