\magnification 1900
\parskip 10pt
\parindent 0pt
\hoffset -0.4truein
\hsize 7truein

\voffset -0.3truein
\vsize 10truein

\def\emph{}

\def\S{\Sigma_{i=1}^n}
\def\Sm{\Sigma_{i=1}^m}

\def\s{{\sigma}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\h{\hskip 10pt}
\def\u{\vskip -8pt}
\def\bh{\hfil \break}




\def\N{{\bf N}}
\def\S{\Sigma_{i=1}^n}

\def\0{{\bf 0}}
\def\Z{{\bf Z}}
\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\a{{\bf a}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\p{\psi}
\def\f{\phi}
\def\s{\sigma}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{ \vskip -5pt }
\def\h{\vskip -5pt \hskip 20pt}


Scalar Line Integrals:

Let $\x: [a, b] \rightarrow \R^n$ be a $C^1$ path.  $f:  \R^n \rightarrow R$, a 
scalar field.


$\Delta s_k = $ length of kth segment of path 

\centerline{= $\int_{t_{k-1}}^{t_k} ||\x'(t)||dt =
||\x'(t_k^{**})||(t_k - t_{k-1}) = ||\x'(t_k^{**})||\Delta t_k$  }

 \rightline{for some $t_k^{**} \in [t_{k-1}, {t_k}] $}

$\int_\x f~ds \sim \S f(\x(t_k^*))\Delta s_k = \S f(\x(t_k^*)) 
||\x'(t_k^{**})||\Delta 
t_k$  
\f
\centerline{Thus $\int_\x f~ds = \int_a^b f(\x(t))||\x'(t)|| dt$}
\vfill

\vskip 5pt
\hrule

Vector Line integrals:

Let $\x: [a, b] \rightarrow \R^n$ be a $C^1$ path.  $F:  \R^n \rightarrow R^n$, a
vector field.

$\x'(t_k^{*}) \sim { \Delta \x_k \over  \Delta  t_k}$

$\int_\x F \cdot ds \sim \S F(\x(t_k^*)) \cdot \Delta \x_k = \S F(\x(t_k^*))  \cdot
\x'(t_k^{*}) \Delta  t_k$
\f
\centerline{Thus $\int_\x F \cdot ds = \int_a^b F(\x(t)) \cdot \x'(t) dt$}

\vfill

\eject
Other Formulations of Vector Line integrals:
\vskip -5pt
The tangent vector to $\x$ at $t$ is $T(t) = {\x'(t) \over ||\x'(t)||}$


$\int_\x F \cdot ds
 = \int_a^b F(\x(t)) \cdot \x'(t) dt 
 = \int_a^b F(\x(t)) \cdot \x'(t)  {||\x'(t)|| \over ||\x'(t)||} dt 
 = \int_a^b F(\x(t)) \cdot T(t) ||\x'(t)||  dt 
 = \int_\x F(\x(t)) \cdot T(t) ds
$

Note  $\int_a^b F(\x(t)) \cdot T(t) ds $ is a scalar line integral of the scalar 
field $F \cdot T:  \R^n \rightarrow \R$ over the path $\x$.
\f
\line{Note:  $F \cdot T$ is the tangential component of $F$ along the path $\x$.}

\vskip 5pt
\hrule
\vskip -5pt

Another notation (differential form):
\vskip -5pt
For simplicity, we will work in $\R^2$, but the following generalizes to any 
dimension.

Let $\x(t) = (x(t), y(t))$.  Let 
$F(x, y) = (M(x, y), N(x, y) )$

$x = x(t), y = y(t)$.  Also $x'(t) = {dx \over dt}$,  $y'(t) = {dy \over dt}$

\vskip -3pt

$$\int_\x F \cdot ds = \int_a^b F(\x(t)) \cdot \x'(t) dt$$
$$= \int_a^b  (M(x, y), N(x, y) ) \cdot  (x'(t), y'(t)) dt$$
$$= \int_a^b  (M(x, y), N(x, y) ) \cdot  (x'(t)dt, y'(t)dt)$$
$$= \int_\x  (M(x, y), N(x, y) ) \cdot  (dx, dy)$$
$$= \int_\x  M(x, y) dx +  N(x, y) dy$$
%%$$= \int_a^b  (M(x, y), N(x, y) ) \cdot  (x'(t), y'(t)) dt$$
%%$$= \int_a^b  M(x, y)x'(t)dt +  N(x, y) y'(t) dt$$
%%$$= \int_\x  M(x, y) dx +  N(x, y) dy$$

\eject

Definitions:


A {\it curve} is the image of  piecewise $C^1$ path $\x: [a, b] \rightarrow \R^n$.

\vfill

A curve is {\it simple} if it has no self-intersections;  that is, $\x$ is 1:1 on the open interval 
$(a, b)$

\vfill

A path is {\it closed} if $\x(a) = \x(b)$

A curve is {\it closed} if there exists a parametrization of the curve such that $\x(a) = \x(b)$

\vfill

$\int_\x F \cdot ds$ is called the {\it circulation} of $f$ along $\x$ if $\x$ is a closed path.

\eject
A {\it parametrization} of a curve $C$ is a path whose image is $C$.  Normally we will require a 
parametrization of a curve to be 1:1 where possible.


\vskip 5pt
\hrule
A piecewise $C^1$ path $\y: [c, d] \rightarrow \R^n$ is a
 {\it reparametrization} of a  piecewise $C^1$ path $\x: [a, b] \rightarrow \R^n$ if there exists a 
bijective $C^1$ function $u:  [c, d] \rightarrow [a, b]$ where the inverse of $u$ is also $C^1$ and 
$\y = \x \circ u$ (i.e., $\y(t) = \x(u(t))$).


Note that either
\bh
1.)  $u(a) = c$ and $u(b) = d$. 
 In this case, we say that $\y$ (and $u$ are orientation-preserving OR

\u
2.)  $u(a) = d$ and $u(b) = c$.
 In this case, we say that $\y$ (and $u$ are orientation-reversing.

\vskip 5pt
\hrule


Given  piecewise $C^1$ path, 
 $\x: [a, b] \rightarrow \R^n$, the {\it opposite path} is $
\x_{opp}: [a, b] \rightarrow \R^n$ 
$\x_{opp} = \x(a + b - t)$ 

That is $\x_{opp}$ is an orientation-reversing  reparametrization of $\x$ where $u [a, b] \rightarrow [a, b]$, $u(t) = a + 
b - t$.


\vskip 5pt
\hrule

Thm:  Let $\x: [a, b] \rightarrow \R^n$ be a  piecewise $C^1$ path and let $\y: [c, d] \rightarrow \R^n$ be a reparametrization of $\x$.  Then 

if $f: \R^n \rightarrow \R$ is continuous,  then $\int_\y f~ds = \int_\x f~ds$ 

if $F: \R^n \rightarrow \R^n$ is continuous,  then

~~~~~~~$\int_\y F \cdot ds = \int_\x F \cdot ds $ if $\y$ is orientation-preserving.

~~~~~~~$\int_\y F \cdot ds = -\int_\x F \cdot ds $ if $\y$ is orientation-reversing.




\end









\end

Given $\nabla f = ( {\partial f \over \partial x}, {\partial f \over 
\partial y})$, find $f({\bf p}) - f({\bf q})$.

%%Suppose $\Delta f = (c_1, c_2)$, a constant gradient field.