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23.3 Practical Comments

While the reduction process here is straightforward, there are many details to handle and it is easy to go astray.

Standard errors are the usual careless ones: loss of a factor between steps, and also failure to differentiate the parametric expression for ρ ⟶ before taking the dot product with w ⟶ , as well as misreading or misinterpreting the question, switching addition and multiplication, etc.

What happens when your path P is presented to you as the solution to two equations, or by a qualitative description?

In order to reduce the path integral to a single integral over one variable, you almost always have to be able to express position on the path (the values of x , y and z or some other set of coordinates on the path) as functions of that variable. This usually requires your being able to produce a parametric representation of the path with your variable as parameter from the given information.

When the path consists of pieces that are straight lines or arcs of circles you should be prepared to represent them parametrically from knowing the beginning and end points of the straight line and the center radius and end angles of the arc.

A straight line segment from ( a , b ) to ( c , d ) can be represented by

x = a + s ( c − a ) y = b + s ( d − b )

for 0 < s < 1 .

Exercise: 23.1 State the analogous result in three dimensions.

A circle with radius u centered at ( b , c ) can be described by the parametric representation

x = b + u cos ⁡ s y = c + u sin ⁡ s

Exercise 23.2 What are appropriate limits on s here?

When the path is defined by equations and you are able to solve them to find all variables in terms of one of them or in terms of some other parameter, you can handle the problem as if the information were given parametrically from the start.

Otherwise, you can always find lots of points on the curve numerically and use a numerical approximation to the resulting integral, by means we shall soon describe.

Of course the easiest way to do such an integral is to observe that the vector w ⟶ or a significant part of it is the gradient of a scalar function. The integration then reduces to evaluations at the end points and we do not have to worry about the path at all.

Exercises:

23.3 Integrate the dot product of w ⟶ with d l ⟶ over the helical path above with w ⟶ given by w ⟶ = ( x , y 2 , z 3 ) .

23.4 Integrate this same integral over the straight line path from ( 1 , 0 , 0 ) to ( 1 , 0 , ( 2 π ) 3 ) .

23.5 Integrate the line integral w ⟶ · d l ⟶ with w ⟶ = ( y , − x , sin ⁡ ( e z y ) ) around the unit circle in the x y plane centered about the point ( 5 , 0 , 0 ) . (Hint: use Stokes theorem.)

23.6 You want to find the arc length of the ellipse defined by a x 2 + b y 2 = c . Express its boundary in parametric form, set up the integral and do it.

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