Paths and Surfaces

Paths

Integrals along paths and surfaces look a little different (but not much, and I’m not convinced this needs its own page but the book has these topics in a separate chapter)

Definition: The path integral of $f:{\mathbb{R}}^2\to \mathbb{R}$ along a curve $c$ where $c:I=[a,b]\to {\mathbb{R}}^3$ is defined as ${\int }_{c}fds={\int }_a^{b}f(x(t),y(t),z(t))\Vert c^{\prime }(t)\Vert dt$

Example: Take the curve $c:\left[0,\frac{\pi }{2}\right]\to {\mathbb{R}}^2$, $c(t)=(30{\cos }^{3}t,30{\sin }^{3}t)$, $f(x,y)=1+\frac{y}{3}$
$\Vert c^{\prime }(t)\Vert =\Vert (-90{\cos }^{2}t\sin t,0-{\sin }^{2}t\cos t)\Vert =90\cos t\sin t$
${\int }_0^{\pi /2}\left(1+\frac{30{\sin }^{3}t}{3}\right)0-\cos t\sin tdt={\int }_0^1(1+10u^3)90udu=270$

Definition: Let $F$ be a vector field, $F:U\subset {\mathbb{R}}^3\to {\mathbb{R}}^3$ continuous along $c:[a,b]\to {\mathbb{R}}^3$, then the line integral of $F$ along $c$ is ${\int }_c\vec{F}\cdot d\vec{s}={\int }_a^b(\vec{F}(c(t))\cdot c^{\prime }(t))dt$

Note that $d\vec{s}=d\vec{x}\mathbf{i}+d\vec{y}\mathbf{j}+d\vec{z}\mathbf{k}$, so a line integral can be written as ${\int }_a^b\left(F_1\frac{dx}{dt}+F_2\frac{dy}{dt}+F_3\frac{dz}{dt}\right)dt$

Definition: Let $h:I\to I_1$ be a $C^1$ real-valued function that maps $[a,b]$ onto $[a_1,b_1]$. Let $\mathbf{c}:I_1\to {\mathbb{R}}^3$ be a piecewise $C^1$ path. Then $\mathbf{p}=\mathbf{c}\circ h:I\to {\mathbb{R}}^3$ is a reparameterization of $\mathbf{c}$.

In other words, $p(t)=c(h(t))$

The reparameterization can either be orientation-preserving or orientation-reversing ($b_1$ maps to $a$, $a_1$ maps to $b$)

Theorem: ${\int }_{p}F\cdot ds=\pm {\int }_{c}F\cdot ds$
With path integrals, the direction of the path does not matter, ${\int }_{c}f(x,y,z)ds={\int }_{p}f(x,y,z)ds$

Theorem: Suppose $f:{\mathbb{R}}^3\to \mathbb{R}$ is of class $C^1$ and that $\mathbf{c}:[a,b]\to {\mathbb{R}}^3$ is a piecewise $C^1$ path. Then ${\int }_c\nabla f\cdot ds=f(\mathbf{c}(b))-f(\mathbf{c}(a))$

Surfaces

While it is tempting to use the definition of a graph as a surface $z=f(x,y)$, this is too limiting. We want a definition that allows for donuts (torus).

Definition: Let $S\subset {\mathbb{R}}^3$ be a surface parameterization of $S$ $\Phi :D\subset {\mathbb{R}}^2\to {\mathbb{R}}^3$ such that $\Phi (D)=S$. If $(u,v)\in D$, then $\Phi (u,v)\in S$. If $\Phi$ is differentiable, then $S$ is called differentiable

This is like twisting and transforming around a 2d plane

The parametric equation of a plane $P$ is $\Phi (u,v)=\alpha u+\beta v+\gamma$

The tangent vectors on the surface at a point $(u_0,v_0)$ are denoted as ${\mathbf{T}}_u=\frac{\partial \mathbf{\Phi }}{\partial u}=\frac{\partial x}{\partial u}(u_0,v_0)\mathbf{i}+\frac{\partial y}{\partial u}(u_0,v_0)\mathbf{j}+\frac{\partial z}{\partial u}(u_0,v_0)\mathbf{k}$ and ${\mathbf{T}}_v=\frac{\partial \mathbf{\Phi }}{\partial v}=\frac{\partial x}{\partial v}(u_0,v_0)\mathbf{i}+\frac{\partial y}{\partial v}(u_0,v_0)\mathbf{j}+\frac{\partial z}{\partial v}(u_0,v_0)\mathbf{k}$

The vector $T_u\times T_v$ is normal to the surface at $(u_0,v_0)$

Definition: $S$ is regular or smooth at $\mathbf{\Phi }(u_0,v_0)$ if ${\mathbf{T}}_u\times {\mathbf{T}}_v\ne 0$ at $(u_0,v_0)$. A surface is regular, if all points are regular

Definition: The tangent plane of a surface at $\mathbf{\Phi }(u_0,v_0)$ is given by $(x-x_0,y-y_0,z-z_0)\cdot \mathbf{n}=0$ where $\mathbf{n}={\mathbf{T}}_u\times {\mathbf{T}}_v\ne 0$

Definition: The surface area of a surface $A(S)={\iint }_D\Vert {\mathbf{T}}_u\times {\mathbf{T}}_v\Vert dudv$, which is independent of any specific parameterization

If $S$ is a graph $z=g(x,y)$ then the equation reduces to $A(S)={\iint }_D\sqrt{{\left(\frac{\partial g}{\partial x}\right)}^2+{\left(\frac{\partial g}{\partial y}\right)}^2+1}dA$

Theorem: Say $\Phi :D\to S\subset {\mathbb{R}}^3$, then the integral of the surface over a scalar function $f$ is ${\iint }_{S}f(x,y,z)dS={\iint }_{S}fdS={\iint }_{D}f(\mathbf{\Phi }(u,v))\Vert {\mathbf{T}}_u\times {\mathbf{T}}_v\Vert dudv$

If $S$ is a graph, this equation reduces to ${\iint }_{D}f(x,y,g(x,y))\sqrt{1+{\left(\frac{\partial g}{\partial u}\right)}^2+{\left(\frac{\partial g}{\partial v}\right)}^2}dudv$

Definition: The surface integral of $F$ over $\Phi$ (which parameterizes $S$) is defined by ${\iint }_{\Phi }F\cdot d\vec{S}={\iint }_{D}F\cdot (T_u\times T_v)dudv$

This is related to the scalar surface integral by the equation ${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_S(\mathbf{F}\cdot \mathbf{n})dS$

We want to define ${\iint }_{S}F\cdot d\vec{S}={\iint }_{\Phi }F\cdot d\vec{S}$ regardless of the specific parameterization $\Phi$ (except perhaps for sign)

Definition: An oriented surface has a positive and a negative side. Each point has two normal vectors $n_1=-n_2$ which are each associated with a side.

It follows that the unit normal vector at a point can be defined as $\pm n(\Phi (u,v))=\frac{T_u\times T_v}{\Vert T_u\times T_v\Vert }$. If the sign is positive at all points then $\Phi$ is orientation-preserving, otherwise it is orientation-reversing.