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Line integrals (10/22/04)
F r  :vector function of position in 3 dimensions.
C r  :space curve
With each point P is associated a differential distance vector

 dr
dr  ds  uˆds
ds
Definition of the line integral ofF r along space curve C, from point P1 to P2.


C F  dr  C F  uˆ ds
Example: F r  represents the force on a particle then the line integral
represents the work done by the force in moving the particle from point P1 to P2.)
If
F  P( x, y, z )iˆ  Q( x, y, z ) ˆj  R( x, y, z )kˆ
and
r  xiˆ  yjˆ  zkˆ; dr  dxiˆ  dyjˆ  dzkˆ
then
 
C F  dr  C Pdx  Qdy  Rdz 
In general, the line integral will depend on the path that is taken
There are special physical cases (conservative fields) for which the
line integral is independent of path.
The line integral around a closed loop  F  ds is called the circulation of the vector
.
Example:
ˆ
ˆ  xzk
Find the work for a particle taken through the force field F  yziˆ  xyj
from (0,0,0) to (1,1,1) along the curve C defined by:
x  y2,
x  1,
z  0 along the xy plane from (0,0,0) to (1,1,0)
y  1,
perpendicular to the xy plane from (1,1,0) to (1,1,1)
Take the force in Newtons and the distances to be in meters.
y
(1,1,0)
(1,1,1)
x
z
F  yziˆ  xyjˆ  xzkˆ
For the first part of the path, relate the distances and differential to each another.
x  y 2 , z  0;
dx  2 ydy, dz  0
 
F  dr  yzdx  xydy  xzdz  0  y3dy  0
(1,1,0) 

 1 3
F  dr   y dy  1 / 4
(0,0,0)
0
For the second part of the curve, x=y=1, dx=dy=0.
(1,1,1) 
 1
 F  dr   zdz  1/ 2
(1,1,0)
0
The final integral is
(1,1,1)
 F  dr  1/ 4 1/ 2  3 / 4 Joules
( 0,0,0)
What if the path were the straight line from (0,0,0) to (1,1,1)?
Then the curve is described by x=y=z, and dx=dy=dz
2ˆ
2ˆ
2ˆ
F  z i  z j  z k;
1
2
 F  dr   3 z dz  1
0
In this example the integral depended on the path chosen.
This is an example of what is known as a non-conservative force.
Special case - Conservative forces
The vector field can be expressed as the gradient of a scalar field
 
F  
 
 ˆ  ˆ
F
iˆ 
j
k
x
y
z
The line integral between two points will not depend on the path taken.
(This can be a representation of a force due to a scalar potential field.)
 F  dr  



dx 
dy 
dz   d  ( P2 )  ( P1 )
x
y
z
The line integral depends only on start and finish point.
If the integral is a closed loop, then
.
   dr  0
Green’s and Stokes theorems
1. Green’s theorem
Let
Q( x, y)
be a function with continuous first partial derivatives in a certain region.
Consider a small rectangle in the (x,y) plane
Surface integral:
1.
d
A
d b Q
d
Q
dxdy   
dxdy   Q(b, y )  Q(a, y )  dy

A x
c a x
c
c
a
Counter-clockwise line integral:
b
 Q( x, y)dy
Only vertical lines contribute since dy – horizontal = 0
2.
d
 Q( x, y )dy   Q(b, y )  Q(a, y )  dy
c
Comparing 1 and 2 gives:
d
Q
dxdy   Q(b, y )  Q(a, y )  dy   Q( x, y )dy


x
A
c
Let
P( x, y)
be a function with continuous first partial derivatives in a certain region.
Repeat the above but interchanging the x and y operations.
b
P
 dxdy    P( x, b)  P( x, a)  dx   P( x, y)dx
A y
a
Combining
 Q P 

 
 dxdy   Q( x, y )dy   P( x, y )dx
y 
A  x
Let P and Q be the x and y components of a vector
F  Piˆ  Qjˆ  Fxiˆ  Fy ˆj
 Fy Fx 

 
 dxdy   Fx ( x, y)dx   Fy ( x, y)dy   F  dr

x

y
A

 (  F ) kˆ dA   F  dr
A
 (  F ) kˆ dA   F  dr
A
This is also true for an arbitrary 3-d surface bounded by a curve
 (  F ) nˆ dA   F  dr
A
Stokes Law
Why is this important in physics applications?
Stay tuned.