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6.3  Vector Line integrals:

Calc 1 review: Suppose $f:  \R \rightarrow \R$, 


Fundamental Theorem of Calculus:  $\int_a^b f'(t) dt = f(b) - f(a)$.

$\int_a^b $(velocity) $dt =$ distance traveled.

$\int_a^b $(rate of change)$ dt = $ total change.


Given $f'(t)$, can find $f(b) - f(a)$.

Given velocity, can find distance traveled.

Given rate of change, can find total change.

Calc III:  Suppose $f:  \R^2 \rightarrow \R$, 

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

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

\eject

6.1 vector field integral review:

Let $F(x, y) = (x, y)$, let $x(t) = $

{\bf Notation 1:  Work definition}

$\int_C F \cdot ds = \int_{C_1} F \cdot ds + \int_{C_2} F \cdot ds $

$= \int_0^1 F(\x_1(t))\cdot \x_1'(t) dt + 
 \int_0^1 F(\x_2(t))\cdot \x_2'(t) dt$

$= \int_0^1 F(t, 0) \cdot (1, 0) dt + 
 \int_0^1 F(1, t) \cdot (0, 1) dt$

$= \int_0^1 (t,0) \cdot (1, 0) dt + 
 \int_0^1 (1, t) \cdot (0, 1) dt$

$= \int_0^1 t dt + 
 \int_0^1  t dt =  2({1 \over 2}t^2)|_0^1 = {1}$


{\bf Notation 2:  differential form}

$\int_C F \cdot ds = \int_{C_1} F \cdot ds + \int_{C_2} F \cdot ds $

$= \int_{C_1} F \cdot (dx, dy) + \int_{C_2} F \cdot (dx, dy) $

$= \int_{C_1} (x, y) \cdot (dx, dy) + \int_{C_2} (x, y) \cdot (dx, dy) $

$= \int_{C_1} (x dx + y dy) + \int_{C_2} (x dx + y dy) $


$= \int_0^1 t dt  + \int_0^1 (1(0) + t dt) = 2({1 \over 2}t^2)|_0^1 = {1}$



{\bf  Note:} Both of the above methods are algebraically equivalent, so it doesn't matter which notation you use.

\eject

{\bf Notation 3:  Tangent vector (circulation)}


$\int_C F \cdot ds = \int_\x F(x, y) \cdot T(x, y) ds$  where $T$ is the unit tangent vector to the path $\x$.

$\x_1(t) = (t, 0)$, 
$\x_1'(t) = (1, 0)$,  
$||\x_1'(t)|| = 1$.~~  

\rightline{
Thus $T_1(x, y) = (1, 0)$}

$\x_2(t) = (1, t)$, 
$\x_2'(t) = (0, 1)$,  $||\x_2'(t)|| = 1$.  

\rightline{
Thus $T_2(x, y) = (0, 1)$}



$\int_C F \cdot ds 
= \int_{\x_1} F \cdot T_1 ds + \int_{\x_2} F \cdot T_2 ds $  

$= \int_{\x_1} F(x, 0) \cdot (1, 0) ds + \int_{\x_2} F(1, y) \cdot (0, 1) ds$  

$= \int_{\x_1} (x, 0) \cdot (1, 0) ds + \int_{\x_2} (1, y) \cdot (0, 1) ds$  

$= \int_{\x_1} x ds + \int_{\x_2} y ds$ \hfill	 [Note:  these are  scalar 
line 
integrals]

$= \int_0^1 t ||x_1'(t)|| dt  + \int_0^1 t ||x_2'(t)|| dt $

$= \int_0^1 t (1) dt  + \int_0^1 t (1)dt  = 2({1 \over 2}t^2)|_0^1 = {1}$



Note:  The algebra for notation 3 is slightly messier than for notations 1 and 2, thus this notation is normally used only when it is possible to determine circulation geometrically.

%%\vskip 5pt
%%\hrule
%%\vskip -5pt

\eject

Can use any parametrization for the path $\x$.  For example:

$\x_1:[0, 1] \rightarrow \R^2$, $\x_1(t) = (t^2, 0)$

$\x_2:[0, 1] \rightarrow \R^2$, $\x_2(t) = (t^3, 0)$


{\bf Notation 1:  Work definition}

$\int_C F \cdot ds = \int_{C_1} F \cdot ds + \int_{C_2} F \cdot ds $

$= \int_0^1 F(\x_1(t))\cdot \x_1'(t) dt + 
 \int_0^1 F(\x_2(t))\cdot \x_2'(t) dt$

$= \int_0^1 F(t^2, 0) \cdot (2t, 0) dt + 
 \int_0^1 F(1, t^3) \cdot (0, 3t^2) dt$

$= \int_0^1 (t^2,0) \cdot (2t, 0) dt + 
 \int_0^1 (1, t^3) \cdot (0, 3t^2) dt$

$= \int_0^1 2t^3 dt + 
 \int_0^1  3t^5 dt =  {1 \over 2} t^4|_0^1 + {1 \over 2}t^6|_0^1 = {1}$


{\bf Notation 2:  differential form}

$\int_C F \cdot ds = \int_{C_1} F \cdot ds + \int_{C_2} F \cdot ds $

$= \int_{C_1} F \cdot (dx, dy) + \int_{C_2} F \cdot (dx, dy) $

$= \int_{C_1} (x, y) \cdot (dx, dy) + \int_{C_2} (x, y) \cdot (dx, dy) $

$= \int_{C_1} [x dx + y dy] + \int_{C_2} [x dx + y dy] $


$= \int_0^1 t^2 (2t)dt  + \int_0^1 1(0) + t^3 3t^2dt =  {1 \over 2} t^4|_0^1 + {1 \over 2}t^6|_0^1 = {1}$

{\bf  Note:} Both of the above methods are algebraically equivalent, so it doesn't matter which notation you use.

\eject

{\bf Notation 3:  Tangent vector (circulation)}


$\int_C F \cdot ds = \int_\x F(x, y) \cdot T(x, y) ds$  where $T$ is the unit tangent vector to the path $\x$.

$\x_1(t) = (t^2, 0)$, 
$\x_1'(t) = (2t, 0)$,  
$||\x_1'(t)|| = 2t$.

\rightline{Thus $T_1(x, y) = (1, 0)$}

$\x_2(t) = (1, t^3)$, 
$\x_2'(t) = (0, 3t^2)$,  $||\x_2'(t)|| = 3t^2$.  

\rightline{Thus $T_2(x, y) = (0, 1)$}



$\int_C F \cdot ds 
= \int_{\x_1} F \cdot T_1 ds + \int_{\x_2} F \cdot T_2 ds $  

$= \int_\x F(x, 0) \cdot (1, 0) ds + \int_\x F(1, y) \cdot (0, 1) ds$  

$= \int_\x (x, 0) \cdot (1, 0) ds + \int_\x (1, y) \cdot (0, 1) ds$  

$= \int_\x x ds + \int_\x y ds$  \hfill [Note:  these are scalar line 
integrals]


$= \int_0^1 t^2 ||x_1'(t)|| dt  + \int_0^1 t^3 ||x_2'(t)|| dt $

$= \int_0^1 t^2 (2t)dt  + \int_0^1 t^3 3t^2dt =  {1 \over 2} t^4|_0^1 + {1 \over 2}t^6|_0^1 = {1}$


 

Note:  The algebra for notation 3 is slightly messier than for notations 1 and 2, thus this notation is normally  used only when it is possible to determine circulation geometrically.

\vfill

{\bf Method 2:  6.2 Green's Theorem}

\h Curve is not closed, so can't use Green's Theorem.

\eject


{\bf Method 3:  6.3 Claim $F$ has path independent line integrals}

Claim $F$ is a gradient field


{\it submethod 1}:\hfil \break
 $F(x, y) = (x, y) = ( {\partial f \over \partial x}, {\partial f \over \partial y}) = \nabla f = \nabla ({1 \over 2}x^2 + {1 \over 2}y^2)$.

\h I.e, $f(x, y) = {1 \over 2}x^2 + {1 \over 2}y^2$.


\h $\int_C F \cdot ds = f(1, 1) - f(0, 0) =  {1 \over 2} + {1 \over 2} - 0 = 1$.


{\it submethod 2}: Claim $(\nabla \times F)(\x) = \0$ for all $\x \in \R^n$

Since $n = 2$, $(\nabla \times F)(\x) = \0$ for all $\x \in \R^n$
iff ${\partial N \over \partial x} = {\partial M \over \partial y}$, where $F = (M, N)$.

 \h ${\partial y \over \partial x} = 0 = {\partial x \over \partial y}$

Thus can choose any path starting at (0, 0) and ending at (1, 1)

Ex:  Let $\x:[0, 1] \rightarrow \R^2$, $\x(t) = (t, t)$.

$\int_C F \cdot ds  
= \int_0^1 F(\x(t))\cdot \x'(t) dt$


$= \int_0^1 F(t, t)\cdot (1, 1) dt$

$= \int_0^1 (t, t)\cdot (1, 1) dt$

$= \int_0^1 2t dt =  t^2|_0^1 = {1}$




\eject

{\bf Closed curve example}

Suppose $\x:[0, 2\pi]\rightarrow \R^n$, $\x(t) = (cos(t), sin(t))$

{\bf Notation 1:  Work definition}

$\int_C F \cdot ds  
= \int_0^{2\pi} F(\x(t))\cdot \x'(t) dt$ 

$= \int_0^{2\pi} F(cos(t), sin(t))\cdot (-sin(t), cos(t)) dt$ 

$= \int_0^{2\pi} (cos(t), sin(t))\cdot (-sin(t), cos(t)) dt$ 

$= \int_0^{2\pi} [-cos(t)sin(t) +  sin(t)cos(t)] dt$ 

$= \int_0^{2\pi} 0 ~dt = 0$ 

{\bf Notation 2:  differential form}

$\int_C F \cdot ds
= \int_{C} F \cdot (dx, dy) $

$= \int_{C} (x, y) \cdot (dx, dy) $

$= \int_{C} [x dx + y dy]$


$= \int_0^{2\pi} [cos(t) (-sin(t))dt + sin(t)cos(t) dt]$

$= \int_0^{2\pi} 0~ dt = 0$ 

\vfill

{\bf Notation 3:  Tangent vector (circulation)}


Circulation = 0.  Thus 
$\int_C F \cdot ds = 0$



\eject
{\bf Method 2:  6.2 Green's Theorem}

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

\h $\int_C F \cdot ds = \int_C Mdx + Ndy 
 =  \int\int_D (\nabla \times f) \cdot \k ~dA$


\h $
= \int\int_D [{\partial N \over \partial x} - {\partial M \over \partial y}]dA
= \int\int_D [{\partial y \over \partial x} - {\partial x \over \partial y}]dA$
$= \int\int_D 0 dA$


\h Use chapter 5 methods to evaluate double integral.

Note this method is algebraically different than using the definition of vector line integrals. Thus if one method is algebraically difficult, try the other method.

{\bf Method 3:  6.3 Claim $F$ has path independent line integrals}

Claim $F$ is a gradient field




{\it submethod 1:} \hfil \break
 $F(x, y) = (x, y) = ( {\partial f \over \partial x}, {\partial f \over \partial y}) = \nabla f = \nabla ({1 \over 2}x^2 + {1 \over 2}y^2)$.

\h I.e, $f(x, y) = {1 \over 2}x^2 + {1 \over 2}y^2$.


\h $\int_C F \cdot ds = 0$ since C is a closed curve.


{\it submethod 2:} Claim $(\nabla \times F)(\x) = \0$ for all $\x \in \R^n$

Since $n = 2$, $(\nabla \times F)(\x) = \0$ for all $\x \in \R^n$
iff ${\partial N \over \partial x} = {\partial M \over \partial y}$, where $F = (M, N)$.

\h ${\partial y \over \partial x} = 0 = {\partial x \over \partial y}$

\h $\int_C F \cdot ds = 0$ since C is a closed curve.




\end