Line Integrals: Learn It 3

Giselle Miller

Line Integrals: Learn It 3

Vector Line Integrals

The second type of line integrals are vector line integrals. Vector line integrals allow us to integrate along a curve through a vector field. Consider a continuous vector field in [latex]{\mathbb{R}}^{3}[/latex]:

[latex]{\textbf{F}}(x,y,z) = P(x,y,z){\textbf{i}} + Q(x,y,z){\textbf{j}} + R(x,y,z){\textbf{k}}[/latex]

Suppose [latex]{\textbf{F}}[/latex] represents a force on a particle, and [latex]C[/latex] is a smooth curve in [latex]{\mathbb{R}}^{3}[/latex] contained in the domain of [latex]{\textbf{F}}[/latex]. How do we compute the work done by [latex]{\textbf{F}}[/latex] in moving a particle along [latex]C[/latex]?

To answer this question, first note that a particle can travel in two directions along a curve: forward or backward. The work done by the vector field depends on which direction the particle moves. We must specify a direction along curve [latex]C[/latex], called an orientation.

The specified direction is the positive direction along [latex]C[/latex], and the opposite direction is the negative direction. When [latex]C[/latex] has been given an orientation, we call it an oriented curve (Figure 1). The work done on the particle depends on the direction along the curve in which the particle is moving.

Two images, labeled A and B. Image A shows a curve C that is an oriented curve. It is a curve that connects two points; it is a line segment with curves. Image B, on the other hand, is a closed curve. It has no endpoints and completely encloses an area. — Figure 1. (a) An oriented curve between two points. (b) A closed oriented curve.

Defining the Vector Line Integral

Let [latex]{\textbf{r}}(t)[/latex] be a parameterization of [latex]C[/latex] for [latex]a \leq t \leq b[/latex] such that the curve is traversed exactly once and the particle moves in the positive direction along [latex]C[/latex]. We’ll build up the work calculation step by step.

Divide the parameter interval [latex][a,b][/latex] into [latex]n[/latex] subintervals [latex][t_{i-1}, t_i][/latex], [latex]0 \leq i \leq n[/latex], of equal width. Denote the endpoints [latex]{\textbf{r}}(t_0), {\textbf{r}}(t_1), \ldots, {\textbf{r}}(t_n)[/latex] by [latex]P_0, \ldots, P_n[/latex]. These points divide [latex]C[/latex] into [latex]n[/latex] pieces.

Let [latex]{\Delta}s_i[/latex] denote the length of the piece from [latex]P_{i-1}[/latex] to [latex]P_i[/latex]. For each [latex]i[/latex], choose a value [latex]{t}^{*}_{i}[/latex] in the subinterval [latex][{{t}_{{i} - {1}}}, {t}_{i}][/latex]. The endpoint of [latex]{\textbf{r}}{({t}^{*}_{i})}[/latex] is a point in the piece of [latex]C[/latex] between [latex]{P}_{{i} - {1}}[/latex] and [latex]{P}_{i}[/latex] (Figure 2).

If [latex]{\Delta}s_i[/latex] is small, then as the particle moves from [latex]P_{i-1}[/latex] to [latex]P_i[/latex] along [latex]C[/latex], it moves approximately in the direction of [latex]{\textbf{T}}(P_i)[/latex], the unit tangent vector at the endpoint of [latex]{\textbf{r}}{({t}^{*}_{i})}[/latex]. Let [latex]{P}^{*}_{i}[/latex] denote the endpoint of [latex]{\textbf{r}}{({t}^{*}_{i})}[/latex].

The work done by the force vector field in moving the particle from [latex]P_{i-1}[/latex] to [latex]P_i[/latex] is [latex]{\textbf{F}}{({P}^{*}_{i})} \cdot {({\Delta}{{s}_{i}}{\textbf{T}}{({P}^{*}_{i})})}[/latex]. Therefore, the total work done along [latex]C[/latex] is:

[latex]{\displaystyle\sum^{n}_{i = 1}} \ {\textbf{F}}{({P}^{*}_{i})} \cdot ({{\Delta}{s}_{i}}{\textbf{T}}{({P}^{*}_{i})}) = {\displaystyle\sum^{n}_{i = 1}} \ {\textbf{F}}{({P}^{*}_{i})} \cdot {\textbf{T}}{({P}^{*}_{i})}{{\Delta}{s}_{i}}[/latex].

An image of a concave down curve – initially increasing, but later decreasing. Several points are labeled along the curve, as are arrowheads along the curve pointing in the direction of increasing P value. The points are: P_0, P_1, P_i-1, P_i starred, P_i, P_n-1, and Pn. Two arrows have their endpoints at P_i. The first is an increasing tangent vector labeled T(P_i starred). The second is labeled F(P_i starred) and points up and to the left. — Figure 2. Curve [latex]C[/latex] is divided into n pieces, and a point inside each piece is chosen. The dot product of any tangent vector in the ith piece with the corresponding vector [latex]F[/latex] is approximated by [latex]{\textbf{F}}{({P}^{*}_{i})} \cdot {\textbf{T}}{({P}^{*}_{i})}[/latex].

Letting the arc length of the pieces of [latex]C[/latex] get arbitrarily small by taking a limit as [latex]n \to \infty[/latex] gives us the work done by the field in moving the particle along [latex]C[/latex]. The work done by [latex]{\textbf{F}}[/latex] in moving the particle in the positive direction along [latex]C[/latex] is:

[latex]W = \displaystyle\int_{C}{\textbf{F}} \cdot {\textbf{T}} ds[/latex]

This integral is called a vector line integral.

vector line integral

The vector line integral of vector field [latex]{\textbf{F}}[/latex] along oriented smooth curve [latex]C[/latex] is:

[latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds} = {\displaystyle\lim_{{n}{\rightarrow}{\infty}}}{\displaystyle\sum^{n}_{i = 1}} \ {\textbf{F}}{({P}^{*}_{i})} \cdot {\textbf{T}}{({P}^{*}_{i})}{{\Delta}{s}_{i}}[/latex]

if that limit exists.

With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once, the value of the line integral is unchanged.

With vector line integrals, orientation is critical. If you hike up a mountain, the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, Earth’s gravitational force does positive work on you. Reversing the path changes the work value from negative to positive.

If [latex]C[/latex] is an oriented curve, then [latex]-C[/latex] represents the same curve but with opposite orientation.

Computing Vector Line Integrals with Parameterization

As with scalar line integrals, it’s easier to compute a vector line integral using the parameterization function [latex]{\textbf{r}}[/latex] and the variable [latex]t[/latex].

To translate the integral [latex]\displaystyle\int_{C}{\textbf{F}} \cdot {\textbf{T}} ds[/latex] in terms of [latex]t[/latex], recall that the unit tangent vector [latex]{\textbf{T}}[/latex] along [latex]C[/latex] is:

[latex]{\textbf{T}} = \frac{{\textbf{r}}'(t)}{\left | {\textbf{r}}'(t) \right |}[/latex]

(assuming [latex]\left | {\textbf{r}}'(t) \right | \neq 0[/latex])

Since [latex]ds = \left | {\textbf{r}}'(t) \right | dt[/latex], as we saw when discussing scalar line integrals, we have:

[latex]{\textbf{F}} \cdot {\textbf{T}} ds = {\textbf{F}}({\textbf{r}}(t)) \cdot \frac{{\textbf{r}}'(t)}{\left | {\textbf{r}}'(t) \right |}\left | {\textbf{r}}'(t) \right | dt = {\textbf{F}}({\textbf{r}}(t)) \cdot {\textbf{r}}'(t) dt[/latex]

This gives us a practical formula for computing vector line integrals.

computing vector line integrals

[latex]\displaystyle\int_{C}{\textbf{F}} \cdot {\textbf{T}} ds = \displaystyle\int^{b}_{a}{\textbf{F}}({\textbf{r}}(t)) \cdot {\textbf{r}}'(t) dt[/latex]

We often use the notation [latex]\displaystyle\int_{C}{\textbf{F}} \cdot d{\textbf{r}}[/latex] for the line integral [latex]\displaystyle\int_{C}{\textbf{F}} \cdot {\textbf{T}} ds[/latex].

[latex]\\[/latex]

If [latex]{\textbf{r}} = \langle x(t), y(t), z(t) \rangle[/latex], then [latex]d{\textbf{r}}[/latex] denotes the vector differential [latex]\langle x'(t), y'(t), z'(t) \rangle dt[/latex].

Find the value of integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex], where [latex]C[/latex] is the semicircle parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {\cos{t}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}[/latex] and [latex]{\textbf{F}} = {\left \langle {-y}, {x} \right \rangle}[/latex].

Find the value of integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {dr}[/latex], where [latex]C[/latex] is the semicircle parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {\cos{({t} + {\pi})}}, {\sin{t}} \right \rangle}, {0} \leq {t} \leq {\pi}[/latex], and [latex]{\textbf{F}} = {\left \langle -{y}, {x} \right \rangle}[/latex].

The previous examples illustrate an important property:

[latex]\displaystyle\int_{-C}{\textbf{F}} \cdot d{\textbf{r}} = -\displaystyle\int_{C}{\textbf{F}} \cdot d{\textbf{r}}[/latex]

Reversing the orientation of a curve changes the sign of a line integral. This makes physical sense when thinking about work—if you reverse direction, the work done changes sign.

Alternative Notation

Another standard notation for integral [latex]\displaystyle\int_{C}{\textbf{F}} \cdot d{\textbf{r}}[/latex] is [latex]\displaystyle\int_{C}P dx + Q dy + R dz[/latex]. In this notation, [latex]P[/latex], [latex]Q[/latex], and [latex]R[/latex] are functions, and we think of [latex]d{\textbf{r}}[/latex] as vector [latex]\langle dx, dy, dz \rangle[/latex].

To justify this convention, recall that [latex]d{\textbf{r}} = {\textbf{T}} ds = {\textbf{r}}'(t) dt = \left \langle \dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt} \right \rangle dt[/latex]. Therefore:

[latex]{\textbf{F}} \cdot d{\textbf{r}} = \langle P, Q, R \rangle \cdot \langle dx, dy, dz \rangle = P dx + Q dy + R dz[/latex]

If [latex]d{\textbf{r}} = \langle dx, dy, dz \rangle[/latex], then [latex]\dfrac{d{\textbf{r}}}{dt} = \left \langle \dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt} \right \rangle[/latex], which implies that [latex]d{\textbf{r}} = \left \langle \dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt} \right \rangle dt[/latex]. Therefore:

[latex]\begin{aligned} {\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}} & = {\displaystyle\int_{C}}{Pdx} + {Qdy} + {Rdz} \\ & = {\displaystyle\int}{\left ( {P}{({\textbf{r}}{(t)})}{\frac{dx}{dt}} + {Q}{({\textbf{r}}{(t)})}{\frac{dy}{dt}} + {R}{({\textbf{r}}{(t)})}{\frac{dz}{dt}} \right )}{dt}. \end{aligned}[/latex]

Find the value of integral [latex]{\displaystyle\int_{C}}{zdx} + {xdy} + {ydz}[/latex], where [latex]C[/latex] is the curve parameterized by [latex]{\textbf{r}}{(t)} = {\left \langle {t^{2}}, {\sqrt{t}},{t} \right \rangle}, {1} \leq {t} \leq {4}[/latex].

Piecewise Smooth Curves

We have learned how to integrate smooth oriented curves. Now suppose that [latex]C[/latex] is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that [latex]C[/latex] is a piecewise smooth curve.

To be precise, curve [latex]C[/latex] is piecewise smooth if [latex]C[/latex] can be written as a union of [latex]n[/latex] smooth curves [latex]C_{1}, C_{2}, \ldots, C_{n}[/latex] such that the endpoint of [latex]C_i[/latex] is the starting point of [latex]C_{i+1}[/latex] (Figure 4). When curves [latex]C_i[/latex] satisfy this condition, we write their union as [latex]C_{1} + C_{2} + \cdots + C_{n}[/latex].

Three curves: C_1, C_2, and C_3. One of the endpoints of C_2 is also an endpoint of C_1, and the other endpoint of C_2 is also an end point of C_3. C_1’s and C_3’s other endpoints are not connect to any other curve. C_1 and C_3 appear to be nearly straight lines while C_2 is an increasing concave down curve. There are three arrowheads on each curve segment all pointing in the same direction: C_1 to C_2, C_2 to C_3, and C_3 to its other endpoint. — Figure 4. The union of [latex]{{C}_{1}}, {{C}_{2}}, {{C}_{3}}[/latex] is a piecewise smooth curve.

The next theorem summarizes several key properties of vector line integrals.

theorem: properties of vector line integrals

Let [latex]{\textbf{F}}[/latex] and [latex]{\textbf{G}}[/latex] be continuous vector fields with domains that include the oriented smooth curve [latex]C[/latex]. Then:

[latex]{\displaystyle\int_{C}}{({\textbf{F}} + {\textbf{G}})} \cdot {d}{\textbf{r}} = {\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}} + {\displaystyle\int_{C}}{\textbf{G}} \cdot {dr}[/latex]
[latex]{\displaystyle\int_{C}}{k}{\textbf{F}} \cdot {d}{\textbf{r}} = {k}{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}[/latex], where [latex]k[/latex] is a constant
[latex]{\displaystyle\int_{-C}}{\textbf{F}} \cdot {d}{\textbf{r}} = -{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}}[/latex]
Suppose instead that [latex]C[/latex] is a piecewise smooth curve in the domains of F and G, where [latex]{C} = {C_1} + {C_2} + ... + {C_n}[/latex] and [latex]{C_1}, {C_2}, ..., {C_n}[/latex]are smooth curves such that the endpoint of [latex]C_i[/latex] is the starting point of [latex]{C}_{{i} + {1}}[/latex]. Then:
[latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{s}} = {\displaystyle\int_{C_1}}{\textbf{F}} \cdot {d}{\textbf{s}} + {\displaystyle\int_{C_2}}{\textbf{F}} \cdot {d}{\textbf{s}} + ... + {\displaystyle\int_{C_n}}{\textbf{F}} \cdot {d}{\textbf{s}}[/latex]

Understanding the Properties

Notice the similarities between these properties and the properties of single-variable integrals.

Properties 1 and 2 say that line integrals are linear, which is true of single-variable integrals as well. Property 3 says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along [latex]C[/latex], then this makes sense—if the particle moves backward rather than forward, the value of the work done has the opposite sign. This is analogous to the equation [latex]{\displaystyle\int^{b}_{a}}{f}{(x)}{dx} = {-}{\displaystyle\int^{a}_{b}}{f}{(x)}{dx}[/latex].

Finally, if [latex][a_1, a_2][/latex], [latex][a_2, a_3][/latex], [latex]\ldots[/latex], [latex][a_{n-1}, a_n][/latex] are intervals, then:

[latex]{\displaystyle\int^{a_n}_{a_1}}{f}{(x)}{dx} = {\displaystyle\int^{a_2}_{a_1}}{f}{(x)}{dx} + {\displaystyle\int^{a_3}_{a_1}}{f}{(x)}{dx} + ... + {\displaystyle\int^{a_n}_{a_{n-1}}}{f}{(x)}{dx},[/latex]

This is analogous to property 4.

Find the value of integral [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{ds}[/latex], where [latex]C[/latex] is the rectangle (oriented counterclockwise) in a plane with vertices [latex](0, 0)[/latex], [latex](2, 0)[/latex], [latex](2, 1)[/latex],and [latex](0, 1)[/latex] and where [latex]{\textbf{F}} = {\left \langle {x} - {2{y}}, {y} - {x} \right \rangle}[/latex] (Figure 5).

A vector field in two dimensions. The arrows following roughly a 90-degree angle to the origin in quadrants 1 and 3 point to the origin. As the arrows deviate from this angle, they point away from the angle ad become smaller. Above, they point up and to the left, and below, they point down and to the right. A rectangle is drawn in quadrant 1 from 0 to 2 on the x axis and from 0 to 1 on the y axis. C_1 is the base, C_2 is the right leg, C_3 is the top, and C_4 is the left leg. — Figure 5. Rectangle and vector field.

Note that curve [latex]C[/latex] is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let [latex]C_1[/latex] represent the side from [latex](0, 0)[/latex] to [latex](2, 0)[/latex], let [latex]C_2[/latex] represent the side from [latex](2, 0)[/latex] to [latex](2, 1)[/latex], let [latex]C_3[/latex] represent the side from [latex](2, 1)[/latex] to [latex](0, 1)[/latex], and let [latex]C_4[/latex] represent the side from [latex](0, 1)[/latex] to [latex](0, 0)[/latex] (Figure 5). Then,

[latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} = {\displaystyle\int_{C_1}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} + {\displaystyle\int_{C_2}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} + {\displaystyle\int_{C_3}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} + {\displaystyle\int_{C_4}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}}[/latex].

We want to compute each of the four integrals on the right-hand side using The Scalar Line Integral Calculation Theorem Equation. Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse [latex]C[/latex] counterclockwise):

[latex]\begin{aligned} {C_1} & : {\langle {t}, {0}\rangle}, {0} \leq {t} \leq {2} \\ {C_2} & : {\langle {2}, {t} \rangle}, {0} \leq {t} \leq {1} \\ {C_3} & : {\langle {2} - {t}, {1}\rangle}, {0} \leq {t} \leq {2} \\ {C_4} & : {\langle {0},{1} - {t}\rangle}, {0} \leq {t} \leq {1}. \end{aligned}[/latex].

Therefore,

[latex]\begin{aligned} {\displaystyle\int_{{C}_{1}}}{\textbf{F}} \cdot {\textbf{T}}{d}{\textbf{r}} & = {\displaystyle\int^{2}_{0}}{\textbf{F}}{({\textbf{r}}({t}))} \cdot {\textbf{r}}{\prime}{(t)}{dt} \\ & = {\displaystyle\int^{2}_{0}}{\left \langle {t} - {2}{(0)}, {0} - {t} \right \rangle} \cdot {\left \langle {1}, {0} \right \rangle}{dt} = {\displaystyle\int^{1}_{0}}{t}{dt} \\ & = {\left [ {\frac{t^{2}}{2}} \right ]}^{2}_{0} = {2}. \end{aligned}[/latex].

Notice that the value of this integral is positive, which should not be surprising. As we move along curve [latex]C_1[/latex] from left to right, our movement flows in the general direction of the vector field itself. At any point along [latex]C_1[/latex], the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along [latex]C_1[/latex], and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

[latex]\begin{aligned} {\displaystyle\int_{{C}_{2}}}{\textbf{F}} \cdot {d}{\textbf{r}} & = {\displaystyle\int^{1}_{0}}{\left \langle {2} - {2{t}}, {t} - {2} \right \rangle} \cdot {\left \langle {0}, {1} \right \rangle}{dt} \\ & = {\displaystyle\int^{1}_{0}}{({t} - {2})}{dt} \\ & = {\left [ {\frac{t^{2}}{2}} - {2{t}} \right ]}^{1}_{0} = -{\frac{3}{2}}, \end{aligned}[/latex]

[latex]\begin{aligned} {\displaystyle\int_{{C}_{3}}}{\textbf{F}} \cdot {\textbf{T}}{ds} & = {\displaystyle\int^{2}_{0}}{\left \langle ({2} - {t}) - {2}, {1} - ({2} - {t}) \right \rangle} \cdot {\left \langle {-1}, {0} \right \rangle}{dt} \\ & = {\displaystyle\int^{2}_{0}}{tdt} = {2}, \end{aligned}[/latex]

and

[latex]\begin{aligned} {\displaystyle\int_{{C}_{4}}}{\textbf{F}} \cdot {d}{\textbf{r}} & = {\displaystyle\int^{1}_{0}}{\left \langle {-2}(1 - t) , {1 - t}\right \rangle} \cdot {\left \langle {0}, {-1} \right \rangle}{dt} \\ & = {\displaystyle\int^{1}_{0}}{({t} - {1})}{dt} \\ & = {\left [ {\frac{t^{2}}{2}} - {t} \right ]}^{1}_{0} = -{\frac{1}{2}}. \end{aligned}[/latex]

Thus, we have [latex]{\displaystyle\int_{C}}{\textbf{F}} \cdot {d}{\textbf{r}} = {2}[/latex].