APEX Calculus: for University of Lethbridge

These integrals are known as line integrals over vector fields. By contrast, the line integrals we dealt with in Section 16.1 are sometimes referred to as line integrals over scalar fields. Just as a vector field is defined by a function that returns a vector, a scalar field is a function that returns a scalar, such as \(z = f(x,y)\text{.}\) We waited until now to introduce this terminology so we could contrast the concept with vector fields.

We formally define this line integral, then give examples and applications.

In Definition 16.3.2, note how the dot product \(\vec F \cdot \vec T\) is just a scalar. Therefore, this new line integral is really just a special kind of line integral found in Section 16.1; letting \(f(s) = \vec F(s)\cdot \vec T(s)\text{,}\) the right-hand side simply becomes \(\int_C f(s)\, ds\text{,}\) and we can use the techniques of that section to evaluate the integral. We combine those techniques, along with parts of Equation (16.3.1), to clearly state how to evaluate a line integral over a vector field in the following Key Idea.

An important concept implicit in this Key Idea: we can use any continuously differentiable parametrization \(\vrt\) of \(C\) that preserves the orientation of \(C\text{:}\) there isn’t a “right” one. In practice, choose one that seems easy to work with.

Notation note: the above Definition and Key Idea implicitly evaluate \(\vec F\) along the curve \(C\text{,}\) which is parametrized by \(\vrt\text{.}\) For instance, if \(\vec F = \langle x+y, x-y\rangle\) and \(\vrt = \langle t^2,\cos t\rangle\text{,}\) then evaluating \(\vec F\) along \(C\) means substituting the \(x\)- and \(y\)-components of \(\vrt\) in for \(x\) and \(y\text{,}\) respectively, in \(\vec F\text{.}\) Therefore, along \(C\text{,}\) \(\vec F = \langle x+y,x-y\rangle = \la t^2+\cos t, t^2-\cos t\ra\text{.}\) Since we are substituting the output of \(\vrt\) for the input of \(\vec F\text{,}\) we write this as \(\vec F\big(\vrt\big)\text{.}\) This is a slight abuse of notation as technically the input of \(\vec F\) is to be a point, not a vector, but this shorthand is useful.

We use an example to practice evaluating line integrals over vector fields.

Line integrals over vector fields share the same properties as line integrals over scalar fields, with one important distinction. The orientation of the curve \(C\) matters with line integrals over vector fields, whereas it did not matter with line integrals over scalar fields.

It is relatively easy to see why. Let \(C\) be the unit circle. The area under a surface over \(C\) is the same whether we traverse the circle in a clockwise or counterclockwise fashion, hence the line integral over a scalar field on \(C\) is the same irrespective of orientation. On the other hand, if we are computing work done by a force field, direction of travel definitely matters. Opposite directions create opposite signs when computing dot products, so traversing the circle in opposite directions will create line integrals that differ by a factor of \(-1\text{.}\)

We demonstrate using these properties in the following example.

We are preparing to make important statements about the value of certain line integrals over special vector fields. Before we can do that, we need to define some terms that describe the domains over which a vector field is defined.

A region in the plane is connected if any two points in the region can be joined by a piecewise smooth curve that lies entirely in the region. In Figure 16.3.19, sets \(R_1\) and \(R_2\) are connected; set \(R_3\) is not connected, though it is composed of two connected subregions.

A region is simply connected if every simple closed curve that lies entirely in the region can be continuously deformed (shrunk) to a single point without leaving the region. (A curve is simple if it does not cross itself.) In Figure 16.3.19, only set \(R_1\) is simply connected. Region \(R_2\) is not simply connected as any closed curve that goes around the “hole” in \(R_2\) cannot be continuously shrunk to a single point. As \(R_3\) is not even connected, it cannot be simply connected, though again it consists of two simply connected subregions.

We have applied these terms to regions of the plane, but they can be extended intuitively to domains in space (and hyperspace). In Figure 16.3.20.(a), the domain bounded by the sphere (at left) and the domain with a subsphere removed (at right) are both simply connected. Any simple closed path that lies entirely within these domains can be continuously deformed into a single point. In Figure 16.3.20.(a), neither domain is simply connected. A left, the ball has a hole that extends its length and the pictured closed path cannot be deformed to a point. At right, two paths are illustrated on the torus that cannot be shrunk to a point.

We will use the terms connected and simply connected in subsequent definitions and theorems.

Recall how in Example 16.3.7 particles moved from \(A = (-1,1)\) to \(B = (1,1)\) along two different paths, wherein the same amount of work was performed along each path. It turns out that regardless of the choice of path from \(A\) to \(B\text{,}\) the amount of work performed under the field \(\vec F = \la y, x\ra\) is the same. Since our expectation is that differing amounts of work are performed along different paths, we give such special fields a name.

When \(\vec F\) is a conservative field, the line integral from points \(A\) to \(B\) is sometimes written as \(\int_A^B\vec F\cdot d\vec r\) to emphasize the independence of its value from the choice of path; all that matters are the beginning and ending points of the path.

How can we tell if a field is conservative? To show a field \(\vec F\) is conservative using the definition, we need to show that all line integrals from points \(A\) to \(B\) have the same value. It is equivalent to show that all line integrals over closed paths \(C\) are 0. Each of these tasks are generally nontrivial.

There is a simpler method. Consider the surface defined by \(z = f(x,y) = xy\text{.}\) We can compute the gradient of this function: \(\nabla f = \la f_x, f_y\ra = \la y, x\ra\text{.}\) Note that this is the field from Example 16.3.7, which we have claimed is conservative. We will soon give a theorem that states that a field \(\vec F\) is conservative if, and only if, it is the gradient of some scalar function \(f\text{.}\) To show \(\vec F\) is conservative, we need to determine whether or not \(\vec F = \nabla f\) for some function \(f\text{.}\) (We’ll later see that there is a yet simpler method). To recognize the special relationship between \(\vec F\) and \(f\) in this situation, \(f\) is given a name.

We now state the Fundamental Theorem of Line Integrals, which connects conservative fields and path independence to fields with potential functions.

We practice using this theorem again in the next example.

The Fundamental Theorem of Line Integrals states that we can determine whether or not \(\vec F\) is conservative by determining whether or not \(\vec F\) has a potential function. This can be difficult. A simpler method exists if the domain of \(\vec F\) is simply connected (not just connected as needed in the Fundamental Theorem of Line Integrals), which is a reasonable requirement. We state this simpler method as a theorem.