The function F( x, y ) = < M(x, y), N(x, y) > is a vector field over a region in the x-y plane. The function F( x, y, z ) = < M( x, y, z ), N( x, y, z ), P( x, y, z ) > is a vector field over some solid in space. Each vector field assigns a vector to each point in the region or solid. The gradient of some function f( x, y ) is an example of a vector field in a plane since gradient of f( x, y ) = < fx( x, y ), fy( x, y ) >. Each point in the plane is assigned a vector pointing in the direction of maximum change of the function z = f( x, y ). Likewise the gradient of a function w = f( x, y, z ) is a vector field over 3-space. We could plot a vector field by plotting the vectors at every integer point, as in the example of the gradient of f( x, y ) = x2 + y2.
More usual, however, is to plot those vectors having the same length as in a contour plot. For example, given the vector field F( x, y ) = < x, 3y >, we could find all those vectors having length 1, plot them, then find all those having length 2, etc. We can simplify matters by having Mathematica do the solving for us with the Solve function to obtain the following plot of the vector field:
Be sure that you see why we have ellipses here. If this vector
field F( x, y ) = < x, 3y > is the gradient for some
function z = f( x, y ) then F( x, y ) is said to be
conservative and the function f( x, y ) is its potential function.
The test for a conservative vector field of F( x, y ) = < M( x, y
), N( x, y ) > is to see if 
.
Since M( x, y ) = x and N( x, y ) = 3y, each partial derivative = 0
and the vector field F( x, y ) = < x, 3y > is indeed
conservative. This means there exists a potential function z = f( x,
y ) that has F( x, y ) as its gradient. How do we find this
potential function? If the gradient of f( x, y ) = < x, 3y >,
then fx( x, y ) = x and fy( x, y ) = 3y. By
integrating fx( x, y ) with respect to x we get f( x, y )
= x2/2 + C( y ) since y is a constant here, and
integrating fy( x, y ) with respect to y, we obtain f( x,
y ) = 3y2/2 + C( x ). Setting these two equal gives us the
potential function f( x, y ) = x2/2 + 3y2/2.
Verify that the gradient of f( x, y ) is the vector field F(
x, y ) = < x, 3y >.
In Mathematica we could do the following to find the potential function for the vector field F( x, y ) = < x + y, x - y >:
Mathematica returns True, proving that the vector field F( x, y ) is conservative and has a potential function. Let's find it.
Div and Curl
For vector fields of three variables the situation is a bit more complex as you would imagine. First the definitions of divergence ( div ) and curl of a vector field F( x, y , z ).
In this context the gradient is considered a differential operator that "operates" on M, N and P.
The curl F( x, y, z ) = < 0, 0, 0 > if and only if F( x, y, z ) is conservative.
Line integrals
The definite integral in Calculus I was defined to be an infinite sum of rectangles with width x and height f( xi ). These rectangles extended to the x-axis in the case of dx or to the y-axis for dy. The double integral of Calculus III was defined over a region in the x-y plane, while the triple integral was defined over a solid in 3-space. The line integral can be considered an infinite sum of rectangles drawn to a curve or path. These rectangles would have a width of Ds, where Ds is a piece of the arc length along the path. If the function f( x, y , z ) represents density of a wire at some point (x, y, z ) along the path, then the line integral would define the mass of the wire. The line integral may be defined in 2-space or 3-space.
For example, if f( x, y ) = x + y along the path y = 
from ( 0, 0 ) to ( 1, 1 ) and back
again along y = x2,
This is going from ( 0, 0 ) to ( 1, 1 ) along the square root path. The numeric answer is 1.5. Coming back down:
This also gives 1.5. Actually the line integral is independent of the parameterization except for orientation.
Line Integral over a vector field
Line integrals may be evaluated over a vector field along a path determined by a vector-valued function. Recall that work ( chapter 6 ) done by a variable force along the x-axis is the integral of F dx. By using the line integral to sum up force vectors projected on the tangent vectors along the path taken, we can evaluate the work done when moving a particle through a vector (or force ) field.
This can be extended to vector fields in 3-space as well. Let's
find the work done in moving a particle in the force field defined by
over the circular path defined by
.
Please note that the force field,
f, is defined in terms of t.
Verify that the work done is exactly 0.
The line integral over a vector field is sometimes written in
differential form, 
.
As afinal example, let's find the work done in moving a particle over the vector field < x, y, 2 > along the helix, r( t ) = < cos( t ), sin( t ), t >.
