We take a path in the P that can be divided into a finite number of pieces each of which resembles a straight line at small distances, and choose a differentiable scalar field f defined on and around it.
We prove the fundamental theorem in this context by the same approach as used in the previous cases. We first consider a very small piece of the path integrated over, which for a differentiable curve looks like a straight line.
The change of f between the endpoints of the line segment that makes up an infinitesimal piece of C is the directional derivative of f in the direction of the line C resembles, multiplied by ds, the length of the segment.
If that segment has unit tangent vector T, then this change is given by (Tf)ds, and we have
f(r + Tds) - f(r) = ds(Tf).
If we sum this statement over the entire path P, we get
This means that the integral of the component of the gradient of f along the path will give the change in f or difference between its values at the front and back ends of P.
And this is the form that the fundamental theorem of calculus takes for line or path integrals in several dimensions.
In Chapter 23 we will show how to make a path integral into an ordinary integral, to which the tools for finding anti-derivatives can be applied.
The fundamental theorem here gives us an easier approach to evaluating definite integrals when the integrand multiplied by ds has the form (Tf)ds which can also be written as (dsf). You just evaluate f at the ends of the path and take their difference!
Notice that this means such integrals do not depend on the path you use, just on the endpoints.
To use this result effectively, you must develop some idea of what vector fields can be recognized as gradients of scalar fields.
One way to do this is to find the gradients of various scalar fields, so you will be familiar with them.
In the next Chapter we will find a test you can use to determine whether your integral has the form appropriate for using this theorem.
Not only that, we will find a formula for relating the integral from a to b along one path to its integral along another, when the integrand does not have the form of a gradient dotted with ds.
Remember that you may be able to recognize part of your integral as a gradient, leaving you with a relatively easy part that requires reduction to ordinary integration.
Find the integral of over
a path from (0, 0, 0) to (1, 2, 3) that goes from the origin up the x-axis to
x = 1, then parallels the y-axis to y =2 and then parallel to the z-axis to
z = 3.
(Hint: ignore the path and use the fundamental theorem.)
21.6 Evaluate the gradients of the following fields: