As a consequence of Stokes' Theorem it seems that the perimeter of a closed curve $C$ can be obtained by choosing $F$ to be the vector field formed by rewriting the unit tangent $T$ of $C$ as a function of position.
Let $T$ be rewritten as a vector field on $\Bbb R^3$:
$$
T = \langle f(x,y,z), g(x,y,z), h(x,y,z) \rangle.
$$
This vector field obtained from $T$ is defined on the surface patch bounded by $C$ if this vector field for $T$ is defined on a subset of $\Bbb R^3$ containing the surface patch bounded by $C$.
Integrating the curl of this vector field over the surface patch whose boundary is $C$, you get the perimeter (length) of the curve.
For example, suppose the surface is the paraboloid $z=x^2+y^2$ and the curve $C$ is the circle of radius $R$ in the $z=R^2$ plane. The unit tangent of $C$ in terms of $x, y$ and $z$ would be $\langle -y/R , x/R , 0 \rangle$. Integrating the curl of this vector field over the surface patch whose boundary is $C$, you get $2\pi R$.
I've verified it works on a lot of examples.
Analogously, Green's theorem can also be used to find perimeter. Simply find $f$ and $g$ for $T = \langle f(x,y), g(x,y) \rangle $.
Is this a common application of Stokes' Theorem / Green's Theorem? Can anyone attest to ever having heard of it before?
The only reasons I have for believing this actually works are that the math loosely seems to make intuitive sense and that it didn't produce any contradictions when I tested it on examples.
And when I say that the math loosely seems to make intuitive sense, I mean that the length of a curve is equal to the line integral of the unit tangent over the curve.
Best Answer
The definitions of the scalar and vector line integrals are
$$\int_Cf\:ds = \int_a^b f(\mathbf{r}(t))\cdot|\mathbf{r}'(t)|\:dt$$
and
$$\int_C\mathbf{F}\cdot\mathbf{dr} = \int_a^b\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\:dt$$
respectively. Notice that when we set $\mathbf{F}(\mathbf{r}(t)) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$, the vector line integral becomes
$$\int_a^b \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}\cdot\mathbf{r}'(t)\:dt = \int_a^b |\mathbf{r}'(t)|\:dt = \int_C ds \equiv \operatorname{Length}(C)$$
In the special case that $C$ is a differentiable (not piecewise) closed curve, Stokes' theorem applies and you can compute the quantity either as a line integral or a surface integral - which is what you found. Because of the strict quantifiers, this is not commonly useful for curves. But you should still be proud for finding this, as the same intuition is much more broadly applicable to the integrals over surfaces rather than line integrals.