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

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 some set $$S \subset \R^3 \ $$ where $$S \ $$ is a surface, and so, by the general Stokes' theorem,

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

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

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

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

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

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 $$U \subset \R^2 \ $$ and so, by the general Stoke's theorem,

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

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

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 $$f'(x)dx \ $$ and so, by 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

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