Active Calculus - Multivariable

In Section 13.2, the idea of a line integral was introduced by looking at the work done by a vector field when traveling along \(C\text{,}\) a path in space. In particular, the line integral measured the accumulated amount of the vector field that is along the path \(C\) (in the direction of travel.) Definition 13.2.9 shows how to use a Riemann sum to measure the accumulation of the vector field in the direction of travel along the given curve. Additionally, Theorem 13.3.6 shows how the line integral \(\int_C \vF\cdot d\vr \) can be efficently calculated in terms of the vector field and the derivative of any paramaterization of \(C\text{.}\) Philosophically, the Riemann sum in Definition 13.2.9 is adding up the scalar function given by the dot product of the vector field and the unit vector in the direction of travel (along the curve).

We will devote the rest of this section to answering these questions about generalizing the ideas of line integrals as fully as possible.

In Preview Activity 13.5.1, you approximated the distance traveled for various paths and multiplied by the density of the copium on each piece of the path. In contrast to the line integral of a vector field, the calculations of the ore mined does not depend on what direction the path was traveled. We will now use these same ideas to give precise meaning to the measurement of the acculumation of a scalar function’s output over a path in space.

Let \(f\) be a contiunuous function of \(x\text{,}\) \(y\text{,}\) and \(z\) for some open set around \(C\text{,}\) a curve from a point \(P\) to a point \(Q\text{.}\) We will begin to approximate the acculumlation of the output of \(f\) over \(C\) by breaking \(C\) into pieces with boundary points \(P=\vr_0, \vr_1, \dots, \vr_n-1, \vr_n=Q\text{.}\) The curve \(C_i\) is the part of \(C\) that goes from \(\vr_{i-1}\) to \(\vr_i\) and \(\Delta \vr_i\) is the displacement vector from \(\vr_{i-1}\) to \(\vr_i\)

The notation for a scalar line integral (\(\int_C f \, ds \)) may not immediately make sense. As with the other types of integration we have done (single variable integration, double integrals, line integrals of vector fields, etc.), the subscript of the integral symbol denotes the region of integration. In the case of a scalar line integral, the region of integration is a collection of points given by a curve in space. The function we are integrating is \(f\text{,}\) a scalar-valued functiton of multiple variables. The differential \(ds\) may seem unusual to you. If you remember from Section 10.3, \(s\) is the arc length of a curve in space. So the differential \(ds\) in the scalar line integral notation means that we are adding up the output of \(f\) over steps in arc length. This should make sense in terms of how we set up our Riemann sums. We did not set up the pieces of our curve as steps in \(x\text{,}\) \(y\text{,}\) or \(z\text{,}\) but rather as steps in arc length (estimated by \(|{\vr_{i+1}-\vr_{i}}|\)). This may feel similar to situations such as double integrals, where we generically used \(dA\) for the differential, but the different contexts called for different values of \(dA\text{.}\) For instance, in polar coordinates, we use \(dA = r\, dr\, d\theta\)

Before delving into the exact computation of line integrals of scalar-valued functions, here is an activity that allows you to practice arguments similar to Example 13.5.6.

Definition 13.5.5 defined \(\int_C f \, ds\) in terms of a limit of a Riemann sum which is often useful for understanding what is being measured and not very useful when it comes to efficiently calculating the value of a given integral. A scalar line integral is presented algebraically in terms of three variables because the curve is given in terms of points in three coordinates and the function to be integrated is dependent on those same coordinate values. Geometrically, the scalar line integral is a one dimensional problem because we only have one dimension to travel; namely, we can travel along the curve in steps of arc length. Remember that a parameterization of a curve in space is a description of how to travel through the points (given as three coordinates) of the curve in terms of a parameter (usually given as \(t\text{.}\)) Parameterizations are very useful converting the three-variable algebra of a scalar line integral problem into a one dimensional integral. Once we have done that, we can use all of the tools of single-variable calculus to evaluate the scalar line integral.

Let’s look at applying a parameterization for \(C\) given by a vector-valued function of one variable \(\vr(t)=\langle f(t), g(t), h(t)\rangle\) for \(t\) in some interval \([a,b]\) to Definition 13.5.5. Instead of thinking in terms of pieces of the curve \(C\text{,}\) the parameterization allows us to break the interval \([a,b]\) into pieces \(a=t_0, t_1, \dots ,t_n=b\) where \(t_i=a+i(\Delta{t})\) and \(\Delta{t}=\frac{b-a}{n}\text{.}\) While these pieces will be equally spaced in terms of the parameter \(t\text{,}\) the corresponding points on the curve \(C\) given by \(r(t_i)\) may not be equally spaced.

With Theorem 13.5.11 established, we now consider a couple of examples that allow us to find the exact value of line integrals of scalar functions considered in Activity 13.5.2.

When working through Activity 13.5.2, you may have found the line integrals involving \(f_3(x,y,z)=x-y\) to be more challenging to reason through without computational tools. We next work through the details of evaluating the line integral of this function along the same curve as in the previous example.

In Activity 13.5.3, it may have been tempting to note that output of the function being integrated is simply the \(x\)-coordinate of the points on the curve to try to make an argument by symmetry. Since we are adding up the \(x\)-values of the points on \(C_1\) and \(C_2\) and the \(x\)-coordinates on \(C_1\) go from \(-1\) to \(0\) and the \(x\)-coordinates on \(C_2\) go from \(0\) to \(1\text{,}\) you may have wanted to argue that the values of \(\int_{C_1} x \, ds \) and \(\int_{C_2} x \, ds \) will cancel out to zero. However, as you found, this is not correct because the scalar line integrals are not just adding up the output of our scalar function. The Riemann sum used in the scalar line integral is adding up the product of the scalar function’s output with the displacement of the segment used. Because the displacement used in \(C_1\) and \(C_2\) will not be symmetric, we cannot make the geometric cancelation argument as in part 13.5.6.a.

Before stating some useful properties of scalar line integrals, we will recall some convenient notation from Properties of Line Integrals. If \(C_1\) and \(C_2\) are oriented curves, with \(C_1\) from a point \(P\) to a point \(Q\) and \(C_2\) from \(Q\) to a point \(R\text{,}\) we denote by \(C_1+C_2\) the oriented curve from \(P\) to \(R\) that follows \(C_1\) to \(Q\) and then continues along \(C_2\) to \(R\text{.}\) Also, if \(C\) is an oriented curve, \(-C\) denotes the same curve but with the opposite orientation. The list below summarizes some other properties of line integrals, each of which has a familiar in definite integrals.

The biggest difference bewteen Properties of Line Integrals and Properties of Scalar Line Integrals is part c. The orientation of the curve does not change the value of the scalar line integral. Activity 13.5.4 will have you make sense of these properties for scalar line integrals.

We will spend the last part of this section talking about a way to try to visualize the scalar line integral as an area under a curve, much as we visualized integrals when we first encountered them. Let’s return to our copium analogy from Preview Activity 13.5.1. In particular, we can look the left side of the mining area.

We could visualize the linear density of copium along the left side of the area using a plot like Figure 13.5.18.

In Figure 13.5.18, the horizontal axis gives the distance traveled along the left side of Figure 13.5.1. Because this is a stright path, we could plot the density above the path of the copium mining plot. In fact, we could plot the density above the plot for all of the sides of the mining plot.

Figure 13.5.19 shows the copium mine plot (in gray) and the paths that are the boundary of the plot in magenta. The curve in green shows the copium Density at each point on the boundary of the mine plot. The area in yellow would be the scalar line integral for the path that is the boundary of the mine plot. In particular, the area in yellow would give the total copium mined from driving our mining machine around the boundary of the mine plot.

Because the curve we are looking at in Figure 13.5.19 involves straight lines and simple heights, there is no confusion when looking at this plot and using an area under the curve analogy. However, what if we looked at the scalar line integral of a function like \(f(x,y,z)=x-y\) along the helix given below?

Suppose now that above the points of our green helix we try to plot a second blue curve where the position of the point on the blue curve is \(f(x,y,z)=x-y\) units above or below the position of the point on the helix. This would be the analogous idea to what we did in Figure 13.5.19, but for a three-dimensional curve. Here we run into issues, however, as our area might intersect other parts of the curve. The plot below shows the confusing plot we would have if we looked at as the height above our curve in blue.