]> 23.1 Making a Line Integral into an Ordinary Integral

23.1 Making a Line Integral into an Ordinary Integral

We seek to evaluate an integral along a path, P , of the dot product w ( x , y , z ) · d l .

To do this we first need a description of the path; (assuming we cannot recognize w as a gradient.)

There are in general three ways to define a path; the one most conducive to evaluating the integral is the parametric representation.

Alternatively the path can be defined qualitatively, (for example: "It is a circle centered about the point ( a , b , c ) of radius d ") or it can be defined by an equation (in two dimensions) or by two equations (in three dimensions).

When the path is defined qualitatively you must provide a parametric representation yourself.

Normally we confine our attention to relatively simple paths, that consist of parts that are arcs of circles or straight lines or portions of a conic section (ellipse, parabola or hyperbola) and there are standard parametric representations of these.

When the path is defined by two equations, you can try to solve these equations for two of the variables x , y and z (or perhaps other variables like those of spherical coordinates) in terms of the third, which can serve as the parameter. The difficulty of this task depends on the nature of the equations, and we will not discuss it further here.

We will assume here that we start with a parametric representation. This means that we have a parameter s (which often represents time t or can often be chosen to be one of the variables x , y or z ) and formulae which give x , y and z as functions of s on P . These can be written as ρ ( s ) , or as ( x ( s ) , y ( s ) , z ( s ) ) , as x ( s ) i ^ + y ( s ) j ^ + z ( s ) k ^ , or we can be given expressions for each of the three components here.

An example is the helix

x ( s ) = cos s , y ( s ) = sin s , z ( s ) = s

In the current notation, s is an arbitrary parameter, not arc-length along the path.

The curve P will also have a beginning parameter value s 0 and an end one, s 1 .

The vector w is presumably given to us as a function of x , y and z , which we can convert into a function of s by using the definitions of x , y and z as functions of it.

Our plan is to reduce w ( x , y , z ) · d l to an integral of something ds from s 0 to s 1 , and we do so as follows

P w ( x , y , z ) · d l = s = s 0 s 1 ( w x d x d s + w y d y d s + w z d z d s ) d s = s = s 0 s 1 w · d l d s d s

Now w ( x , y , z ) · d l can be written out as w x d x + w y d y + w z d z , and we can replace d x by d x d s d s with similar replacements for y and z . The middle expression here is an ordinary integral and is the reduction sought here.

So how do we construct it?

We perform the following procedure:

Step 1: Express x , y and z in terms of s . That is, write down the parametric representation of P .

Step 2: Differentiate each of x , y and z with respect to the parameter s .

Step 3: Take the dot product of the resulting vector with w ( x ( s ) , y ( s ) , z ( s ) ) .

Step 4: Integrate the resulting function with respect to s from s 0 to s 1 .