Line Integrals of Scalar Functions

In , the idea of a line integral was introduced by looking at the work done by a vector field when traveling along \(C\), 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.) 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, shows how the line integral \(\int_C \vF\cdot d\vr\) can be efficiently calculated in terms of the vector field and the derivative of any parameterization of \(C\). Philosophically, the Riemann sum in 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.

• How can we measure the accumulation of a scalar valued function along a curve in space?
• What would that accumulation measure physically?
• What is the most efficient way to calculate this accumulation?

In , 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 accumulation of a scalar function's output over a path in space. As you might guess, we will use a classic calculus approach of 1) making an approximation, 2) describe how this approximation changes under smaller scales, and 3) use a limit to find the exact value. In the preceeding discussion, we will briefly cover this CCA and return to the details of step 2 later in this section.

Let \(f\) be a continuous function of \(x\), \(y\), and \(z\) for some open set around \(C\), a curve from a point \(P\) to a point \(Q\). We will begin to approximate the accumulation of the output of \(f\) over \(C\) by breaking \(C\) into pieces with boundary points \(P=\vr_0, \vr_1, \dots, \vr_n-1, \vr_n=Q\). 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 \(\vr_i\)

For Step 1 of the CCA, we can approximate the accumulation of \(f\) over \(C\) with the following sum \[\begin{aligned}\end{aligned}\], where \(f(r_i^{*})\) is the output of \(f\) for some \(\vr_i^{*} \in C_i\). Similar to our work on line integrals of a vector field, as this sum uses more pieces and all of the lengths of the pieces goes to zero (i.e. \(\vecmag{\Delta \vr_i} \rightarrow 0\)), we would expect that the sum will approach the actual accumulation of \(f\) over \(C\). Notice that it won't matter how we select the point \(\vr_i^*\) that is used in each piece to evaluate the output of \(f\) since evaluating the limit as the length of \(C_i\) gets smaller will ensure that the output value chosen will be within a shrinking error from the average value on each piece. Evaluating the limit of the sum above as the size of all of the pieces goes to zero will transform our Riemann sum into an integral that will measure the accumulation of the output of \(f\) over \(C\).

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 .

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

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 \(t\).) Parameterizations are very useful converting the three-variable (\(x,y,z\)) 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. To understand this conversion, we will look at the details of the CCA using a parameterization.

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 . Instead of thinking in terms of pieces of the curve \(C\), 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 \(\Delta{t}=\frac{b-a}{n}\). While these pieces will be equally spaced in terms of the parameter \(t\), the corresponding points on the curve \(C\) given by \(r(t_i)\) may not be equally spaced.

To simplify the notation of our function evaluation, we will use the following: \[\begin{aligned}\end{aligned}\] where \(t_{i-1}\leq t_i^{*}\leq t_i\). Remember that the parameterization will allow us to write all parts of the scalar line integral as a function of \(t\), so we will simplify \(\vecmag{\Delta \vr_i}\) as \[\begin{aligned}\end{aligned}\] Applying our parameterization and corresponding points to gives \[\begin{aligned}\int_C f \, ds \amp =\lim_{n\rightarrow \infty} \sum_{i=1}^{n} f(\vr_i^{*}) \vecmag{\vr_{i}-\vr_{i-1}} \\ \amp =\lim_{n\rightarrow \infty} \sum_{i=1}^{n} f(t_i^{*}) \frac{\vecmag{r(t_{i-1}+\Delta t)-r(t_{i-1})}}{\Delta{t}}\Delta{t}\end{aligned}\]. This Riemann sum corresponds to the definite integral of a scalar function of \(t\), specifically \(f(t) v(t)\) where \(v(t)=\vecmag{\vr'(t)}\). We may think of \(v(t)\) as the speed of the parameterization \(\vr\).

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Before stating some useful properties of scalar line integrals, we will recall some convenient notation from . 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 \(R\), 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 \(R\). 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.

For a constant scalar \(k\), scalar valued functions\(f\) and \(g\), and oriented curves \(C\), \(C_1\), and \(C_2\), the following properties hold:

1. \(\displaystyle \int_C (k f) \, ds = k \int_C f \, ds\)

2. \(\displaystyle \int_C (f+g) \, ds = \int_C f \, ds + \int_C g \, ds\)

3. \(\displaystyle \int_{-C} f \, ds = \int_C f \, ds\)

4. \(\displaystyle \int_{C_1+C_2} f \, ds = \int_{C_1} f \, ds + \int_{C_2} f \, ds\)

The biggest difference between and is part. The orientation of the curve does not change the value of the scalar line integral. will have you make sense of these properties for scalar line integrals.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

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 . 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 .

In , the horizontal axis gives the distance traveled along the left side of . 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.

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 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 , 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.