Prob 1.33

This is an interesting problem because it involves a line integral. It is important that we correctly remember how to do line integrals. This was something that I needed to review, so that will be my focus for this problem. We are moving to a higher level of mathematical skill and we must be sure we understand clearly the methods we will be employing to study E&M. I have discover that juniors and seniors do not know how to setup and do integrals.

Consider the line integral of

with F a constant force (magnitude = f) from

A------------B. s = length A to B.

The integral from A to B should be +fs.

The integral from B to A should be -fs.

Before reading further set up this integral defining the vectors and the limits of integration and see what you get. Don't make the definitions so that the sign is correct. Be sure you set up the definitions using your understanding of vector calculus and then see what sign you get.

You must be careful in order to get the sign correct. One pitfall is to define

The correct definition is

In the case of the integral from B to A dx becomes negative based on the integration limits. dl is a general step along a line and the limits determine the sign.

However one might take the approach that dl should be definable as an arbitrary variable when you set up the problem. Then you might put a vector dl in the negative x direction. In this case you are making a change of variables and the relationship between dl and dx provides a second negative because dl=-dx (Based on l=B-x). The combination of the two minus signs cancel.

with l an arbitrary variable introduced to solve the problem.

Let us consider problem 1.33

(dx, dy, dz will be + or - based on the limits of integration.)

We will stick with the dl defined wrt the coordinate system we chose. This definition is fixed as above.

in this case dx=0 and we can either eliminate z in terms of y or vice versa.

case 1: (Start at the right and integrate up and to the left.)

dl goes up dz and the back along y.

We will integrate from z = 0 to z = 2.

case 2: (Start at the right and integrate up and to the left. Note dl indicates that you step in the positive y direction then go down in z. The limits of integration will reverse this direction.)

dl goes along + y and then down in the z direction.

We will integrate from y = 2 to y = 0.

because 2 is the lower limit you get an additional minus sign.

For the area integral