Notes - Development of the Line Integral
In the first 3 animations we use the vector field F = -y i + x j.
In the first animation the path is the unit square.
In the second animation the path starts at the origin, The particle moves along the parabola
y = x2 to the point (1,1) then moves back to the origin along y = x.
In the third animation the path is a triangle with vertices at (0,0), (1,0), and (1/2,1/2).
In the animations 4 through 6 we change the vector field to F
= y i + x j but use the
same paths as in the first 3.
Animation 1 - Unit Square
Animation 2 - Parabolic Path
Animation 3 - Triangular Path
Animation 4 - Unit Square
Animation 5 - Parabolic Path
Animation 6 - Triangular Path
Solutions to Animations 1-3
Solutions to Animations 4-6
In the previous examples we proceeded as if the parameterization of a particular
curve was unimportant. The following download demonstrates that indeed this is the case.
In this download we also discuss what happens if we reverse the orientation along a given path.
Notes - Independence of Parameterization
Path Independence and Conservative Vector Fields
Show that if F is a force field with constant magnitude k pointing outward from the
then the work done as a particle travels along the smooth curve y = f(x) as x varies
from a to b
is k ((b2 +f(b)2)1/2 -
Solution to Radial Force Problem
A 3-D example using direct computation and Stoke's Thm
Another Example in 3-D with direct computation and Stoke's Theorem
For more examples involving Stokes Theorem
see the page on Flux Integrals