Surfaces and Stokes Theorem

Parameterized Surfaces:

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A function $ϕ : D \subset R^{2} \to R^{3}$ where the corresponding surface is $S = ϕ (D)$
$ϕ : D \subset R^{2} \to R^{3}$ such that $ϕ = [\begin{matrix} x (u, v), y (u, v), t (u, v) \end{matrix}]$

Consider only piecewise regular surfaces that are unions of images of parametrized surfaces i : $D_{i} \to R^{3}$ for which:

(i) $D_{i}$ is an elementary region in the plane;

(ii) $ϕ_{i}$ is of class $C^{1}$ and one-to-one, except possibly on the boundary of $D_{i}$ ; and

(iii) $S_{i}$ , the image of $ϕ_{i}$ , is regular, except possibly at a finite number of points.

Graph as a Surface

\begin{matrix} f : R^{n} \to R \\ graph (f) = (x_{1}, x_{2}, \dots, x_{n}, f (x_{1}, \dots ., x_{n})) \end{matrix}

Plane as a Surface

$a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$

since $n \cdot [\begin{matrix} (x - x_{0}) \\ (y - y_{0}) \\ (z - z_{0}) \end{matrix}] = 0$

$n = [a, b, c]$ is the norm

where n is the norm of the plane
can solve for points to get parametric form

parameterization

Φ (u, v) = α u + β v + γ .

$γ$ is the point (anchor).
$α$ and $β$ are the direction vectors.
$u$ and $v$ are scalar parameters.

If you have $3$ points on a plane then you get $v, w$ by finding the direction vectors.
Project points on x,y plane and use that for parameters

Tangent Vector of a Surface

\begin{matrix} ϕ : D \subset R^{2} \to R^{3} \\ ϕ (u, v) = [x (u, v,), y (u, v), z (u, v,)] \end{matrix}

$ϕ (u, v_{0})$ fixes $v$ and traces curve in one direction
$ϕ (u_{0}, v)$ does the opposite

when $u$ is free we have the tangent:

T_{u} = [\frac{\partial x}{\partial u} (u, v), \frac{\partial y}{\partial u} (u, v), \frac{\partial z}{\partial u} (u, v)]

when $v$ is free we have:

T_{v} = [\frac{\partial x}{\partial v} (u, v), \frac{\partial y}{\partial v} (u, v), \frac{\partial z}{\partial v} (u, v)]

The span of the tangent vectors is the tangent plane of a surface

Norm and Tangent Plane of a Surface

if $ϕ$ is regular at $(u_{0}, v_{0})$ the norm of the tangent plane at $u_{0}, v_{0}$ is
$n = T_{u} \times T_{v}$
There tangent plane is therefore the points

[\begin{matrix} x - x_{0}, y - y_{0}, z - z_{0} \end{matrix}] \cdot n = 0

Def: regular differential surface

A differential surface $ϕ (u, v)$ is regular at $(u_{0}, v_{0})$ if
$T u \times T v \neq \vec{0}$

Smooth/Regular if this is true for all $(u, v)$

Area of a Surface

Define $ϕ (u, v) st ϕ (D) = S$

A (D) = \int \int_{D} | | T_{u} \times T_{v} | | d u d v

if we have $z = f (x, y)$

Surface Area = \iint_{D} \sqrt{1 + {(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2}} d A

A (S) = \iint_{D} ∥ \nabla F ∥ d A .

Integral of a Function Over a Surface

We have a surface $S$ as a real-valued continuous function defined on $S$
The integral of $f$ over $S$ is

\int \int_{S} f (x, y, x) d S = \int \int_{S} f d S = \int \int_{D} f (ϕ (u, v)) | | T_{u} \times T_{v} | | d u d v

we can think of this as summing over up small areas:

| | T_{u} \times T_{v} | | = | | T_{v} | | | | T_{v} | | \sin (θ)

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Example:

Let $S$ be the surface determined by the graph of the function $f : [0, 1] \times [- 1, 1] \to R .$ defined by $f (x, y) = x^{2} + y .$

$(u, v) \overset{ϕ}{\mapsto} {[\begin{matrix} u, v, u^{2} + v \end{matrix}]}^{T}$

$T_{u} \times T v = [- 2 u, - 1, 1] ⟹ | | T_{u} \times T_{v} | | = \sqrt{2 + 4 u^{2}}$

$\int \int_{D} x d S = \int_{- 1}^{1} \int_{0}^{1} u \sqrt{2 + 4 u^{2}} d u d v = \frac{1}{6} (6^{3 / 2} - 2^{3 / 2})$

Integral Over Graphs

$ϕ (x, y, x) = z - g (x, y) = 0$

\int \int_{S} f (x, y, g (x, y)) d S = \int \int_{D} f (x, y, g (x, y)) (\frac{| | N | |}{N \cdot \hat{k}}) d x d y

where $N = - \frac{\partial g}{\partial x} \hat{i} - \frac{\partial g}{\partial y} \hat{i} + \hat{k}$

so $d S = \frac{d x d y}{n \cdot \hat{k}}$

Integral of a Vector Field Over a Surface

$F : R^{3} \to R^{3}$

$\int \int_{ϕ} F \cdot d S = \int \int_{D} F (ϕ (u, v)) \cdot (T_{u} \times T_{v}) d u d v$

Stoke's Theorem

Line integral of a boundary of a surface in $R^{3}$ . $F : R^{3} \to R^{3}$

Let S be an oriented surface defined by a one-to-one parametrization : $D \subset R^{2}$ → S, where D
is a region to which Green’s theorem applies. Let $\partial S$ denote the oriented boundary
of S and let F be a $C_{} 1$ vector field on S.

** $\int_{\partial S} F \cdot d s = \int \int_{S} curl (F) d S = \int \int_{S} (\nabla \times F) d S = \int \int_{D} \nabla \times f (ϕ (u, v)) \cdot (T_{u} \times T_{v}) d u d v$

If S has no boundary, and this includes surfaces such as the sphere, then the integral
on the left is zero

Complete Formula for Implicit Surfaces

$F (x, y, z) = 0$

After choosing a convenient projection, the surface integral becomes

\int_{S} (\nabla \times F) \cdot n d S = \int_{Projected Domain} (\nabla \times F) \cdot \frac{\nabla F}{∥ \nabla F ∥} d A

Example:

$\int_{c} F \cdot d S = \int_{c} - y^{3} x d x + y^{3} d y - z^{3} d z$

$C$ is the intersection of $x^{2} + y^{2} = 1 and x + y + z = 1$

So $C$ is the intersection of a cylinder and a plane

We want $ϕ : D \subset R^{2} \to R^{3}$
we can find this graph simply as $z = 1 - x - y$
So $f (x, y) = 1 - z - y$ and $ϕ$ = Graph( $D$ ) = $(x, y, f (x, y))$

$D = {(x, y) : x + y \leq 1}$

$ϕ (x, y) = [\begin{matrix} x \\ y \\ 1 - x - y \end{matrix}] .$

$F (x, y, z) = [\begin{matrix} - y^{3} x \\ y^{3} \\ - z^{3} \end{matrix}]$ curl( $F$ ) = $[\begin{matrix} 0 \\ 0 \\ 3 x^{2} + 3 y^{2} \end{matrix}]$ we get $\int \int_{x^{2} + y^{2} \leq 1} [\begin{matrix} 0 \\ 0 \\ 3 x^{2} + 3 y^{2} \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] d x d y$

using polar co-ordinates (Change of Variables) we get:

$x = r \cos θ$
$y = r \sin θ$

$\int_{0}^{1} \int_{0}^{2 π} 3 r^{2} r d θ d r = \frac{3 π}{2}$

Idea of Stokes's Theorem uses chain rule and green's theorem.
S is a graph, we want a curve in a plane to integrate over. $S = [\begin{matrix} x \\ y \\ z (x, y) \end{matrix}]$

\begin{aligned} \int_{\partial S} F_{1} d x + F_{2} d y + F_{3} d z = \int_{\partial S} (F_{1} \frac{d x}{d t} + F_{2} \frac{d y}{d t} + F_{3} \frac{d y}{d t}) d t \\ d z = \frac{d z}{d x} \frac{d x}{d t} + \frac{d z}{d y} \frac{d y}{d t} \\ = \int_{\partial D} (F_{1} + F_{3} \frac{\partial z}{\partial x}) \frac{d x}{d t} + (F_{2} + F_{3} \frac{\partial z}{\partial y}) \frac{d y}{d t} = \int \int_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial x}) d A \end{aligned}