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Chapter 9: Vector Calculus
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Section 9.5: Line Integrals
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Essentials


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Table 9.5.1 lists some salient points about line integrals.
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A line integral is a summative process along a path that is not the horizontal coordinate axis.

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The partition along the path is taken with respect to arc length $s$.

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Line integrals are definite integrals.

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The line integral can be of a scalar, the tangential or the normal component of a vector field.

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Mechanical work is defined as the line integral of the tangential component of the force field doing the work.

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Circulation of a fluid's velocity field along a path is the line integral (along the path) of the tangential component of the field.

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The line integral of the normal component of a planar field is the flux (i.e., flow) of the field through the curve.


Table 9.5.1 Salient points relevant to line integrals



From elementary physics comes the definition of mechanical work: the product of a force and a displacement. If the force varies along a trajectory, or the trajectory is not uniform, the process of "force times distance" is discretized and the "bits" of work added by integration. The component of force along the trajectory is the component doing the mechanical work. Hence, the line integral of the tangential component of a force acting along a curve gives the mechanical work associated with the process.
If the vector field is the velocity field of a laminar flow, the line integral of the tangential component around a closed curve is defined as the circulation of the flow.
The "flow" of a planar vector field through a curve is called the flux of the field through the curve. It is defined as the line integral of the normal component of the field along the curve. The flow of a spatial vector field through a surface is also the flux of the field, but this case can only be treated after a discussion of surface integration in Section 9.6
Table 9.5.2 lists how paths of integration can be defined in Cartesian coordinates.
In 2D

In 3D

Explicitly: $y\=y\left(x\right)$
Parametrically: $x\=x\left(t\right)\,y\=y\left(t\right)$
Parametrically, as a position vector: $\mathbf{R}\=\left[\begin{array}{c}x\(t\)\\ y\(t\)\end{array}\right]$

Parametrically: $x\=x\left(t\right)\,y\=y\left(t\right)\,z\=z\left(t\right)$
Parametrically, as a position vector: $\mathbf{R}\=\left[\begin{array}{c}x\(t\)\\ y\(t\)\\ z\(t\)\end{array}\right]$

Table 9.5.2 Representing paths of integration for line integrals



In the plane, a curve defined explicitly by $y\=y\left(x\right)$ can also be given parametrically by the equations $x\=t\,y\=y\left(t\right)$, or even as the position vector $\mathbf{R}\=x\mathbf{i}plus;y\left(x\right)\mathbf{j}$.
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Table 9.5.3 details ${\int}_{C}f\mathrm{ds}$, the line integral of the scalar function $f\left(x\,y\right)$ along the curve $C$.
Representation of $C$

Implementation of the line integral ${\int}_{C}f\mathrm{ds}$

$y\=y\left(x\right)$

${\int}_{x\=a}^{x\=b}f\left(x\,y\left(x\right)\right)\sqrt{1plus;{\left(y\prime \right)}^{2}}\mathrm{dx}$

$x\=x\left(t\right)\,y\=y\left(t\right)$
$\mathbf{R}\=\left[\begin{array}{c}x\(t\)\\ y\(t\)\end{array}\right]$

${\int}_{t\=a}^{t\=b}f\left(x\left(t\right)\,y\left(t\right)\right)\sqrt{{\stackrel{period;}{x}}^{2}plus;{\stackrel{period;}{y}}^{2}}\mathrm{dt}$

$x\=x\left(t\right)\,y\=y\left(t\right)\,z\=z\left(t\right)$
$\mathbf{R}\=\left[\begin{array}{c}x\(t\)\\ y\(t\)\\ z\(t\)\end{array}\right]$

${\int}_{t\=a}^{t\=b}f\left(x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right)\sqrt{{\stackrel{period;}{x}}^{2}plus;{\stackrel{period;}{y}}^{2}plus;{\stackrel{period;}{z}}^{2}}\mathrm{dt}$

Table 9.5.3 Line integrals of scalar fields



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Of course, the astute reader will recognize that $\sqrt{1\+{\left(y\prime \right)}^{2}}\mathrm{dx}$, $\sqrt{{\stackrel{\.}{x}}^{2}\+{\stackrel{\.}{y}}^{2}}\mathrm{dt}$, and $\sqrt{{\stackrel{\.}{x}}^{2}\+{\stackrel{\.}{y}}^{2}\+{\stackrel{\.}{z}}^{2}}\mathrm{dt}$ in Table 9.5.3 are representations of $\mathrm{ds}$, the element of arc length.
Table 9.5.4 contains information pertaining to the line integral of the tangential component of a vector field F.
The line integral of the tangential component of a vector field $\mathbf{F}\=f\mathbf{i}plus;g\mathbf{j}plus;h\mathbf{k}$ will typically be represented by any of the following three notations
${\int}_{C}\mathbf{F}\xb7\mathbf{dr}$ = ${\int}_{C}\mathbf{F}\xb7\mathbf{T}\mathrm{ds}$ = ${\int}_{C}f\mathrm{dx}plus;g\mathrm{dy}plus;h\mathrm{dz}$
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where $C$ is the path for the integration, $s$ is arc length along that path, T is a unit tangent vector along $C$, and r is a positionvector representation of the parametrically given $C$. Some texts will write dr for the differential of the vector r, and some will write dr. In either event, the symbol represents the differential of the vector $\mathbf{r}\left(t\right)$, so it is a derivative times an increment: $\mathbf{dr}\=\stackrel{\.}{\mathbf{r}}\left(t\right)\mathrm{dt}$ = $\left(\stackrel{\.}{x}\left(t\right)\mathbf{i}plus;\stackrel{period;}{y}\left(t\right)\mathbf{j}plus;\stackrel{period;}{z}\left(t\right)\mathbf{k}\right)\mathrm{dt}$ = $\mathrm{dx}\mathbf{i}plus;\mathrm{dy}\mathbf{j}plus;\mathrm{dz}\mathbf{k}$.
If the curve $C$ is parametrized by arc length, then $\mathbf{r}\prime \left(s\right)\=\mathbf{T}\left(s\right)$ is automatically a unit tangent vector, and $\mathbf{dr}\=\mathbf{T}\mathrm{ds}$. It should now be clear why the three forms of the line integral of the tangential component of F are equivalent.

Table 9.5.4 Line integral of the tangential component of F



The following remarks pertain to the line integral of the normal component of a vector field. In particular, they clarify the relation between V, a vector in the plane, and one of the two possible vectors orthogonal to V.
If $\mathbf{V}\mathbf{\=}\left[\begin{array}{c}u\\ v\end{array}\right]$ is a vector in the plane, then $\mathbf{N}\=\left[\begin{array}{c}v\\ u\end{array}\right]$ is clearly orthogonal to V. Moreover, N lies "to the right" of V in the sense that were one to face in the direction of V and extend the right arm to the side, then the arm would point in the direction of N. Be advised, however, this process of interchanging the components of V and negating the second one need not result in the principal normal along a curve, since the principal normal typically points towards the center of curvature.
Counterexample: The circle $x\=\mathrm{cos}\left(t\right)\,y\=\mathrm{sin}\left(t\right)$ is traced counterclockwise as $t$ increases from $t\=0$ where the tangent vector $\mathbf{T}\=\mathbf{j}$ would therefore point upwards. Interchanging components and negating the second component results in the vector i, and not the principal normal $\mathbf{i}$, which points towards the center of the circle.
Table 9.5.5 contains information pertaining to the calculation of flux through a curve, that is, to the line integral of the normal component of a vector field F.
Flux of the field $\mathbf{F}\=f\left(x\,y\right)\mathbf{i}plus;g\left(xcomma;y\right)\mathbf{j}$ through the plane curve $C$, given by $\mathbf{r}\=x\left(t\right)\mathbf{i}plus;y\left(t\right)\mathbf{j}$, is represented by the line integral of the normal component of F along C, for which either of the following two notations suffice.
${\int}_{C}\mathbf{F}\xb7\mathbf{N}\mathrm{ds}$ = ${\int}_{C}f\mathrm{dy}g\mathrm{dx}$
Indeed, if r is parametrized by arc length $s$, then $\mathbf{T}\=x\prime \left(s\right)\mathbf{i}plus;y\prime \left(s\right)\mathbf{j}$ is a unit tangent vector along C, and a normal vector to the right of T is $\mathbf{N}\=y\prime \left(s\right)\mathbf{i}x\prime \left(s\right)\mathbf{j}$. Consequently, $\mathbf{F}\xb7\mathbf{N}\mathrm{ds}equals;f\mathrm{dy}g\mathrm{dx}$.
A mnemonic for remembering the flux integral is the word "flux" itself. Its first letter is the first letter of the integrand; its last, the last letter of the integrand. These two letters indicate where the function $f\left(x\,y\right)$ and the differential $\mathrm{dx}$ go; the function $g\left(x\,y\right)$ and the differential $\mathrm{dy}$ have to fall into place on either side of the minus sign.
Finally, note that for the line integral of the normal component of F, there is no analog for the $\mathbf{F}\xb7\mathbf{dr}$ appearing in the line integral of the tangential component.

Table 9.5.5 Flux through a plane curve, the line integral of the normal component of a field F.



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Line Integrals in Maple


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Line integrals in Maple can always be formulated via the basic commands Int and int and their typeset equivalents. However, line integrals can much more easily be formulated and evaluated with the Context Panel system, several Task Templates, and three commands in the VectorCalculus packages.
Table 9.5.6 details the three commands available in the VectorCalculus packages. The yellow cells show the predefined curves that the commands recognize. Each command from either package recognizes the same set of curves. For these curves, the commands can return both the unevaluated integral and the value of the integral. In addition, for the curves shown in color (i.e., Circle, Line, LineSegments, Path), the Student versions admit an additional optional return, namely, a graph of the field arrows, the curve, and either a tangent or normal vector, as appropriate. In the scalar case, therefore, no graphs are drawn.
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Scalar

Tangential Component

Normal Component

Student: PathInt
Main: PathInt

Student: LineInt
Main: LineInt

Student: Flux
Main: Flux

Arc
Ellipse
Circle
Line
LineSegments
Path

Arc
Ellipse
Circle
Line
LineSegments
Path

Arc
Ellipse
Circle
Line
LineSegments
Path

Table 9.5.6 Commands for line integrals



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The hyperlinks in Table 9.5.6 lead to the appropriate help pages. Although the help pages for the Student version of the LineInt and Flux commands lists only those curves for which graphs can be drawn, the Student versions of these commands do recognize the same curves as the parallel commands in the VectorCalculus package.
Table 9.5.7 details the task templates available for implementing line integrals.
Tools≻Tasks≻Browse: Calculus  Vector≻Integration≻


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There are no task templates implementing the PathInt command


Line Integrals

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There are nine task templates implementing the LineInt command for the following options:


Flux

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There are four task templates implementing the Flux command in the plane.

(There are seven others for flux through surface.)

Table 9.5.7 Task templates for line integrals



The nine task templates implementing the LineInt command display that command. Those for the Flux command are constructed from embedded components, and are therefore "syntaxfree." The task templates for calculating flux through plane curves are illustrated in Examples 9.5.(1723).
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With the Student VectorCalculus package installed, the Context Panel for a vector field contains the option "Line Integral", the selection of which opens the dialog shown in Figure 9.5.1.

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With this dialog, the line integral of the vector field can then be integrated over any path supported by the LineInt command.

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The dialog returns the unevaluated integral, the value of that integral, or, where applicable, a graph of the field, curve, and tangent vector(s) along the curve.

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Use of this functionality is made in Examples 9.5.(1016).

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The Context Panel for a scalar contains the options "Line Integral (2D)" and "Line Integral (3D)". Selecting the first opens the dialog shown in Figure 9.5.2; the second, in Figure 9.5.3.



Figure 9.5.1 Lineintegral dialog in the Context Panel






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Figure 9.5.2 Line integral of a scalar, 2D





Figure 9.5.3 Line integral of a scalar, 3D






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Examples


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Example 9.5.1

Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken along the line segment from $\left(1\,2\right)$ to $\left(3\,5\right)$.

Example 9.5.2

Obtain the line integral of the scalar function $f\left(x\,y\,z\right)\=xyz$, taken along the line segment from $\left(1\,2\,3\right)$ to $\left(5\,3\,2\right)$.

Example 9.5.3

Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken along the polygonal line connecting the points $\left(1\,2\right)$, $\left(3\,5\right)$, $\left(4\,0\right)$, $\left(2\,7\right)$, in that order.

Example 9.5.4

Obtain the line integral of the scalar function $f\left(x\,y\,z\right)\=xyz$, taken along the polygonal line connecting the points $\left(1\,2\,3\right)$, $\left(5\,3\,2\right)$, and $\left(0\,1\,4\right)$, in that order.

Example 9.5.5

Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken around the circle whose center is $\left(3\,5\right)$ and whose radius is 2.

Example 9.5.6

Let $C$ be the circle whose center is $\left(3\,5\right)$ and whose radius is 2. Let $c$ be the arc on $C$ subtended by a central angle of $\mathrm{\π}\/4$ radians measured counterclockwise from the right half of a horizontal diagonal. Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken along $c$.

Example 9.5.7

Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken around the ellipse whose center is $\left(3\,4\right)$ and whose semimajor and semiminor axes, parallel to the coordinate axes, are 2 and 1, respectively.

Example 9.5.8

Let $C$ be the ellipse ${x}^{2}\+4{y}^{2}equals;1$ and let $c$ be the arc subtended by an angle of $\mathrm{\π}\/4$ radians measured counterclockwise from the positive $x$axis. Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken along $c$.

Example 9.5.9

Let $C$ be that portion of the parabola $y\={x}^{2}$ from the origin to the point $\left(1\,1\right)$. Obtain the line integral of the scalar function $f\left(x\,y\right)\=xy$, taken along $C$.

Example 9.5.10

Calculate the work that the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ does on a unit positive charge as the charge moves along a line from $\left(1\,2\right)$ to $\left(3\,5\right)$.

Example 9.5.11

Calculate the work that the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ does on a unit positive charge as the charge moves along the polygonal line connecting the points $\left(1\,2\right)$, $\left(3\,5\right)$, $\left(4\,0\right)$, $\left(2\,7\right)$, in that order.

Example 9.5.12

Calculate the circulation of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ on the circle whose center is $\left(3\,5\right)$ and whose radius is 2.

Example 9.5.13

Let $C$ be the circle whose center is $\left(3\,5\right)$ and whose radius is 2. Let $c$ be the arc on $C$ subtended by a central angle of $\mathrm{\π}\/4$ radians measured counterclockwise from the right half of a horizontal diagonal. Calculate the work that the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ does on a unit positive charge as the charge moves counterclockwise along $c$.

Example 9.5.14

Calculate the circulation of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ on the ellipse whose center is $\left(3\,4\right)$ and whose semimajor and semiminor axes, parallel to the coordinate axes, are 2 and 1, respectively.

Example 9.5.15

Let $C$ be the ellipse ${x}^{2}\+2{y}^{2}equals;1$ and let $c$ be the arc subtended by an angle of $\mathrm{\π}\/4$ radians measured counterclockwise from the positive $x$axis. Calculate the work that the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ does on a unit positive charge as the charge moves counterclockwise along $c$.

Example 9.5.16

Let $C$ be that portion of the parabola $y\={x}^{2}$ from the origin to the point $\left(1\,1\right)$. Calculate the work that the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ does on a unit positive charge as the charge moves away from the origin along $C$.

Example 9.5.17

Calculate the flux of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ through the line from $\left(1\,2\right)$ to $\left(3\,5\right)$. Take the normal to the right of the tangent along the line.

Example 9.5.18

Calculate the flux of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ through the polygonal line connecting the points $\left(1\,2\right)$, $\left(3\,5\right)$, $\mathrm{OmegaR}\u2254\left[\mathrm{dx1}\,{\ⅇ}^{\mathrm{x1}}\left(\mathrm{x3}\+\mathrm{x1}\mathrm{x3}\+\mathrm{x2}\right)\mathrm{dx1}\+\mathrm{dx2}\,\mathrm{x3}{\ⅇ}^{\mathrm{x1}}\mathrm{dx1}\+\mathrm{dx3}\right]$, $\left(7\,6\right)$, in that order. Take the normal to the right of the tangent along the path.

Example 9.5.19

Calculate the flux of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ through the circle whose center is $\left(3\,5\right)$ and whose radius is 2. Take the outward normal along the circle.

Example 9.5.20

Calculate the flux of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ through the ellipse whose center is $\left(3\,4\right)$ and whose semimajor and semiminor axes, parallel to the coordinate axes, are 2 and 1, respectively. Take the outward normal along the ellipse.

Example 9.5.21

Let $C$ be that portion of the parabola $y\={x}^{2}$ from the origin to the point $\left(1\,1\right)$. Calculate the flux of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ through $C$. Take the normal to the right of the tangent along the path.

Example 9.5.22

Let $C$ be the circle whose center is $\left(3\,5\right)$ and whose radius is 2. Let $c$ be the arc on $C$ subtended by a central angle of $\mathrm{\π}\/4$ radians measured counterclockwise from the right half of a horizontal diagonal. Calculate the flux of the field $\mathbf{F}\=xy\mathbf{i}plus;\left(xplus;y\right)\mathbf{j}$ through $c$. Take the normal to the right of the tangent along $c$.




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