# Implications of Stokes' Theorem

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## Theorem

Stokes' Theorem implies all of the following results:

## Proof

### Classical Stokes' Theorem

We note that given $f_1, f_2, f_3: \R^3 \to \R$, $f_1 \rd x + f_2 \rd y + f_3 \rd z$ 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:

\(\ds \oint_{\partial S} f_1 \rd x + f_2 \rd y + f_3 \rd z\) | \(=\) | \(\ds \iint_S \map \rd {f_1 \rd x + f_2 \rd y + f_3 \rd z}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \iint_S \frac {\partial f_1} {\partial x} \rd x \wedge \d x + \frac {\partial f_1} {\partial y} \rd y \wedge \d x + \frac {\partial f_1} {\partial z} \rd z \wedge \d x\) | ||||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac {\partial f_2} {\partial x} \rd x \wedge \d y + \frac {\partial f_2} {\partial y} \rd y \wedge \d y + \frac {\partial f_2} {\partial z} \rd z \wedge \d y\) | |||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac {\partial f_3} {\partial x} \rd x \wedge \d z + \frac {\partial f_3} {\partial y} \rd z \wedge \d y + \frac {\partial f_3} {\partial z} \rd z \wedge \d z\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \iint \paren {\frac {\partial f_2} {\partial x} - \frac {\partial f_1} {\partial y} } \rd x \wedge \d y + \paren {\frac {\partial f_3} {\partial y} - \frac {\partial f_2} {\partial z} } \rd y \wedge \d z + \paren {\frac {\partial f_1} {\partial z} - \frac{\partial f_3}{\partial x} } \rd z \wedge \d x\) |

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:

- $\d y \wedge \d z$ replacing $\mathbf e_1$
- $\d z \wedge \d x$ replacing $\mathbf e_2$

and:

- $\d x \wedge \d y$ replacing $\mathbf e_3$.

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### Green's Theorem

We note that given $A, B: \R^2 \to \R$, $A \rd x + B \rd y$ 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 Stokes' theorem:

\(\ds \oint_{\partial U} \paren {A \rd x + B \rd y}\) | \(=\) | \(\ds \iint_U \map \rd {A \rd x + B \rd y}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \iint_U \paren {\frac {\partial A} {\partial x} \rd x \wedge \d x + \frac {\partial A} {\partial y} \rd y \wedge \d x + \frac {\partial B} {\partial x} \rd x \wedge \d y + \frac {\partial B} {\partial y} \rd y \wedge \d y}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \iint_U \paren {0 - \frac {\partial A} {\partial y} \rd y \wedge \d x + \frac {\partial B} {\partial x} \rd x \wedge \d y + 0}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \iint_U \paren {\frac {\partial B} {\partial x} - \frac{ \partial A} {\partial y} } \rd x \rd y\) |

$\Box$

### Divergence Theorem

We note that the expression $\mathbf F \cdot \mathbf n \rd S$ is a $2$-form, since it takes two vectors ($F$ and $n$) and outputs a scalar.

Less formally:

- $\mathbf F \cdot \mathbf n \rd S = $ (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:

\(\ds \iint_{\partial U} \mathbf F \cdot \mathbf n \rd S\) | \(=\) | \(\ds \iiint_U \map \rd {\mathbf F \cdot \mathbf n \rd S}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \iiint_U \map \rd {\mathbf F \cdot \mathbf n} \rd S + \mathbf F \cdot \mathbf n \rd^2 S\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \iiint_U \map \rd {\mathbf F \cdot \mathbf n} \rd S\) |

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### 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:

- $\map {f'} x \rd x$

Since the ordinary differential is a $1$-form, and we are integrating on a $1$-manifold, we may use the general Stokes' theorem:

- $\ds \int_I \map f x \rd x = \int_{\partial I} \map f x$

By Induced Orientations on Boundaries, $\partial I = \set {x_0', x_1}$ for some points $x_0, x_1$ where $y'$ indicates a point with "negative" orientation.

We then have

- $\ds \int_{\partial I} \map f x = -\map f {x_0} + \map f {x_1} = \map f {x_1} - \map f {x_0}$

$\Box$

### Cauchy's Residue Theorem

Let $z = x + iy$ and $\map f z = \map u {x, y} + i \map v {x, y}$ for two real functions $u, v$.

Let $D$ be a disk and its boundary $\partial D = C$ a circle.

Then by Stokes' Theorem:

- $\ds \int_C \map f z \rd z = \int_C \map u z \rd x + \map v z \rd y = \int_D \frac {\partial u} {\partial y} - \frac {\partial v} {\partial x}$

which is zero if $f$ satisfies the Cauchy-Riemman conditions.

$\blacksquare$