Line Integrals: Learn It 2

Scalar Line Integrals

A scalar line integral is defined similarly to a single-variable integral, except the integrand is a function of more than one variable and the domain of integration is a curve rather than an interval on the [latex]x[/latex]-axis.

Riemann Sums for Single-Variable Integrals

For [latex]\int_a^b g(x) dx[/latex], we:

  1. Partition [latex][a,b][/latex] into subintervals of width [latex]\Delta x[/latex]
  2. Choose sample points [latex]x_i^*[/latex] in each subinterval
  3. Form the sum [latex]\sum g(x_i^*) \Delta x[/latex]
  4. Take the limit as [latex]\Delta x \to 0[/latex]

Scalar line integrals follow the same pattern, but over curves instead of interval

For a scalar line integral, we start with a smooth curve [latex]C[/latex] and a function [latex]f[/latex] whose domain includes [latex]C[/latex]. Here’s the process:

  1. Chop the curve into small pieces
  2. For each piece, choose a point [latex]P[/latex] in each piece and evaluate [latex]f(P)[/latex]
  3. Multiply [latex]f(P)[/latex] by the arc length [latex]\Delta s[/latex] of that piece
  4. Sum all products [latex]f(P)\Delta s[/latex]
  5. Let the arc lengths shrink to zero by taking a limit

The result is the scalar line integral of the function over the curve.

Formal Definition

Let [latex]C[/latex] be a smooth curve in space given by the parameterization [latex]\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle[/latex], [latex]a \leq t \leq b[/latex]. Let [latex]f(x,y,z)[/latex] be a function whose domain includes curve [latex]C[/latex].

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 [latex]i[/latex]th interval [latex][t_{i-1}, t_i][/latex].

Denote the endpoints [latex]\mathbf{r}(t_0), \mathbf{r}(t_1), \ldots, \mathbf{r}(t_n)[/latex] by [latex]P_0, \ldots, P_n[/latex]. These points divide curve [latex]C[/latex] into [latex]n[/latex] pieces [latex]C_1, C_2, \ldots, C_n[/latex] with lengths [latex]\Delta s_1, \Delta s_2, \ldots, \Delta s_n[/latex]. Let [latex]{P}^{*}_{i}[/latex] denote the endpoint of [latex]{\textbf{r}}({t^{*}_{i}})[/latex] for [latex]{1} \leq {i} \leq {n}[/latex].

A diagram of a curve in quadrant one. Several points and segments are labeled. Starting at the left, the first points are P_0 and P_1. The segment between them is labeled delta S_1. The next points are P_i-1, P_i, and P_i+1. The segments connecting them are delta S_i and delta S_j+1. Point P_i starred and point P_i+1 starred are located on each segment, respectively. The last two points are P_n-1 and P_n, connected by segment S_n.
Figure 1. Curve [latex]C[/latex] has been divided into [latex]n[/latex] pieces, and a point inside each piece has been chosen.

Now we evaluate function [latex]f[/latex] at point [latex]{P}^{*}_{i}[/latex] foreach [latex]i[/latex]. Multiply [latex]{f}{({P}^{*}_{i})}[/latex] by the length [latex]\Delta s_i[/latex] of [latex]C_i[/latex], which gives the area of a “sheet” with base [latex]C_i[/latex] and height [latex]f(P_i^*)[/latex]. This is analogous to using rectangles to approximate area in single-variable integrals.

Form the sum [latex]\sum_{i=1}^n f(P_i^*) \Delta s_i[/latex]. As with Riemann sums, we define the integral by letting the width of the pieces shrink to zero. The result is the scalar line integral of [latex]f[/latex] along [latex]C[/latex].

Unlike single-variable integrals where we partition into equal lengths, the arc lengths [latex]\Delta s_1, \Delta s_2, \ldots, \Delta s_n[/latex] aren’t necessarily equal. However, this difference doesn’t affect the limit—as arc lengths shrink to zero, their values become close enough that small differences become irrelevant.

scalar line integral

Let [latex]f[/latex] be a function with a domain that includes the smooth curve [latex]C[/latex] parameterized by [latex]\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle[/latex], [latex]a \leq t \leq b[/latex]. The scalar line integral of [latex]f[/latex] along [latex]C[/latex] is:

[latex]\int_C f(x,y,z) ds = \lim_{n \to \infty} \sum_{i=1}^n f(P_i^*) \Delta s_i[/latex]

if this limit exists.

For a planar curve with parameterization [latex]x = x(t)[/latex], [latex]y = y(t)[/latex], [latex]a \leq t \leq b[/latex]:

[latex]\int_C f(x,y) ds = \lim_{n \to \infty} \sum_{i=1}^n f(P_i^*) \Delta s_i[/latex]

If [latex]f[/latex] is continuous on smooth curve [latex]C[/latex], then [latex]\int_C f \, ds[/latex] always exists. The continuity of [latex]f[/latex] guarantees the existence of the limit, just as [latex]\int_a^b g(x) dx[/latex] exists when [latex]g[/latex] is continuous over [latex][a,b][/latex].

Geometric Interpretation

Before learning how to compute line integrals, let’s examine the geometry they capture.

Suppose [latex]f(x,y) \geq 0[/latex] for all points [latex](x, y)[/latex] on a smooth planar curve [latex]C[/latex]. Imagine projecting curve [latex]C[/latex] “up” to the surface defined by [latex]f(x,y)[/latex], creating a new curve [latex]C'[/latex] that lies on the graph of [latex]f(x,y)[/latex] (Figure 2). Now drop a “sheet” from [latex]C'[/latex] down to the [latex]xy[/latex]-plane. The area of this sheet is [latex]\int_C f(x,y) ds[/latex].

If [latex]f(x,y) \leq 0[/latex] for some points in [latex]C[/latex], then [latex]\int_C f(x,y) ds[/latex] equals the area above the [latex]xy[/latex]-plane minus the area below the [latex]xy[/latex]-plane. This is analogous to single-variable integrals [latex]\int_a^b g(x) dx[/latex].

A diagram in three dimensions. The original curve C in the (x,y) plane looks like a parabola opening to the left with vertex in quadrant 1. The surface defined by f(x,y) is shown always above the (x,y) plane. A curve on the surface directly above the original curve C is labeled as C’. A blue sheet stretches down from C’ to C.
Figure 2. The area of the blue sheet is [latex]{\displaystyle\int_{C}}{f}{({x},{y})}{ds}[/latex].
From this geometry, we see that line integral [latex]\int_C f(x,y) ds[/latex] does not depend on the parameterization [latex]\mathbf{r}(t)[/latex] of [latex]C[/latex]. As long as the curve is traversed exactly once, the area of the sheet formed by the function and curve is the same. This geometric argument extends to three-dimensional curves as well.
Find the value of integral [latex]{\displaystyle\int_{C}}{2}{ds}[/latex], where [latex]C[/latex] is the upper half of the unit circle.

Computing Scalar Line Integrals

Scalar line integrals integrate with respect to arc length [latex]s[/latex], which can make calculations difficult. We can simplify by translating [latex]\int_C f \, ds[/latex] to an integral with variable [latex]t[/latex].

Let [latex]\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle[/latex] for [latex]a \leq t \leq b[/latex] be a parameterization of smooth curve [latex]C[/latex]. Since [latex]C[/latex] is smooth, [latex]\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle[/latex] is continuous for all [latex]t[/latex] in [latex][a,b][/latex].

According to the arc length formula:

[latex]\text{length}(C_i) = \Delta s_i = \int_{t_{i-1}}^{t_i} ||\mathbf{r}'(t)|| dt[/latex]

If width [latex]\Delta t_i = t_i - t_{i-1}[/latex] is small, then [latex]||\mathbf{r}'(t)||[/latex] is nearly constant over interval [latex][t_{i-1}, t_i][/latex]. Therefore:

[latex]\int_{t_{i-1}}^{t_i} ||\mathbf{r}'(t)|| dt \approx ||\mathbf{r}'(t_i^*)|| \Delta t_i[/latex]

This gives us:

[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].

A segment of an increasing concave down curve labeled C. A small segment of the curved is boxed and labeled as delta t_i. In the zoomed-in insert, this boxed segment of the curve is almost linear.
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.

Taking the limit as [latex]n \to \infty[/latex]:

[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].

This gives us our main computational theorem.

evaluating a scalar line integral

Let [latex]f[/latex] be a continuous function with a domain that includes the smooth curve [latex]C[/latex] with parameterization [latex]{\textbf{r}}{(t)}, {a} \leq {t} \leq {b}[/latex]. Then:

[latex]{\displaystyle\int_C}{f}{d}{s} = {\displaystyle\int^{b}_{a}}{f}{({\textbf{r}}{(t)})}{\left \| {\textbf{r}}^{\prime}{(t)} \right \|}{dt}[/latex].

In our derivation, we wrote [latex]\int_{t_{i-1}}^{t_i} ||\mathbf{r}'(t)|| dt \approx ||\mathbf{r}'(t_i^*)|| \Delta t_i[/latex]. This is an approximation, not an equation. The left-hand side approaches the right-hand side as [latex]n \to \infty[/latex]. In other words, letting the widths of the pieces shrink to zero makes the right-hand sum arbitrarily close to the left-hand sum.

Since [latex]||\mathbf{r}'(t)|| = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}[/latex], we obtain a practical formula for computation.

scalar line integral calculation

Let [latex]f[/latex] be continuous with domain including smooth curve [latex]C[/latex] with parameterization [latex]\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle[/latex], [latex]a \leq t \leq b[/latex]. Then:

[latex]{\displaystyle\int_{C}}{f}{({x},{y},{z})}{ds} = {\displaystyle\int^{b}_{a}}{f}{({\textbf{r}}{(t)}}){\sqrt{{({x}^{\prime}{(t)})}^{2} + {({y}^{\prime}{(t)})}^{2} + {({z}^{\prime}{(t)})}^{2}}}{dt}[/latex].

For a planar curve:

[latex]{\displaystyle\int_{C}}{f}{({x},{y})}{ds} = {\displaystyle\int^{b}_{a}}{f}{({\textbf{r}}{(t)}}){\sqrt{{({x}^{\prime}{(t)})}^{2} + {({y}^{\prime}{(t)})}^{2}}}{dt}[/latex]

A key consequence of this theorem is the equation [latex]ds = ||\mathbf{r}'(t)|| dt[/latex]. In other words, the change in arc length can be viewed as a change in the [latex]t[/latex] domain, scaled by the magnitude of vector [latex]\mathbf{r}'(t)[/latex].

Find the value of integral [latex]{\displaystyle\int_{C}}{({x^2} + {y^2} + {z})}{ds}[/latex], where [latex]C[/latex] is part of the helix parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {\cos {t}}, {\sin {t}}, {t} \right \rangle}, {0} \leq {t} \leq {2}{\pi}[/latex].

Find the value of integral [latex]{\displaystyle\int_{C}}{({x^2} + {y^2} + {z})}{ds}[/latex], where [latex]C[/latex] is part of the helix parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {\cos}{(2t)}, {\sin}{(2t), {2}{t}} \right \rangle}, {0} \leq {t} \leq {\pi}[/latex]. [latex]\\[/latex]Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast.

Arc Length as a Line Integral

Line integrals provide a way to calculate arc length. If [latex]f(x,y,z) = 1[/latex], then:

[latex]\begin{aligned} {\displaystyle\int_{C}}{f}{({x}, {y}, {z})}{ds} & = {\displaystyle\lim_{{n}{\rightarrow}{\infty}}}\displaystyle{\sum^{n}_{i = 1}}{f}{({t}^{*}_{i})}{\Delta}{{s}_{i}} \\ & = \displaystyle{\lim_{{n}{\rightarrow}{\infty}}}{\displaystyle\sum^{n}_{i = 1}}{\Delta}{{s}_{i}} \\ & = {\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{\text{length}}{(C)} \\ & = {\text{length}}{(C)}. \end{aligned}[/latex]

Therefore, [latex]{\displaystyle\int_{C}}{1}{ds}[/latex] is the arc length of [latex]C[/latex].

A wire has a shape that can be modeled with the parameterization [latex]{\textbf{r}}{(t)} = {\left \langle {\cos{t}}, {\sin{t}}, {\frac{2}{3}}{t}^{3/2} \right \rangle}, {0} \leq {t} \leq {4}{\pi}[/latex]. Find the length of the wire.