Implications of Stokes' Theorem

= Theorem =

Stokes' Theorem implies all of the following results:


 * Classical Stokes' Theorem


 * Green's Theorem


 * Divergence Theorem


 * Fundamental Theorem of Calculus

= Proof =

Classical Stokes' Theorem
We note that given $f_1,f_2,f_3 :\R^3 \to \R$, $f_1dx + f_2dy +f_3dz$ is a 1-form defined on a surface. Since we are integrating over the boundary of a surface, which is a 1-manifold, we may use the general Stokes' theorem:

$\oint_{\partial S} f_1dx+f_2dy+f_3dz = \iint_S d(f_1dx+f_2dy+f_3dz) =$

$\iint_S \frac{\partial f_1}{\partial x} dx \wedge dx + \frac{\partial f_1}{\partial y} dy \wedge dx + \frac{\partial f_1}{\partial z} dz \wedge dx$

$+ \frac{\partial f_2}{\partial x} dx \wedge dy + \frac{\partial f_2}{\partial y} dy \wedge dy + \frac{\partial f_2}{\partial z} dz \wedge dy$

$+ \frac{\partial f_3}{\partial x} dx \wedge dz + \frac{\partial f_3}{\partial y} dy \wedge dz + \frac{\partial f_3}{\partial z} dz \wedge dz$

$ = \iint \left({\frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y} }\right) dx \wedge dy + \left({\frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} }\right) dy \wedge dz + \left({\frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x} }\right) dz \wedge dx$

If we define $\mathbf{F}:\R^3 \to \R^3$ as $\mathbf{F} = f_1 \mathbf{e}_1 + f_2 \mathbf{e}_2 + \mathbf{e}_3$, we recognize this expression as $\nabla \times \mathbf{F}$, with $dy\wedge dz$ replacing $\mathbf{e}_1, dx \wedge dz$ replacing $\mathbf{e}_2$, and $dx \wedge dy$ replacing $\mathbf{e}_3$.

Green's Theorem
We note that given $A,B :\R^2 \to \R$, $Adx + Bdy$ is a 1-form defined on some open set $U \subset \R^2$. Since we are integrating a 1-form on $\partial U$, a 1-manifold, we may use the general Stoke's theorem:

$\oint_{\partial U} \left({ Adx+Bdy }\right) = \iint_U d \left({Adx+Bdy }\right) = \iint_U \left({ \frac{\partial A}{\partial x} dx \wedge dx + \frac{\partial A}{\partial y} dy \wedge dx + \frac{\partial B}{\partial x} dx \wedge dy + \frac{\partial B}{\partial y} dy \wedge dy }\right) \ $

$= \iint_U \left({ 0 - \frac{\partial A}{\partial y} dx \wedge dy + \frac{\partial B}{\partial x} dx \wedge dy + 0 }\right) = \iint_U \left({ \frac{\partial B}{\partial x} - \frac{ \partial A}{\partial y} }\right) dxdy \ $

Divergence Theorem
We note that the expression $\mathbf{F}\cdot\mathbf{n}dS$ is a 2-form, since it takes two vectors ($F$ and $n$) and outputs a scalar; or, less formally, $\mathbf{F}\cdot\mathbf{n}dS = $ (scalar)(area 2-form) = 2-form. Since we are integrating this expression over the boundary of a region in three space, which is a 2-manifold, we may use the general Stokes' theorem:

$\iint\limits_{\partial U} \mathbf{F} \cdot \mathbf{n}\ dS = \iiint_U d \left({ \mathbf{F} \cdot \mathbf{n}\ dS }\right) = \iiint_U d(\mathbf{F}\cdot\mathbf{n})dS + \mathbf{F}\cdot\mathbf{n}d^2S = \iiint_U d(\mathbf{F}\cdot\mathbf{n})dS \  $

Fundamental Theorem of Calculus
Given a function $f:\R \to \R$, integrable on some open set $I$, we can recognize such a function a 0-form - that is, defined at a point and taking no vectors as inputs. Hence the exterior derivative is just the ordinary differential $f'(x)dx$. Since the ordinary differential is a 1-form, and we are integrating on a 1-manifold, we may use the general Stokes' theorem:

$\int_I f'(x)dx = \int_{\partial I} f(x)$

By Induced Orientations on Boundaries, $\partial I = \left\{{ x_0', x_1 }\right\}$ for some points $x_0, x_1$ where $y'$ indicates a point with "negative" orientation; we then have

$\int_{\partial I} f(x) = -f(x_0) + f(x_1) = f(x_1)-f(x_0)$