A line integral gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. There are two types of line integrals: scalar line integrals and vector line integrals. Scalar line integrals are integrals of a scalar function over a curve in a plane or in space. Vector line integrals are integrals of a vector field over a curve in a plane or in space. Let’s look at scalar line integrals first.

A scalar line integral is defined just as a single-variable integral is defined, except that for a scalar line integral, the integrand is a function of more than one variable and the domain of integration is a curve in a plane or in space, as opposed to a curve on the [latex]x[/latex]-axis.

For a scalar line integral, we let [latex]C[/latex] be a smooth curve in a plane or in space and let [latex]f[/latex] be a function with a domain that includes [latex]C[/latex]. We chop the curve into small pieces. For each piece, we choose point [latex]P[/latex] in that piece and evaluate [latex]f[/latex] at [latex]P[/latex]. (We can do this because all the points in the curve are in the domain of [latex]f[/latex].) We multiply [latex]{f}{(P)}[/latex] by the arc length of the piece [latex]{\Delta}{s}[/latex], add the product [latex]{f}{(P)}{\Delta}{s}[/latex] over all the pieces, and then let the arc length of the pieces shrink to zero by taking a limit. The result is the scalar line integral of the function over the curve.

For a formal description of a scalar line integral, let [latex]C[/latex] be a smooth curve in space given by the parameterization [latex]{\textbf{r}}(t) = {\left \langle {x}{(t)}, {y}{(t)}, {z}{(t)} \right \rangle}, {a} \leq {t} \leq {b}[/latex]. Let [latex]f(x, y, z)[/latex] be a function with a domain that includes curve [latex]C[/latex]. To define the line integral of the function [latex]f[/latex] over [latex]C[/latex], we begin as most definitions of an integral begin: we chop the curve into small pieces. Partition the parameter interval [latex][{a},{b}][/latex] into [latex]n[/latex] subintervals [latex][{{t}_{i-1}},{t_i}][/latex] of equal width for [latex]{1} \leq {i} \leq {n}[/latex], where [latex]{{t}_{0}} = {a}[/latex] and [latex]{{t}_{n}} = {b}[/latex] (Figure 1). Let [latex]{t}_{i}^{*}[/latex] be a value in the ith interval [latex][{{t}_{i-1}},{t_i}][/latex]. Denote the endpoints of [latex]{\textbf{r}}{(t_0)}, {\textbf{r}}{(t_1)},…,{\textbf{r}}{(t_n)}[/latex] by [latex]{P}_{0},…{P}_{n}[/latex]. Points [latex]P_i[/latex] divide curve [latex]C[/latex] into [latex]n[/latex] pieces [latex]{C}_{1},{C}_{2},…,{C}_{n}[/latex], with lengths [latex]{\Delta}{{s}_{1}}, {\Delta}{{s}_{2}},…, {\Delta}{{s}_{n}}[/latex], respectively. Let [latex]{P}^{*}_{i}[/latex] denote the endpoint of [latex]{\textbf{r}}({t^{*}_{i}})[/latex] for [latex]{1} \leq {i} \leq {n}[/latex]. Now, we evaluate the function [latex]f[/latex] at point [latex]{P}^{*}_{i}[/latex] for [latex]{1} \leq {i} \leq {n}[/latex]. Note that [latex]{P}^{*}_{i}[/latex] is in piece [latex]C_1[/latex], and therefore [latex]{P}^{*}_{i}[/latex] is in the domain of [latex]f[/latex]. Multiply [latex]{f}{({P}^{*}_{i})}[/latex] by the length [latex]{\Delta}{{s}_{1}}[/latex] of [latex]C_1[/latex], which gives the area of the “sheet” with base [latex]C_1[/latex], and height [latex]{f}{({P}^{*}_{i})}[/latex]. This is analogous to using rectangles to approximate area in a single-variable integral. Now, we form the sum [latex]{\displaystyle\sum^{n}_{i = 1}}{f}{({P}^{*}_{i})}{\Delta}{s_{i}}[/latex]. Note the similarity of this sum versus a Riemann sum; in fact, this definition is a generalization of a Riemann sum to arbitrary curves in space. Just as with Riemann sums and integrals of form [latex]{\displaystyle\int^{b}_{a}}{g}{(x)}{dx}[/latex], we define an integral by letting the width of the pieces of the curve shrink to zero by taking a limit. The result is the scalar line integral of [latex]f[/latex] along [latex]C[/latex].s

Figure 1. Curve [latex]C[/latex] has been divided into [latex]n[/latex] pieces, and a point inside each piece has been chosen.

You may have noticed a difference between this definition of a scalar line integral and a single-variable integral. In this definition, the arc lengths [latex]{\Delta}{{s}_{1}}, {\Delta}{{s}_{2}},…, {\Delta}{{s}_{n}}[/latex] aren’t necessarily the same; in the definition of a single-variable integral, the curve in the [latex]x[/latex]-axis is partitioned into pieces of equal length. This difference does not have any effect in the limit. As we shrink the arc lengths to zero, their values become close enough that any small difference becomes irrelevant.

If [latex]f[/latex] is a continuous function on a smooth curve [latex]C[/latex], then [latex]{\displaystyle\int_C}{f}{d}{s}[/latex] always exists. Since [latex]{\displaystyle\int_C}{f}{d}{s}[/latex] is defined as a limit of Riemann sums, the continuity of [latex]f[/latex] is enough to guarantee the existence of the limit, just as the integral [latex]{\displaystyle\int^{b}_{a}}{g{(x)}}{dx}[/latex] exists if [latex]g[/latex] is continuous over [latex][{a},{b}][/latex].

Before looking at how to compute a line integral, we need to examine the geometry captured by these integrals. Suppose that [latex]{f}{({x},{y})} \geq {0}[/latex] for all points [latex](x, y)[/latex] on a smooth planar curve [latex]C[/latex]. Imagine taking curve [latex]C[/latex] and projecting it “up” to the surface defined by [latex]f(x, y)[/latex], thereby creating a new curve [latex]{C}^{\prime}[/latex] that lies in the graph of [latex]f(x, y)[/latex] (Figure 2). Now we drop a “sheet” from [latex]{C}^{\prime}[/latex] down to the [latex]xy[/latex]-plane. The area of this sheet is [latex]{\displaystyle\int_{C}}{f}{({x},{y})}{ds}[/latex]. If [latex]{f}{({x},{y})} \leq {0}[/latex] for some points in [latex]C[/latex], then the value of [latex]{\displaystyle\int_{C}}{f}{({x},{y})}{ds}[/latex] is the area above the [latex]xy[/latex]-plane less the area below the [latex]xy[/latex]-plane. (Note the similarity with integrals of the form [latex]{\displaystyle\int^{b}_{a}}{g}{(x)}{dx}[/latex].)

Figure 2. The area of the blue sheet is [latex]{\displaystyle\int_{C}}{f}{({x},{y})}{ds}[/latex].

From this geometry, we can see that line integral [latex]{\displaystyle\int_{C}}{f}{({x},{y})}{ds}[/latex] does not depend on the parameterization [latex]{\textbf{r}}{(t)}[/latex] of [latex]C[/latex]. As long as the curve is traversed exactly once by the parameterization, the area of the sheet formed by the function and the curve is the same. This same kind of geometric argument can be extended to show that the line integral of a three-variable function over a curve in space does not depend on the parameterization of the curve.

Example: finding the value of a line integral

Find the value of integral [latex]{\displaystyle\int_{C}}{2}{ds}[/latex], where [latex]C[/latex] is the upper half of the unit circle.

Show Solution

The integrand is [latex]f(x, y)=2[/latex]. Figure 3 shows the graph of [latex]f(x, y)=2[/latex], curve [latex]C[/latex], and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width [latex]\pi[/latex] and length 2. Therefore, [latex]{\displaystyle\int_{C}}{2}{ds} = {2}{\pi}[/latex].

Figure 3. The sheet that is formed by the upper half of the unit circle in a plane and the graph of [latex]{f}{({x},{y})} = {2}[/latex].

To see that [latex]{\displaystyle\int_{C}}{2}{ds} = {2}{\pi}[/latex] using the definition of line integral, we let [latex]{\textbf{r}}{(t)}[/latex] be a parameterization of [latex]C[/latex]. Then, [latex]{f}{({\textbf{r}}{(t_{i})})} = {2}[/latex] for any number [latex]{t_{i}}[/latex] in the domain of r. Therefore,

[latex]\begin{aligned} {\displaystyle\int_{C}}{f}{ds} & = {\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{{\displaystyle\sum}^{n}_{i = 1}}{f}{({\textbf{r}}{({t}^{*}_{i})})}{\Delta}{s_{i}} \\ & = {\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{{\displaystyle\sum}^{n}_{i = 1}}{2}{\Delta}{s_{i}} \\ & = {2}{\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{{\displaystyle\sum}^{n}_{i = 1}}{\Delta}{s_{i}} \\ & = {2}{\text{(length of C)}} \\ & = {2}{\pi}. \end{aligned}[/latex]

Note that in a scalar line integral, the integration is done with respect to arc length [latex]s[/latex], which can make a scalar line integral difficult to calculate. To make the calculations easier, we can translate [latex]{\displaystyle\int_{C}}{f}{ds}[/latex] to an integral with a variable of integration that is [latex]t[/latex].

Let [latex]{\textbf{r}}{(t)} = {\left \langle {x}{(t)}, {y}{(t)}, {z}{(t)} \right \rangle}[/latex] for [latex]{a} \leq {t} \leq {b}[/latex] be a parameterization of [latex]C[/latex]. Since we are assuming that [latex]C[/latex] is smooth, [latex]{\textbf{r}}^{\prime}{(t)} = {\left \langle {x}{\prime}{(t)}, {y}{\prime}{(t)}, {z}{\prime}{(t)} \right \rangle}[/latex] is continuous for all [latex]t[/latex] in [latex][{a},{b}][/latex]. In particular, [latex]{x}{\prime}{(t)}, {y}^{\prime}{(t)}[/latex], and [latex]{z}^{\prime}{(t)}[/latex] exist for all [latex]t[/latex] in [latex][{a},{b}][/latex]. According to the arc length formula, we have

[latex]{\text{length}}{({C}_{i})} = {\Delta}{{s}_{i}} = {\displaystyle\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{d}{t}[/latex].

If width [latex]{\Delta}{{t}_{i}} = {{t}_{i}} - {{t}_{{i} - {1}}}[/latex] is small, then function [latex]{\displaystyle\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\left \| {\textbf{r}}^{\prime}{(t)} {\left \| {d}{t}{\approx} \left \| {\textbf{r}}^{\prime}{({t^{*}_{i}})} \right \| {\Delta}{{t}_{i}} \right \|} {\textbf{r}}^{\prime}{(t)} \right \|}[/latex] is almost constant over the interval [latex]\left [ {{t}_{{i} - {1}}},{{t}_{i}} \right ][/latex]. Therefore,

[latex]{\displaystyle\int^{{t}_{i}}_{{t}_{{i} - {1}}}}{\left \| {\textbf{r}}^{\prime}{(t)} {\left \| {d}{t}{\approx} \right \|} {\textbf{r}}^{\prime}{({t^{*}_{i}})} \right \|}{\Delta}{{t}_{i}},[/latex]

and we have

[latex]{\displaystyle\sum^{n}_{{i} = {1}}}{f}{({\textbf{r}}{({t^{*}_{i}})})}{\Delta}{{s}_{i}} = {\displaystyle\sum^{n}_{{i} = {1}}}{f}{({\textbf{r}}{({t^{*}_{i}})})}{\left \| {\textbf{r}}^{\prime}{({t^{*}_{i}})} \right \|}{\Delta}{{t}_{i}}[/latex].

See Figure 4.

Figure 4. If we zoom in on the curve enough by making [latex]{\Delta}{{t}_{i}}[/latex] very small, then the corresponding piece of the curve is approximately linear.

Note that

[latex]{\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{\displaystyle\sum^{n}_{{i} = {1}}}{f}{({\textbf{r}}{({t^{*}_{i}})})}{\left \| {\textbf{r}}^{\prime}{({t^{*}_{i}})} \right \|}{\Delta}{{t}_{i}} = {\displaystyle\int^{b}_{a}}{f}{({\textbf{r}}{(t)})}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{dt}[/latex].

In other words, as the widths of intervals [latex][{{t}_{{i} - {1}}},{{t}_{i}}][/latex] shrink to zero, the sum [latex]{\displaystyle\sum^{n}_{{i} = {1}}}{f}{({\textbf{r}}{({t^{*}_{i}})})}{\left \| {\textbf{r}}^{\prime}{({t^{*}_{i}})} \right \|}{\Delta}{{t}_{i}}[/latex] converges to the integral [latex]{\displaystyle\int^{b}_{a}}{f}{({\textbf{r}}{(t)})}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{dt}[/latex]. Therefore, we have the following theorem.