The Idea of a Line Integral

Section 4.2 The Idea of a Line Integral

Motivating Questions

What is an oriented curve and how can we represent one algebraically?
What is the meaning of the line integral of a vector-valued function along a curve and how can we estimate if its value is positive, negative, or zero?
What are important properties of the line integral of a vector-valued functions along a curve?

As we discussed in Section 4.1, vector fields are often useful as representations of forces such as gravity, wind, and flowing water. We learned in Section 1.3 that the dot product of a force vector and a displacement vector tells us how much work the force did on the object as it moved from the tail of its displacement vector to the tip. However, things get more complicated when an object’s movement is not in a straight line and when the force is not uniform throughout the area in which the object moves. For example, how much work is done by a wind of 30 ^mi⁄_h toward the northwest on an airplane that’s flying 500 mi due north? What if the wind gets weaker the farther north the plane gets? In this section, we begin investigating how integration can be used to calculate the work a force field does in such circumstances.

Preview Activity 4.2.1.

Recall from Section 1.3 that the work done by a force \(\vF\) on an object that moves with displacement vector \(\vv\) is \(\vF\cdot \vv\text{.}\) In this Preview Activity, we consider the work done by a wind blowing due east at 45 ^mi⁄_h on an airplane at various stages of its journey.

(a)

A pilot flies for an hour and finds that he is 300 mi from where he started at a heading of 20 ° degrees east of due north. Find the work the wind has done on the airplane during the flight.

Answer.

\(3078.18\)

Solution.

Here \(\vF = 45\vi\text{.}\) The displacement vector of the airplane is \(\vv_1 = \langle 200\cos(70^\circ),200\sin(70^\circ)\rangle \approx \langle 68.404,187.94\rangle\text{.}\) Thus, the work done by the wind is \(\vF\cdot \vv_1 = 45\cdot 68.404 = 3078.18\text{.}\)

(b)

An hour later, the pilot determines that he is 275 mi due north of where he previously checked his position. Find the work done by the wind on the airplane during the second hour of the flight.

Answer.

\(0\)

Solution.

The displacement vector is \(\vv_2 = 275\vj\text{.}\) Thus, the work done by the wind is \(\vv_2 \cdot \vF = 0\text{.}\)

(c)

Find the pilot’s displacement from his original position after two hours of flying and use that to find the work done by the wind on the airplane during the first two hours of flight.

Answer.

\(\langle 68.40, 462.94\rangle\text{;}\) \(3078.18\)

Solution.

The displacement is \(\vv = \vv_1 + \vv_2 = \langle 68.40, 462.94\rangle\text{.}\) The work done by the wind is thus \(\vv\cdot \vF = 3078.18\text{.}\)

(d)

How does your answer to the previous part connect to the answers to the first two parts?

Answer.

Sum.

Solution.

It is the sum of the answers to the first two parts.

(e)

Suppose that the pilot then flies 45 ° west of due north for 200 mi. Find the work done by the wind on the airplane during this part of the journey.

Answer.

\(-6363.96\)

Solution.

The displacement vector is \(\vv_3 = \langle 200\cos(135^\circ),200\sin(135^\circ)\rangle = \langle -141.42, 141.42\rangle\text{.}\) The work is therefore \(\vv_3 \cdot \vF = -6363.96\text{.}\)

Subsection 4.2.1 Orientations of Curves

Given our motivation for calculating the work that a force field does on an object as it moves through the field, it is natural to concern ourselves with how the object moves. In particular, in many circumstances it will be different if an object moves from the point \((0,1)\) to the point \((4,3)\) by first going up the \(y\)-axis to \((0,3)\) and then moving horizontally to \((4,3)\) than if the object moves along the line segment from \((0,1)\) directly to \((4,3)\text{.}\) Similarly, given a fixed force field, we would expect the work done to be different (in fact, opposite) if the object moves from \((4,3)\) to \((0,1)\) directly along a line segment. We say that a curve in \(\R^2\) or \(\R^3\) is oriented if we have specified the direction of travel along the curve. When a curve is given parametrically (including as a vector-valued function), our convention will be that the orientation follows from the smallest allowable value of the parameter to the largest.

Activity 4.2.2.

For each curve below, find a parametrization of the curve. Ensure that each curve’s orientation matches the one specified.

(a)

The line segment in \(\R^3\) from \((0,1,-2)\) to \((3,-1,2)\text{.}\)

Answer.

\(\vr(t) = 3t\vi + (1-2t)\vj + (-2+4t)\vk\) for \(0\leq t\leq 1\)

Solution.

The direction vector for the line segment is \(\vv = \langle 3-0, -1-1, 2-(-2)\rangle = \langle 3,-2,4\rangle\text{.}\) The line segment starts at \(\vr_0 = \langle 0,1,-2\rangle\text{.}\) Thus, the line segment given as the vector-valued function

\begin{equation*} \vr(t) = 3t\vi + (1-2t)\vj + (-2+4t)\vk\qquad\text{for }0\leq t\leq 1\text{.} \end{equation*}

(b)

The line segment in \(\R^3\) from \((3,-1,2)\) to \((0,1,-2)\text{.}\)

Answer.

\(\vr(t) = (3-3t)\vi + (-1+2t)\vj + (2-4t)\vk\) for \(0\leq t\leq 1\)

Solution.

The direction vector for the line segment is \(\vv = \langle 0-3, 1-(-1), -2-2\rangle = \langle -3,2,-4\rangle\text{.}\) The line segment starts at \(\vr_0 = \langle 3,-1,2\rangle\text{.}\) Thus, the line segment given as the vector-valued function

\begin{equation*} \vr(t) = (3-3t)\vi + (-1+2t)\vj + (2-4t)\vk\qquad\text{for }0\leq t\leq 1\text{.} \end{equation*}

(c)

The circle of radius \(3\) (in \(\R^2\)) centered at the origin, beginning at the point \((0,-3)\) and proceeding clockwise around the circle.

Answer.

\(\vr(t) = \langle 3\cos(t),-3\sin(t)\rangle\) for \(\pi/2\leq t \leq 5\pi/2\)

Solution.

We can parametrize the circle in question in a clockwise direction as \(\vr(t) = \langle 3\cos(t),-3\sin(t)\rangle\text{.}\) To get the correct starting point, we require \(\pi/2\leq t \leq 5\pi/2\text{.}\)

(d)

In \(\R^2\text{,}\) the portion of the parabola \(y^2 = x\) from the point \((4,2)\) to the point \((1,-1)\text{.}\)

Answer.

\(\langle t^2,-t\rangle\) for \(-2\leq t\leq 1\)

Solution.

Here we can let \(y(t) = -t\) and \(x(t) = t^2\) for \(-2\leq t\leq 1\text{.}\)

Notice that there are, in general, many ways to parametrize an oriented curve. With line segments, it is common to have the parameter range from \(0\) to \(1\text{,}\) although there are sometimes good reasons to choose another method. For circles and ellipses, you may find it useful to interchange the placement of \(\cos(t)\) and \(\sin(t)\) to change the orientation, but then careful attention may need to be paid to the start and end points.

The applet below is an example of a parametric curve plotted in GeoGebra. Right click in the box and move your mouse to view the curve from different angles. The vector valued function \(\vr(t) = \langle \cos(5t), 2\sin(3t), 3-0.5 t\rangle\) is graphed, as well as a point \(P\) that traverses \(\vr(t)\) from \(t=0\) to \(t=8\text{.}\) Click here to view in GeoGebra

www.geogebra.org/calculator/vt3ywd8t

. From GeoGebra, you can click on the calculator labeled "Algebra" on the left hand bar to open the input. Try changing the input and seeing how the graph changes!

Instructions.

You can view a larger version of this here

www.geogebra.org/calculator/vt3ywd8t

Figure 4.2.1. A GeoGebra applet to explore oriented curves. View in GeoGebra

geogebra.org/calculator/vt3ywd8t

Subsection 4.2.2 Line Integrals

Just as when we differentiated a vector-valued function \(\vr(t)\) to find a tangent vector, we begin by dividing a curve \(C\) oriented from a point \(P\) to a point \(Q\) into \(n\) small, straight pieces. Each of these pieces is in an area where the vector field \(\vF\) is nearly constant, provided we use enough pieces. In Figure 4.2.2, we show this situation. Each \(\vr_i\) is the tip of a vector that traces out the curve, which makes the vectors \(\Delta\vr_i = \vr_{i+1}-\vr_i\) (shown in blue) approximate the curve \(C\text{.}\) The green vectors are the vectors in the vector field \(\vF\) at each of the designated points along the curve.

described in detail following the image — Figure 4.2.2. A curve \(C\) oriented from the point \(P\) to the point \(Q\text{.}\) The tips of the vectors \(\vr_i\) that trace out the curve are shown as points. The blue vectors are the \(\Delta\vr_i\text{,}\) while the green vectors are the vectors associated to each of the points by a vector field \(\vF\text{.}\)

If we are trying to determine how much a wind current helps or hinders an aircraft flying along a path determined by the curve, then calculating the dot product \(\vF(\vr_i)\cdot \Delta\vr_i\) makes sense for the local amount of help or hindrance. This is because if the vector \(\vr_i\) along the curve and the force field vector \(\vF(\vr_i)\) point in similar directions, the dot product will be positive.

⁴

We are abusing notation here a tiny bit, since technically the domain of \(\vF\) is points in \(\R^2\) or \(\R^3\text{,}\) and \(\vr_i\) is a vector. By \(\vF(\vr)\text{,}\) we mean \(\vF(r_1,r_2)\text{,}\) where \(\vr= \langle r_1,r_2\rangle\text{.}\)

On the other hand, if the angle between them is obtuse, the dot product will be negative and we also would note that the force field is hindering the aircraft’s progress. Taking the sum over \(i=0,\dots,n-1\text{,}\) we have a Riemann sum that approximates the work done by the vector field on the aircraft as it flies along \(C\text{:}\)

\begin{equation*} \sum_{i=0}^{n-1}\vF(\vr_i)\cdot\Delta \vr_i\text{.} \end{equation*}

This suggests the following definition.

Definition 4.2.3.

Let \(C\) be an oriented curve and \(\vF\) a vector field defined in a region containing \(C\text{.}\) The line integral of \(\vF\) along \(C\) is

\begin{equation*} \int_C \vF\cdot d\vr = \lim_{|\Delta\vr_i|\to 0} \sum_{i=0}^{n-1}\vF(\vr_i)\cdot\Delta\vr_i\text{,} \end{equation*}

provided the limit exists.

The limit in Definition 4.2.3 exists provided that \(\vF\) is a continuous vector field, by which we mean that each component function of \(\vF\) is continuous as a function of \(2\) or \(3\) variables, and that \(C\) is a piecewise smooth curved traced out from its initial point to its terminal point without retracing any portion of the curve.

Because the dot products in the definition of the line integral \(\int_C\vF\cdot d\vr\) can each be viewed as the work done by \(\vF\) as an object moves along the (very small) vector \(\Delta\vr_i\text{,}\) the line integral gives the total work done by the vector field on an object that moves along \(C\) (in the direction of its orientation).

Activity 4.2.3.

Shown in Figure 4.2.4 are two vector fields, \(\vF\) and \(\vG\) and four oriented curves, as labeled in the plots. For each of the line integrals below, determine if its value should be positive, negative, or zero. Do this by thinking about if the vector field is helping or hindering a particle moving along the oriented curve, rather than by doing calculations.

(a)

\(\displaystyle\int_{C_1}\vF\cdot d\vr\)

Answer.

\(0\)

Solution.

The portion of \(C_1\) that lies in the third quadrant is oriented in the opposite direction of the vector field. On the other hand, the portion of \(C_1\) in the first quadrant is oriented in the same direction as \(\vF\text{.}\) By symmetry, this forces \(\displaystyle\int_{C_1}\vF\cdot d\vr=0\text{.}\)

(b)

\(\displaystyle\int_{C_2}\vF\cdot d\vr\)

Answer.

Negative

Solution.

The angle between the vectors of \(\vF\) and \(C_2\) are all obtuse. Thus, the line integral is negative.

(c)

\(\displaystyle\int_{C_3}\vG\cdot d\vr\)

Answer.

Negative

Solution.

The angle between the vectors of \(\vG\) and \(C_3\) are all obtuse (except for \(x=0\text{,}\) where there is no angle), and thus the line integral is negative.

(d)

\(\displaystyle\int_{C_4}\vG\cdot d\vr\)

Answer.

Solution.

The vector field \(\vG\) is always orthogonal to \(C_4\text{.}\) Thus, the line integral is \(0\text{.}\)

The next several sections will be devoted to determining ways to calculate line integrals. Although we do not yet have techniques for calculating the exact value of line integrals, we will see methods to determine if a line integral is positive, negative, or zero. We will also be able to order line integrals from greatest to least.

Subsection 4.2.3 Properties of Line Integrals

In the Preview Activity, we implicitly made use of the idea that if \(C\) can be broken up into two curves \(C_1\) and \(C_2\) such that the terminal point of \(C_1\) is the initial point of \(C_2\text{,}\) then the line integral of \(\vF\) along \(C\) is the sum of the line integrals of \(\vF\) along \(C_1\) and along \(C_2\text{.}\) Recalling the property for definite integrals that tells us

\begin{equation*} \int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx\text{,} \end{equation*}

the ability to work with line integrals in this way is probably not surprising.

Before stating some useful properties of line integrals, we will establish some convenient notation. 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 \(R\text{,}\) 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 \(R\text{.}\) 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 analogue amongst the properties of definite integrals.

Properties of Line Integrals.

For a constant scalar \(k\text{,}\) vector fields \(\vF\) and \(\vG\text{,}\) and oriented curves \(C\text{,}\) \(C_1\text{,}\) and \(C_2\text{,}\) the following properties hold:

\(\displaystyle \displaystyle\int_C (k\vF)\cdot d\vr = k\int_C\vF\cdot d\vr\)
\(\displaystyle \displaystyle\int_C(\vF+\vG)\cdot d\vr = \int_C \vF\cdot d\vr + \int_C\vG\cdot d\vr\)
\(\displaystyle \displaystyle\int_{-C}\vF\cdot d\vr = -\int_C \vF\cdot d\vr\)
\(\displaystyle\int_{C_1+C_2} \vF\cdot d\vr = \int_{C_1}\vF\cdot d\vr + \int_{C_2}\vF\cdot d\vr\text{.}\)

Activity 4.2.4.

Figure 4.2.5 shows a vector field \(\vF\) as well as six oriented curves, as labeled in the plot.

(a)

Is \(\int_{C_5}\vF\cdot d\vr\) positive, negative, or zero? Explain.

Answer.

Negative

Solution.

The portion of the vector field that produces acute angles with \(C_5\) has small magnitude than the portion that produces obtuse angles. Thus, the line integral must be negative.

(b)

Let \(C = C_1+C_2+C_3+C_4\text{.}\) Determine if \(\displaystyle\int_C\vF\cdot d\vr\) is positive, negative, or zero.

Answer.

Negative

Solution.

Notice that \(\vF\) is orthogonal to \(C_1\) and \(C_3\text{.}\) Thus, \(\displaystyle\int_{C_1}\vF\cdot d\vr = \int_{C_3}\vF\cdot d\vr = 0\text{.}\) Where the magnitude of the vectors in \(\vF\) is larger, \(\vF\) points in the opposite direction as the orientation of \(C\text{.}\) Thus, \(\int_C\vF\cdot d\vr\) is negative.

(c)

Order the line integrals below from smallest to largest.

\begin{equation*} \int_{C_1}\vF\cdot d\vr\quad \int_{C_2}\vF\cdot d\vr\quad \int_{C_3}\vF\cdot d\vr\quad \int_{C_4}\vF\cdot d\vr\quad \int_{C_5}\vF\cdot d\vr\quad \int_{C_6}\vF\cdot d\vr \end{equation*}

Answer.

\(\)

Solution.

Notice that \(\displaystyle\int_{C_4}\vF\cdot d\vr\) is the only one of the six line integrals that is positive, so it is the largest. Since \(\vF\) is orthogonal to \(C_1\) and \(C_3\text{,}\) \(\displaystyle\int_{C_1}\vF\cdot d\vr = \int_{C_3}\vF\cdot d\vr = 0\text{.}\) Note that while \(C_6\) is twice as long as \(C_2\text{,}\) the vectors in the vector field that are hindering \(C_2\) are more than twice as long as those hindering \(C_6\) Thus, \(\displaystyle\int_{C_2}\vF\cdot d\vr \lt \displaystyle\int_{C_6}\vF\cdot d\vr\text{.}\) We do not yet have all the tools necessary to calculate the exact value of \(\displaystyle\int_{C_5}\vF\cdot d\vr\text{.}\) However, it is greater than \(\displaystyle\int_{C_6}\vF\cdot d\vr\text{.}\) Thus, the ordering is

\begin{equation*} \int_{C_4}\vF\cdot d\vr > \int_{C_1}\vF\cdot d\vr = \int_{C_3}\vF\cdot d\vr > \int_{C_5}\vF\cdot d\vr > \int_{C_6}\vF\cdot d\vr > \int_{C_2}\vF\cdot d\vr\text{.} \end{equation*}

Subsection 4.2.4 The Circulation of a Vector Field

If an oriented curve \(C\) ends at the same point where it started, we say that \(C\) is closed. The line integral of a vector field \(\vF\) along a closed curve \(C\) is called the circulation of \(\vF\) around \(C\text{.}\) To emphasize the fact that \(C\) is closed, we sometimes write \(\oint_C \vF\cdot d\vr\) for \(\int_C \vF\cdot d\vr\text{.}\) Circulation serves as a measure of a vector field’s tendency to rotate in a manner consistent with the orientation of the curve.

Activity 4.2.5.

Determine if the circulation of the vector field around each of the closed curves shown in Figure 4.2.6 is positive, negative, or zero.

Answer.

Circular curve has positive circulation. Larger rectangular curve has positive circulation. Smaller rectangular curve has negative circulation.

Solution.

The circular curve is oriented consistently with the vector field. Thus, the circulation is positive. The larger rectangular curve is either orthogonal to the vector field or oriented consistently with the vector field. Thus, the circulation is positive. The smaller rectangular curve is orthogonal to the vector field or has the opposite orientation. Thus, its circulation is negative.

Subsection 4.2.5 Summary

An oriented curve is a vector-valued function of one variable \(\vr(t)\) where we interpret the initial and terminal values of the domain of \(\vr\) as giving an orientation to the curve. A curve that ends at the same point where it started is said to be closed.
A line integral measures of a vector field along an oriented curve measures the extent to which the vector field points in a direction consistent with the orientation of the curve.
Line integrals have many properties that are analogous to those of definite integrals of functions of a single variable. The line integral of a vector field along a closed curve is called the circulation of the vector field along the curve.

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