Vector Fields and Line Integrals field...242 26. Vector Fields and Line Integrals Introduction...

Module 26

Vector Fields and Line Integrals

Figure 26.1: Air flow over a wing is describedwith vector fields. Picture made available byChaoqun Liu and used with permission.

6?

666 66??? ??

N

S

Figure 26.2: Magnetic fields are vector fields.

Prerequisites.The minimum prerequisites for Module 26, “Vector Fields and Line Integrals” are

• An introduction to vectors such as in Module 20,

• An introduction to multivariable functions such as in Section 21.2,

• An introduction to the dynamics of space curves such as in Module 24,

• Single variable integration,

• Multivariable differentiation and the gradient as introduced for example inModule 25. If we are willing to require the gradient is continuous in a neigh-borhood ofC in Theorem 26.3.1, we only need Sections 25.1 and 25.2.

Figure 26.3: Gravitation is a vector field.

Learning Objectives.After successfully completing this module the student will be able to

1. Sketch the graph of a vector field.

2. Sketch the flow lines of a vector field.

3. State examples of vector fields that occur in nature.

4. Compute the line integral of a vector field along a curve

• directly,

• using the fundamental theorem for line integrals.

5. Estimate line integrals of a vector field along a curve from a graph of thecurve and the vector field.

6. Compute the gradient vector field of a scalar function.

7. Compute the potential of a conservative vector field.

8. Determine if a vector field is conservative and explain why by using deriva-tives or (estimates of) line integrals.

241

schroder

Surface integrals, the Divergence Theorem and Stokes' Theorem are treated in Module 28 "Vector Analysis"

242 26. Vector Fields and Line Integrals

IntroductionScalar quantities (like temperature, pressure, density, etc.) can vary in space

and time (cf. Section 21.2). Similarly, vectorial quantities (like velocities or forces)can also vary in space and time. We will concentrate mostly on the variation inspace in this text. For example,

• The gravitational field of the earth exerts a position-dependent force on anymass. The force is directed towards the center of the earth, and the farther amass is away from the earth, the smaller is the attractive force.

• The flow velocity of air along the wing of an aircraft is larger on top of thewing than it is on the bottom of it. This difference causes lower air pressure atthe top of the wing, which creates the “lift” an airplane needs to stay airborne.

While the above two examples could still be described with scalar functions(provided we “know” the direction), such an easy way out becomes impossiblewhen more complex situations are considered.

Figure 26.4: Simulation of air flow over an air-plane wing. Picture made available by ChaoqunLiu and used with permission.

• The sum of the gravitational forces exerted on a mass also contains compo-nents aside from the gravitation of earth. Computations of the path of space-borne vehicles must take the gravitation of other celestial bodies (sun, moon,planets) into account. All these forces affect the direction of the gravitationalforce acting on the vehicle.

• Air flow is not always completely smooth (or “laminar”). Turbulent air flowscan contain eddies of various sizes that rotate at different speeds. These tur-bulences at the tip of an airplane wing make the plane less efficient, becausethey cause drag. Figure 26.4 shows flow instability on a delta wing. Thefarther to the right we are, the more turbulent the air flow gets.

Vector fields describe how vector valued quantities vary in space and time.Further examples include the electrical field that surrounds charges, the magneticfield that surrounds currents (moving charges) or permanent magnets (cf. Figure26.5) as well as “fields” that at first have no good explanation, such as the Coriolisforce that acts on any moving body on earth. Various types of vector fields aredescribed in Section 26.1.

Figure 26.5: Though they are conceptual enti-ties, the flow lines of vector fields can be madevisible. Iron shavings align themselves with theflow lines of a magnetic field. The above pic-tures show the magnetic fields around a regularmagnet and around a wire in which current isflowing.

Vector fields are detected and measured through their effect on their surround-ings.

• The effect of the gravitational field is omnipresent on earth. For example,to move up a hill or a flight of stairs, we have to work against gravity. Bymeasuring the force that pulls us towards the earth, we measure the effects ofthe gravitational field.

• The effect of a velocity field is motion (of a fluid). It can be seen or felt andthen measured with the appropriate instruments.

• Iron shavings and compass needles align themselves in a magnetic field.

To model the effect of a vector field on its surroundings, we must understandhow the field interacts with objects that move through the field. Basic physical lawsdictate the interaction of a homogeneous field with a body that is at rest or movingat constant velocity. The line integral (discussed in Sections 26.2 and 26.3) is usedto describe the interaction of a general vector field with a moving body.

26.1. Vector Fields 243

26.1 Vector Fields

The mathematical definition of a vector field is straightforward. We assign a vectorto every point in our domain.

Definition 26.1.1 If D is a region inRn, then avector field on D is a functionEF that assigns each(x1, . . . , xn) in D an n-dimensional vectorEF(x1, . . . , xn).

As always, we state the general definition so that we will not need to re-state thedefinition for each dimension. As usual, we avoid unnecessary trouble that stemsfrom distinguishing a point from its position vector.

Notation 26.1.2 For a point a with position vectorEa we will also write EF (Ea)

instead ofEF(a).This is done because we will often evaluate vector fields along curvesEc(t), or,in Module 28, on a surfaceEs(u, v).

Let us now consider some examples.

Example 26.1.3Sketch the vector fieldEF(x, y) =

(−x

x2+y2

−yx2+y2

)and explain its ap-

pearance.Sketching a vector field is similar to sketching a direction field for antideriva-

tives (cf. SectionIND.A). Choose a grid of points in the domain. ComputeEF(a) foreach pointa in the grid and then draw a vectorEF(a) whose starting point is ata.Unlike for direction fields, the vectors will not all have the same length. A com-puter plot of our vector field is given in Figure 26.6. All vectors ofEF(x, y) pointtowards the origin and the magnitude of the vectors increases as we approach theorigin.

Figure 26.6: The vector fieldEF(Er ) = −1

r

Er

r,

of Example 26.1.3. The window is [−1, 1] ×

[−1, 1]. The lengths of the arrows are scaledproportionally to fit the picture.

To explain the appearance we note that

EF(x, y) =

(−x

x2+y2

−yx2+y2

)= −

1

x2 + y2

(xy

).

The vector

(xy

)points radially outwards from the origin, so the vector−

(xy

)attached at(x, y) points towards the origin. The factor

1

x2 + y2is the square of the

magnitude of−

(xy

). This means the magnitude ofEF at (x, y) is

1√x2 + y2

. This

magnitude is large near the origin (it goes to infinity as(x, y) → (0, 0)) and itdecays to zero asx2

+ y2 gets large.

Qualitatively, the fieldEF in Example 26.1.3 behaves like the gravitational fieldof a mass at the origin in two dimensional space. Indeed, because vector fieldsare hard to plot and visualize, people often use two dimensional plots of “cross-sections” to visualize a vector field. For example, Figure 26.5 depicts a cross-section of a three dimensional field.

But is the field EF in Example 26.1.3really a cross section of a gravitationalfield? To investigate this question we use notation that is closer to the way physics


describes vector fields. Denote byr :=√

x2 + y2 the distance of the point(x, y)

to the origin and letEr =

(xy

). Then

EF(x, y) = −1

x2 + y2

(xy

)= −

1

r 2Er = −

1

r

Er

r= EF(Er ).

The magnitude ofEF is1

rand

Er

ris a unit vector. Gravity obeys an inversesquare

law however. So whileEF is acentral vector field (a vector field in which all vectorspoint to the center), it does not represent a gravitational field. Still, pictures such asFigure 26.6 are often used to depict gravitational fields of point masses. While thisis strictly speaking not accurate, there is a good reason to use such a picture.

Figure 26.7: The vector fieldEF(Er ) = −1

r 2

Er

r,

which is the essence of the gravitational field(cf. Example 26.1.4) as well as of the electricalfield surrounding a negative charge (cf. Example26.1.6). The window is [−1, 1] × [−1, 1].

Example 26.1.4The gravitational field of a mass. The massM located at the

origin attracts the massm located atEr with a force of magnitudeF (Er ) =GmM

r 2,

and the force points towards the origin.G ≈ 6.67× 10−11Nm2

kg2 is the gravitational

constant.

This means the corresponding force is

EF(Er ) = −GmM

r 2

Er

r= −

GmM

r 3Er = −

GmM(√x2 + y2

)3

(xy

),

wherer = |Er |. Aside from the scaling factorsG, m and M , the gravitational

field behaves like the fieldEF(Er ) = −1

r 2

Er

r(cf. Figure 26.7). As is customary in

mathematics and science, for a qualitative analysis we will often drop scaling con-stants to be able to graph and analyze a field. This step removes some “numericalclutter” and leaves just the object and its properties to be analyzed.

We note that the field in Figure 26.7 looks different from the fieldEF(Er ) = −1

r

Er

rin Figure 26.6. Both fields point towards the origin. However, the magnitude ofthe vector field in Figure 26.7 shrinks faster as we move away from the origin.Any depiction will show a few discernible vectors near the origin and then a largenumber of vectors that are barely long enough to be recognized.

This is why, for sake of visualization, some vector fields are depicted with thewrong magnitude. The accurate picture of the gravitational field in Figure 26.7explains the concept not nearly as well as a picture such as Figure 26.6.

Finally, note that for a reasonable theory, we need to make our field expressionindependent of the test massm. This is done in Section 26.1.5.

The discussion so far has implicitly shown that there are several ways to repre-sent and describe vector fields.


Remark 26.1.5 Vector fields can be described in any of the following ways.

1. Symbolically with a componentwise expression

EF(x, y, . . .) =

P(x, y, . . .)

Q(x, y, . . .)...

. In this description, every com-

ponent of the vector field is a multivariable function.

2. Verbally, by describing the magnitude and direction of the field at everypoint.

3. Symbolically by giving an expression that is the product of a magnitude

and a unit vector. In this description we often useEr =

x1...

xn

for the

position to reduce the number of symbols.

For sufficiently simple vector fields and for calculations we will use the com-ponentwise expression in 1. The descriptions in 2 and 3 are more prevalent in thesciences. Realistic vector fields, such as fields that obey an inverse square law havecomplicated components. Descriptions 2 and 3 can be more effective than the com-ponentwise descriptions. Especially description 3 can be exact with fewer symbolsthan 1. For example, we can describe our two-dimensional slice of the gravitationalfield in Example 26.1.4 in at least three ways.

Figure 26.8: Light strings align themselves withthe flow lines of the electrical field. The first pic-ture shows an uncharged van de Graaf generator,the second a charged one.

1. The componentwise description isEF(x, y) =

−x

(x2+y2)32

−y

(x2+y2)32

.

2. The verbal description says the following. The vector fieldEF points towardsthe origin and its magnitude is the reciprocal of the square of the distance ofthe point from the origin.

3. Using symbolic magnitude and direction we obtainEF(Er ) = −1

r 2

Er

r.

26.1.1 Electrical Fields and Their Superposition

Another example of a central force field is the electrical field of a point charge.

Example 26.1.6The electrical field of a charge. The chargeQ located at the

origin attracts the chargeq located atEr with the force EF(Er ) =Qq

4πε0r 2

Er

r, where

r = ‖Er ‖ and ε0 = 8.8542· 10−12 As

Vmis the permittivity constant. Figure 26.7

gives an accurate picture how the field scales and Figure 26.6 gives a visualizationwith vectors that are more easily seen.

Since the field ofQ should only depend onQ, we need to make our fieldexpression independent of the test chargeq. This is done in Section 26.1.5.

In gravity, as well as in electricity and other situations, things get interestingwhen several vector fields overlap. Vector fields can be added at each point justlike vectors. This is why we will still say that the electrical field of a point charge isgiven by an expression as in Example 26.1.6, even when another charge is present.


The individual fields of the charges have not changed, but what we observe is thesum of the two fields.

Figure 26.9: The vector field in Example 26.1.7.The window is [−1, 1] × [−0.5, 2.5].

Example 26.1.7Figure 26.9 depicts the vector field

EF(Er ) = −1

r 2

Er

r+

1∥∥∥∥Er −

(02

)∥∥∥∥2

Er −

(02

)∥∥∥∥Er −

(02

)∥∥∥∥ .

This field is the sum of two fields. The fieldEE1(Er ) = −1

r 2

Er

rcan be visualized as

the field generated by a negative chargeq1 that is located at the origin. (We assumeeverything scales in such a way that the constants drop out.) Similarly, the field

EE2(Er ) =1∥∥∥∥Er −

(02

)∥∥∥∥2

Er −

(02

)∥∥∥∥Er −

(02

)∥∥∥∥ can be visualized as the field generated by

a positive chargeq2 = −q1 that is located at the point(0, 2).

Figure 26.10: A depiction of the vector field inExample 26.1.7 with longer vectors and centersmarked. The magnitudes of the vectors do notscale right, but the vectors are more easily visi-ble. The window is [−1, 1] × [−0.5, 2.5].

Figure 26.10 shows the field with exaggerated vector lengths and with the cen-ters of the two fields marked as charges. The overlap of the fields changes theoverall picture. The vectors now have a tendency to point away from(0, 2) andtowards the origin. This tendency varies depending how close to either point weare.

26.1.2 Magnetic Fields and Their Superposition

Not all vector fields are central fields of superpositions thereof.

Example 26.1.8The magnetic field that surrounds a current in a straight (in-finitely long) wire. Wires that carry a current are surrounded by magnetic fields.To get a start on analyzing this phenomenon we consider the idealized situation ofan infinitely long, straight wire. Ifr is the distance of a point to the wire, then the

magnitude of the field at that point isB =I

2πε0c2r=

I µ0

2πr, whereI is the current,

c is the speed of light in vacuum andµ0 :=1

ε0c2= 1.2566· 10−6 Vs

Amis theper-

meability constant. The magnetic field is tangential to concentric circles aroundthe wire. The direction is obtained via a right-hand rule. If the thumb of your righthand points in the direction in which the current flows, then the natural curve ofyour right hand gives the direction of the magnetic field (cf. Figure 26.11). Thefield does not have any component parallel to the current. Eliminating the constantsand assuming the current flows in the third dimension (vertically out of the paper),we obtain the field depicted in Figure 26.12.

Figure 26.11: The right hand rule for the mag-netic field that surrounds a current. The thumbgives the direction of the current and the naturalcurve of the right hand gives the direction of thecircular magnetic field surrounding the current.

Example 26.1.9Find a symbolic expression for the vector field that surrounds acurrent I that flows up along the z axis.

The field is described verbally in Example 26.1.8. We note the following.

1. The vector field has noz-component. This means that for depictions we cansketch a vector field in two dimensions with thez-axis coming vertically outof the sketch.


2. Since the currentI flows in the direction of the positivez-axis, the distanceto the “wire” is the distance to thez-axis, which isr =

√x2 + y2. In a two-

dimensional picture in which thez-axis points up, this is the distance fromthe origin.

Figure 26.12: The vector field

EF(Er ) =1

r 2〈−y, x〉, which is the essence

of the magnetic field in Example 26.1.8. Thewindow is [−2, 2] × [−2, 2].

3. The vectors will be tangent vectors to the concentric circles around the origin.

With the vectorEr =

(xy

)being a radius vector for a such a circle, the vec-

tors Er =

(−yx

)andEr =

(y

−x

)are tangent vectors that point in opposite

directions.

4. By the right hand rule, the magnetic field points in the mathematically posi-

tive direction. This means that the vectorEr =

(−yx

)gives the direction of

the field.

Overall, the magnetic field is

EB(Er ) =I

2πε0c2r

1

r

−yx0

=µ0I

2πr

1

r

−yx0

.

Figure 26.12 shows how this field looks in thexy-plane.

Figure 26.13: Vortices in fluids can be describedwith vector fields. This picture is a satelliteimage of hurricane Andrew off the coast ofLouisiana in 1992. Use of image from the Na-tional Oceanic and Atmospheric Administrationweb site [33] consistent with appropriate NOAApolicy.

“Rotating” fields as in Figure 26.12 also describe the motion of vortices in fluidflows. Figure 26.13 shows an image of hurricane Andrew in 1992. The rotationalmotion of the air and the clouds follows a pattern similar to that of Figure 26.12.

Just like central force fields, rotational fields can overlap.

Example 26.1.10The vector field in Figure 26.14 can be interpreted as the mag-netic field that surrounds two vertical currents. One current flows up thez-axis(which is vertical to the paper). The other current flows down the vertical linethrough(0, 1, 0).

Figure 26.14: The vector field which is theessence of the magnetic field in Example26.1.10. The window is [−1, 1] × [−0.5, 1.5].The vertical wires intersect thexy-plane at(0, 0)

(current goes up) and at(0, 1) (current goesdown).

The reader will determine the equation of this vector field in Exercise 10.

26.1.3 Flow Lines

We have seen that vector fields occur in a number of settings and that they can betricky to visualize using plotted arrows. Another way to visualize a vector field isthrough its flow lines (cf. Figure 26.15).

Definition 26.1.11 Let EF be a vector field. The vector valued functionEc is aflow line (or streamline) of the vector fieldEF if and only if for all t we havethat Ec′(t) is parallel to EF (Ec(t)). That is, at every point of the curve describedby Ec the vector field is tangential to the curve.

Plotting the flow lines essentially means that we “follow the arrows” in a vectorfield plot. This is similar to sketching an antiderivative in a direction field forintegration (cf. ExampleIND.B) or sketching the solution of an initial value probleminto the direction field of a differential equation (cf. ExamplesQDE.A andQDE.B).


Example 26.1.12Sketch an appropriate number of flow lines for the vector fieldin Figure 26.10.

The sketch is given in Figure 26.16. Each flow line starts at the point(0, 2) (thelocation of the positive chargeq2) and ends at the point(0, 0) (the location of thenegative chargeq1). The flow lines are tangential to vectors of the vector field.

There is some judgment to be exercised when using flow lines to visualize avector field. Too many lines will clutter the picture, too few won’t reveal all theinformation.

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666 66??? ??

N

S

Figure 26.15: The magnetic field of a permanentmagnet usually is not depicted using arrows. In-stead, flow lines obtained by following the ar-rows are used.

t

t

6

6� I

z 9

9 z

I �

? ?? ???

q2

q1

Figure 26.16: The flow lines for the vector fieldin Figure 26.10.

Flow lines sometimes are easier to sketch than the field, while carrying the sameinformation content.

Example 26.1.13Sketch the magnetic flow lines that surround a current that flowsup the z-axis.

Figure 26.17: The flow lines of the vector fieldin Example 26.1.13. Higher density of flow linesindicates a stronger field.

From the description of Example 26.1.8 we can infer that the flow lines areconcentric circles around thez-axis with a positive orientation of rotation. Figure26.17 shows the picture.

In Figure 26.17 as well as in many depictions of flow lines, the closer the flowlines are together, the stronger the field is in the respective region.

Flow lines have realistic meaning and they can be visualized. In a magneticfield, dropped iron shavings will outline the flow lines. In a velocity field (suchas the field of particle velocities in a fluid flow), a particle that is set adrift willfollow the flow lines of the velocity field. This visualization of a particle carriedby a moving fluid is probably the most helpful for sketching flow lines. Similarlyin a force field, a particle that is started at rest at a given point will follow the flowline that starts at that very point. So, for example, positive charges that are left tomove in the electrical field in Figure 26.16 will travel towards the negative chargeq1 along the flow line that goes through the charge’s starting position. Negativecharges will follow the flow line to the positive chargeq2.

Remark 26.1.14 The particle paths described by flow lines of velocity and forcefields can be quantified. Particle paths are solutions of differential equations thatreflect the laws governing the motion. These differential equations are easily setup if we know the field. For a velocity fieldEFv, the differential equation that gives

us a particle path isd

dtEc(t) = Ev(t) = EFv(Ec(t)). This is because in a velocity field,

the velocity of the particle is equal to the value of the field at the current posi-tion. For a force fieldEF , the differential equation that gives us a particle path is

md2

dt2Ec(t) = mEa(t) = EF(Ec(t)). This is because in a force field, the acceleration

of the particle times its mass is equal to the value of the force field at the currentposition.

Note that the above are examples of systems of differential equations. We havevectorial quantities on each side of the equal sign. For an example how magneticfields connect to systems of differential equations, cf. SectionEXB.F.

26.1.4 Gradient Fields

The gradientE∇ f (a) of a function f is a vector that depends on the positiona (cf.Section 25.3). Thus, the gradient field is a symbolic example of a vector field.


Definition 26.1.15 For the multivariable function f(x1, x2, . . . , xn), wherex1, . . . , xn are interpreted as space variables, thegradient vector field of f is

given bygrad( f )(x1, . . . , xn) := E∇ f (x1, . . . , xn) :=

∂ f∂x1∂ f∂x2...

∂ f∂xn

For the manifold interpretations of the gradient, cf. Section 25.3.4. Computinga gradient field is simple differentiation.

Example 26.1.16Find the gradient field of the function f(x, y) = 9 − x2− y2.

Figure 26.18: The contour map of Example26.1.16 with some vectors of the gradient fieldsuperimposed (vectors not drawn to scale). Thegradient field points towards the center and thevectors get shorter as we get closer to the origin.

E∇ f (x, y) =

(∂∂x

(9 − x2

− y2)

∂∂y

(9 − x2

− y2) ) =

(−2x−2y

).

A contour plot of the function with some gradients marked is given in Figure26.18. The figure confirms that the gradient points in the direction of steepest ascentand that it is perpendicular to the level curves.

There is an obvious advantage to gradient fields. Once the functionf is known,gradient fields can be described just using the functionf . This function usually isless complicated than the vector field expressionEF . Example 25.3.8 shows that

the central force fieldEF = −1

r 2

Er

ris the gradient of the scalar functionf =

1

r. In

particular, this means the following.

Theorem 26.1.17The electric field generated by a point charge is a gradientfield. The same is true for the gravitational field of a point mass.

With the gradient being a “multidimensional derivative” that retains many prop-erties of the one dimensional derivative (cf. Section 29.1) we should be able torecover the functionf via integration.

Example 26.1.18If possible, find a function f such thatE∇ f =

(6xy + 1

3x2

).

Since each component is a partial derivative, we must undo the partial derivativeby using what may well be called a “partial integral”.∫

6xy + 1 ∂x = 3x2y + x + c(y)∫3x2 ∂y = 3x2y + c(x).

Verify that E∇f=(6xy+1

3x2)!

We obtain two different expressions. However, the integration constants areonly constantwith respect to their integration variable. Therefore the “constant”in the first integral can still depend ony and the “constant” in the second integralcan still depend onx. (Hence the notationc(y) andc(x).) Choosingc(y) = 0and c(x) = x makes both expressions look the same. We obtain the functionf (x, y) = 3x2y + x. It is easy to check that this function really is such that

E∇ f =

(6xy + 1

3x2

).

Not all vector fields can be represented as the gradient of a scalar function.


Example 26.1.19If possible, find a function f such thatE∇ f =

(−yx

).

The same process of “partial integration” as in Example 26.1.18 leads to

∫−y ∂x = −xy + c(y)∫

x ∂y = xy + c(x).

This time, however, becausec(y) cannot depend onx andc(x) cannot dependon x, there is no choice for these expressions that will lead to the same function for

both integrals. Therefore we are unable to find a function such thatE∇ f =

(−yx

).

This outcome is a little unsatisfying. We were unable to find a function such

that E∇ f =

(−yx

). But why is that? Is our process not sophisticated enough or

does such a function not exist at all?

We can argue as follows. If there was such a function, then the integrals wederived would have to describe the function. That is we must have−xy + c(y) =

xy + c(x). The only expressions that depend onx andy are−xy andxy. So, if afunction as stated did indeed exist, we would have that−xy = xy for all x andy,which is nonsense. Thus, such a function cannot exist.

The argument at the end of Example 26.1.19 shows that there are vector fieldsthat are not gradient fields. Any time the results of our partial integration cannotbe made to match up by choosing our constants appropriately, the field is not agradient field. Graphical insight can also be used to rule out fields from beinggradient fields.

Example 26.1.20Explain graphically why there is no function f such that the

gradient satisfiesE∇ f =

(−yx

).

Figure 26.19: The vector field in Examples26.1.19, 26.1.20 and 26.1.22. Consider Figure 26.19, which depictsEF(x, y) =

(−yx

). The field appears to

describe a vortex (an eddy) rotating around the origin. Indeed, its flow lines areclosed circles around the origin. Thus we could follow the field along one of thesecircles and eventually come back to our starting point. This is impossible in agradient field. Following the gradient means constantly going up. If we constantlygo up, we cannot be back where we started at the end of the journey.

Thus the field EF(x, y) =

(−yx

)is not a gradient field, because of the eddy

around the origin.

We will see more along the lines of the graphical explanation in Section 26.3.However, as we have seen, pictures of vector fields may not be scaled right and canthus lead to the wrong impression. If a field is “almost” a gradient field, then itspicture may not differ significantly from that of a gradient field. Thankfully thereis an analytical condition for when a vector field is not a gradient field.


Proposition 26.1.21Let EF =

F1...

Fn

be a vector field with continuous second

partial derivatives. If EF is a gradient field, then for all indices i6= j we have

∂Fi

∂x j=

∂F j

∂xi.

Proof. By Theorem 25.1.10 we know that as long as all second partial deriva-tives are continuous, it does not matter in which order the mixed partial derivativesare taken. Therefore, ifEF is a gradient fieldEF = E∇ f , then for all indicesi 6= j wehave

∂

∂x jFi =

∂

∂x j

∂

∂xif =

∂

∂xi

∂

∂x jf =

∂

∂xiF j

as was to be proved.

Theorem 28.4.2 and Corollary 28.4.5 will show that in two and three dimen-sions the above criterion is an equivalent characterization of gradient fields. This isalso true in higher dimensions, but we will not be able to prove it in this text.

The analytic criterion in Proposition 26.1.21 is easily applied.

Example 26.1.22Prove that there is no function f such thatE∇ f =

(−yx

).

According to Proposition 26.1.21 we need to compare the partial derivativeof the first component with respect toy with the partial derivative of the secondcomponent with respect tox. We see that

∂

∂y(−y) = −1 6= 1 =

∂

∂xx,

which proves thatE∇ f =

(−yx

)cannot be a gradient field.

For vector fields in more than two dimensions all arguments are similar. If thefield is a gradient field, we simply have to compute more integrals. If not, moreequations may need to be checked until we find a pair of corresponding partialderivatives that are not equal. Some three dimensional vector fields will be analyzedin the exercises.

We conclude with a short caution on the language regarding gradient fields. Inphysics and in other engineering and science applications, one is often not inter-ested in the gradient, but in the direction of steepest descent. The reason is thatobjects tend to move in the direction of steepest descent. (Water flows downhill,not uphill). The difference is a negative sign.

Definition 26.1.23 The vector field−E∇ f is also called thepotential field ofthe potential function f . At every point the potential field points in the di-rection of steepest descent for the potential function f . Physically, potentialfields point in the direction in which an object that is drawn from higher tolower potential would travel. (Such objects include masses in a gravitationalfield or positive charges in an electric field.)


People often use “potential field” and “gradient field” interchangeably. As thisseems to be unavoidable, one should be aware of the underlying concept (a gradientof a scalar function is involved) and then let the context determine if people meantto throw in the added negative sign. In this text language will be consistent withDefinitions 26.1.15 and 26.1.23.

26.1.5 Defining the Gravitational and Electrical Fields

In Examples 26.1.4 and 26.1.6 we have described the forces that act betweenmasses and between charges. Formally our discussion then described the forcesthat a certain test mass or charge would experience if it were placed in the field ofthe mass or charge under investigation. Physically, this makes sense because fieldsare only measurable through their effects on their surroundings.

The fields themselves are usually defined as force per test unit. That is, the fieldstrength will be proportional to the force that acts on a given test mass or charge.

Definition 26.1.24 Defining gravitational and electrical fields.

1. At every point in space, the gravitational field is the quotient of the grav-itational force EFg acting on the test mass m and the mass m itself. That

is, E0 :=EFg

m.

2. At every point in space, the electrical field is the quotient of the electro-static force EFe acting on the test charge q and the charge q itself. That

is, EE :=EFe

q.

Example 26.1.25By Example 25.3.8, the electric field of a point charge at the

origin is the potential field (negative gradient) off (Er ) =Q

4πε0 ‖Er ‖.

Similarly, the gravitational field of a massM at the origin is the potential field

of f (Er ) =GM

‖Er ‖.

It turns out that even in general both the gravitational field as well as the elec-trostatic field (no moving charges that could create changing magnetic fields) aregradient fields (cf. Theorem 28.4.3). Proving that this is the case will take a littlewhile however because we will need to ramp up from central force fields to generalfields.

��

��

--

��

Figure 26.20: Solids can sustain shear stress.

In Example 26.1.8 we did not talk about the force that the magnetic field exertson a test object. This is because the magnetic field is not a force field like the grav-itational and electric fields. The magnetic fieldEB is defined through its effect onchargesq moving at velocityEv, which is the forceEFm = qEv × EB. Any attempt todefine the magnetic field through its effect on, say, permanent magnets is consid-ered too complicated. This is because the magnetic north and south poles alwaysoccur together. They cannot be separated like positive and negative charges.

26.1.6 Fluid Flow

��

��

--

��

��

��

Figure 26.21: Fluids cannot sustain shear stress.

Fluids are defined as substances that cannot sustain shear stress (cf. Figures 26.20and 26.21). That means, the term “fluid” includes liquidsand gases. The vector


field usually investigated in fluid flows is the fieldEv(x, y, z) that gives the flowvelocity at position(x, y, z). While there are certain standard fluid flow shapes,fluid flows are not constructed as a superposition of standard flows. It is ironicthat the most intuitive interpretation of a vector field as a velocity field ultimatelyrequires the most complicated analysis to actually describe the situation. Thuswe will use the idea of a fluid flow to interpret some vector fields, but leave thetheoretical analysis of fluid flows to Section 28.5.

In the following we interpret a few standard vector fields as flows.

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Figure 26.22: In a homogeneous (flow) vectorfield all particles move at the same speed in thesame direction.

Example 26.1.26A homogeneousvector field is a vector field that has the samevalue at every point (cf. Figure 26.22). Because the interpretation of a vector fieldis usually dynamic and involves motion, it is not considered appropriate to callthese vector fields “constant”.

The flow described by a homogeneous vector field is a steady stream in whichall particles flow in the same direction at the same speed. Small sections of thewater flow in a river or in a pipe can be considered homogeneous, though largersections will not be homogeneous because of friction (cf. Example 26.1.27).

It should also be said that the electrical field in a plate capacitor is consideredto be homogeneous. Moreover, the gravitational field on earth can be consideredhomogeneous as long as we do not analyze too large a region of space.

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Figure 26.23: The water flow in a pipe or a riveris fastest in the middle and tends to zero on thesides. Ideally all particles still flow in the samedirection.

Example 26.1.27Figure 26.23 gives an idealized depiction of the fluid flow ina river or a pipe. Fluid particles tend to adhere to walls and internal cohesion(viscosity) of the fluid prevents the flow velocity from changing drastically. As aconsequence, in these flows, the fluid near the wall moves slowly and the fastestflow is in the middle. All particles can still be considered as flowing in the samedirection.

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

C

D

6 6 6 6 6 6 6 6 6

6 6 6 6 6 6 6 6 6

Figure 26.24: A vector field that compresses (cf.circle C) and decompresses (cf. circleD) andwhich thus has non-zero divergence.

Example 26.1.28In the vector field in Figure 26.24 all particles flow in the samedirection, but they do so at different speeds. Just as in a traffic flow, this leads tocompression (higher density) in regions where faster particles are behind slowerparticles (see circleC). Conversely, (see circleD) in regions in which slowerparticles are behind faster particles we experience a decompression (a decrease indensity).

Some of the vector fields we have considered so far can also be interpreted asfluid flows. This re-interpretation of electric, magnetic and other fields as flows thatactually move matter has been very helpful in the analysis of these fields.

254 Module 26: Vector Fields and Line Integrals

1. The vector field in Figure 26.12 describes a vortex that spins fastestnear the center. This type of field is a first step towards describingwater draining out of a bathtub as well as more destructive phenom-ena like tornadoes and hurricanes (cf. Figure 26.13).

2. The vector field in Figure 26.19 describes a vortex in which the linearvelocities increase as we move away from the center. This type ofbehavior is associated with rotating rigid bodies such as compactdiscs or DVDs. To keep up with the motion, the particles that arefarther away from the center of rotation must move faster than thosethat are close to the center.

3. The vector field in Figure 26.10 can be interpreted as a field in whichthere is a source of fluid, say, a spigot at the position ofq1 and a sink(like the drain in a bathtub) at the position ofq2. The flow here isstraight outward from the source and inward into the sink, whichmeans as a fluid flow it could not be very deep (otherwise we wouldobserve the characteristic vortex of deep water going into a drain).

The reader will investigate more examples in Exercise 6.

Exercises

1. Match the vector field plots in Figure 26.27 with their equationsgiven below. (The matching will also be useful in Exercise 2of Section 26.3, Exercise 2 in Section 28.2 and Exercise 2 inSection 28.3.)

(a) EF(x, y) =

(11

)

(b) EF(x, y) =

( x√x2+y2

y√

x2+y2

)

(c) EF(x, y) =

( y√

x2+y2

−x√x2+y2

)

(d) EF(x, y) =

(10

)(e) EF(x, y) =

( yx2+y2

−xx2+y2

)

(f) EF(x, y) =

(y

−x

)(g) EF(x, y) =

(−x0

)(h) EF(x, y) =

( −yx2+y2 +

−y(x−1)2+y2

xx2+y2 +

x−1(x−1)2+y2

)

(i) EF(x, y) =

x(√

x2+y2)3

y(√x2+y2

)3

2. Among the vector field plots in Figure 26.27 identify the plot

that is most closely described by each of the following.

(a) The vector field that for each point on a rotating CD givesthe velocity of the particle at that point.

(b) The vector field that for each point in an eddy in a fluidgives the velocity of a small volume of fluid at that point.

(c) The magnetic field generated by two vertical wires withparallel upwards currents.

(d) The electrical field of a positive charge at the origin.

(e) A homogeneous vector field, that is, a vector field whichdoes not depend on the position of(x, y).

3. Sketch the flow lines of each of the following vector fields.

(a) The vector field in Figure 26.27, part 1)

(b) The vector field in Figure 26.27, part 2)

(c) The vector field in Figure 26.27, part 3)

(d) The vector field in Figure 26.27, part 4)

(e) The vector field in Figure 26.27, part 5)

(f) The vector field in Figure 26.27, part 6)

(g) The vector field in Figure 26.27, part 7)

(h) The vector field in Figure 26.27, part 8)

(i) The vector field in Figure 26.27, part 9)

(j) The vector field in Figure 26.14

4. Consider the vector field in Figure 26.25.

(a) Determine where the field is strongest.

(b) Determine where the field is weakest.

(c) Now interpret the field as a flow field. As we follow theflow along a flow line, is the flow compressing or decom-pressing?

Section 26.1: Exercises 255

666O �M �K �

Figure 26.25: The vector field for Exercise 4.

5. Interpret the vector field in Figure 26.26 as a fluid flow and mark

(a) The region(s) where the flow is compressing,

(b) The region(s) where the flow is decompressing,

(c) The region(s) where the flow is strongest,

(d) The region(s) where the flow is weakest.


6. Interpret each vector field as a fluid flow and determine

(i) Where the field compresses or decompresses,

(ii) Where there are sinks or sources in the field.

(a) The vector field in part 2) of Figure 26.27

(b) The vector field in part 3) of Figure 26.27

(c) The vector field in part 5) of Figure 26.27

(d) The vector field in part 8) of Figure 26.27

7. Sketch the given vector field.

(a) EF(x, y) =

(xy

)(b) EF(x, y) =

(11

)(c) EF(x, y) =

(yx

)

(d) EF(x, y) =

(x

−y

)(e) EF(x, y) =

1√x2 + y2

(xy

)(f) The vector field whose flow lines are given in Figure

26.15

8. Write a symbolic expression for the vector field that is verballydescribed.

(a) The field points away from the origin and its magnitudeis equal to the square root of the distance from the origin.

(b) The field is circular around the origin in the mathemati-cally negative direction and its magnitude is constant.

9. Consider the vector field of a point charge at the origin (cf. Ex-ample 26.1.6).

(a) State the first (x−) component of the vector field forpoints in thexy-plane. That is, you will have a functionof x andy.

(b) State the first component of the field for all points inspace, that is, as a function ofx, y, andz.

(c) Will the expressions for the second and third componentslook significantly different?

10. Find the equation of the vector field in Example 26.1.10.

Hint. The equation of the magnetic field surrounding a cur-rent is given in Example 26.1.9. The field forq2 in Example26.1.7 shows how a field that is centered at the origin can bere-centered using shifts similar to the shifts in thex-directionfor functions of one variable (cf. SectionFUN.4).

11. Compute the gradient vector field of the given function.

(a) f (x, y) =1

x2 + y2

(b) f (x, y) = xy

(c) f (x, y, z) =1√

x2 + y2 + z2

12. For each of the following vector fieldsEF

• Find a scalar functionf such thatE∇ f = EF , or

• Show thatEF is not a gradient field.

(a) EF(x, y) =

(2x2y

)(b) EF(x, y) =

(2xy + y2

x2+ 2xy

)(c) EF(x, y) =

(cos(x)

sin(y)

)

(d) EF(x, y, z) =

(2x + yzxzxy

)

(e) EF(x, y, z) =

(2x + y2

x2+ 2yz2

)(f) The magnetic field that surrounds a current (graphical ex-

planation suffices)


13. Compute the given composition. In each case, the vector fielddepends onx andy and the functionsEc andEs providex andyin their components.

(a) EF(x, y) =

(x2

y2+ x

), Ec(t) =

(sin2(t)cos(t)

),

computeEF(Ec(t))

(b) EF(x, y) =

(x2

+ y2

−x

), Es(u, v) =

(u + v

u − v

)computeEF(Es(u, v))

14. A 1kg mass on the surface of the earth experiences an acceler-ation of 9.81m

s2 because of the earth’s gravity. With the earth’scircumference estimated at 40, 000km, determine the size of apoint mass at the center of the earth that would cause the samegravitational force.

We will revisit this idea in Exercise 18 in Section 28.2.

15. Thegeostationary orbit is the height at which a satellite takesexactly one day to circle the earth. A satellite at this heightabove the equator, moving in the mathematically positive direc-tion when viewed from above the North Pole, will not changeits relative position with respect to any location on earth. Manycommunications satellites are positioned in the geostationaryorbit. In this exercise we shall compute its height.

(a) Corollary 24.2.20 gives the force that must act on a massm that moves at velocityv to maintain a circular orbitof radiusr . Explain why for any circular orbit of ra-

dius r this force must be equal toF =GmM

r 2, where

M ≈ 5.9763× 1024kg is the mass of the earth and

G ≈ 6.6720· 10−11Nm2

kg2is the gravitational constant.

(b) Express the velocity that a satellite on the geostationaryorbit must have in terms of the radius of the orbit.

(c) Substitute the expression from Exercise 15b into the equa-tion from Exercise 15a and solve forr . This is the radiusof the geostationary orbit. (Mind the units.)

(d) The height of the geostationary orbit is usually reportedas 35, 786km above mean sea level. If this number isdifferent from the value that you have computed, pleaseexplain the difference.

(e) At what speedv must a satellite fly to maintain its posi-tion on the geostationary orbit?

16. The direction field of a functionf (x).

(a) Explain why for a functionf (x) of one variable the vec-

tor field D(x, y) =1√

1 + ( f (x))2

(1

f (x)

)is the direc-

tion field as defined in SectionIND.A.

(b) Identify the vector fieldD(x, y) =1√

x2 + 4x + 5

(1

x + 2

)in FigureIND.I.

Hint. This is not half as hard as it looks. Consider Exer-cise 16a.


1) 2) 3)

4) 5) 6)

7) 8) 9)

Figure 26.27: Vector field plots for Exercises 1 and 2.


26.2 Line Integrals in Vector Fields

Vector fields manifest themselves through their interactions with objects. Theo-rists use this interaction to detect the field and determine more of its properties.Practitioners put the vector field to use (for example, in hydroelectric power plantsgravitation pulls the water across the turbine that generates electricity) or determinehow much of an adversary it is in an application (for example, gravity is an obstacleto space travel).

To start our analysis, consider a force field, such as a gravitational or an elec-trical field. Energy can be gained by moving an object in the direction of the field.Conversely, if we are moving against the direction of the field, energy is needed tomove the object. To compute this energy we start with first principles. Work is thedot product of force and distance traveledW = EF · Ed (cf. Section 20.3). Becausewe are moving through a force field, the force depends on the present position. Thismeans we cannot simply compute one scalar product.

u�

�

*-

r

K

6

yEc(a)

Ec(ti )

Ec(ti−1)

1 EW = EF(Ec(t∗i−1

))· 1Ec

≈ EF(Ec(t∗i−1

))· Ec′

(t∗i −1

)1t

1Ec

-EF(Ec(ti−1

))

Figure 26.28: The work needed to move througha field along a curve is computed in small incre-ments1W to account for variations in the fieldand the bending of the curve.

SectionsDEF.B and API.F show how to handle a variable force that is parallel tothe direction of motion. The same type of analysis works in higher dimensions.The natural law is stated under the assumption that all quantities involved are con-stant. By investigating small increments we derive laws for more dynamic situ-ations. Let our path beEc(t) with a ≤ t ≤ b. Chooset0 = a, t1, . . . , tn = bsuch that the step1Ec := Ec(ti ) − Ec(ti −1) is so short that the force is approximatelyequal to EF(ti −1) along this stretch ofEc. (This is possible if the force fieldEF iscontinuous.) Then the work performed by going fromEc(ti −1) to Ec(ti ) is approx-imately 1Wi := EF (Ec(ti −1)) · (Ec(ti ) − Ec(ti −1)) . The overall amount of work then

is W ≈

n∑i =1

1Wi =

n∑i =1

EF (Ec(ti −1)) · (Ec(ti ) − Ec(ti −1)) . The above looks like a Rie-

mann sum with scalar products instead of multiplication of numbers and a vectorquantity1Ec := Ec(ti )−Ec(ti −1) instead of the scalar1x = xi −xi −1. Thus we shouldbe able to express the work as an integral. We will do so by reducing the wholeexpression to an integral of a scalar function.

For sufficiently small1Ec = Ec(ti ) − Ec(ti −1) we have thatEc′(t) is approximatelyequal toEc′(ti −1) from ti −1 to ti . Hence1Ec = Ec(ti ) − Ec(ti −1) ≈ Ec′(ti −1)(ti − ti −1)

and the work becomesW ≈

n∑i =1

1Wi =

n∑i =1

EF (Ec(ti −1)) · Ec′(ti −1) (ti − ti −1). This

expression is a Riemann sum of the real-valued functionEF(Ec(t)) · Ec′(t) over the in-terval [a, b]. As we letn → ∞, this sum will converge (under mild technical condi-

tions) and we obtain the work performed as the integralW =

∫ b

a

EF (Ec(t)) · Ec′(t) dt.

Because of the above derivation of work performed by a force field, we definethe line integral of a vector field as follows.\Integration is a universal tool

that allows the analysis ofphenomena in which quantitiesvary using tools designed forsituations in which thesequantities are constant."

Definition 26.2.1 If the vector field EF is continuous on the curve C withparametrizationEc(t) and the parametrizationEc is differentiable, then we definethe line integral of the vector field EF along the curveC as∫

C

EF · dEc :=∫ b

a

EF (Ec(t)) · Ec′(t) dt

Formally we have defined the line integral using one parametrization of thecurveC. It would be strange indeed if the work needed to move along a given

26.2. Line Integrals in Vector Fields 259

curveC depended on the parametrization of the curve. Naturally, the line inte-gral is independent of the parametrization as Proposition 26.2.2 shows. Exercise 6illustrates Proposition 26.2.2 with an example.

Proposition 26.2.2 Let EF be a continuous vector field, let C be parametrizedby the differentiable functionsEc(t), a ≤ t ≤ b andEs(u), d ≤ u ≤ e. Then∫ b

a

EF (Ec(t)) · Ec′(t) dt =

∫ e

d

EF (Es(u)) · Es′(u) dt

Proof (sketch). Since Ec and Es both parametrize the same curve, there is adifferentiable function that maps [a, b] to [d, e] such thatEc(t) = Es(g(t)), g(a) = dandg(b) = e. (This is the hard part of the proof, which we skip. Formally we haveto prove thatg exists.) Now∫ b

a

EF (Ec(t)) · Ec′(t) dt =

∫ b

a

EF (Es(g(t))) · (Es ◦ g)′(t) dt

Differentiate the compositionEs ◦ g withthe chain rule.

=

∫ b

a

EF (Es(g(t))) · Es′(g(t))g′(t) dt

By the substitution rule (cf. TheoremFTC.D) this is exactly the integral we werelooking for.

=

∫ e

d

EF (Es(u)) · Es′(u) dt

Line integrals for which the parametrization of the curve is known are easy toset up.

Example 26.2.3Compute the line integral of the vector fieldEF(x, y) =

(−yx

)along the pathEc(t) =

(cos(t)sin(t)

)with 0 ≤ t ≤ π .

Substituting the path into the vector field we obtain

EF (Ec(t)) = EF(x, y)

∣∣∣x=cos(t),y=sin(t)

=

(−yx

)∣∣∣∣x=cos(t),y=sin(t)

=

(− sin(t)cos(t)

).

Figure 26.29: The vector field and the path inExample 26.2.5.

This leads to the integral

∫C

EF · dEc =

∫ π

0

(− sin(t)cos(t)

)·

(− sin(t)cos(t)

)dt =

∫ π

0sin2(t) + cos2(t) dt

=

∫ π

01 dt = π

Figure 26.29 confirms the sign of our result. The direction of motion is largelyparallel to the direction of the vector field, which means a particle in a force fieldwould gain energy along this trajectory.


As long as we interpret our vector field as a force field, the sign of the lineintegral tells if the particle moving along the given trajectory gains energy from thefield as in Example 26.2.3 or if work has to be done to move the particle against thedirection of the field as below.

Example 26.2.4Compute the line integral of the vector fieldEF(x, y) =

(−yx

)along the pathEk(t) =

(cos(t)

− sin(t)

)with 0 ≤ t ≤ π .Side calculation.

EF(Ek(t))= EF(x,y)lx=cos(t),y=-sin(t)=

( -yx)∣∣∣∣x=cos(t),y=-sin(t)

=( sin(t)cos(t)

)

After substituting the parametrization of the curve into the vector field, theintegral is∫

C

EF · dEk =

∫ π

0

(sin(t)cos(t)

)·

(− sin(t)− cos(t)

)dt =

∫ π

0− sin2(t) − cos2(t) dt

=

∫ π

0−1 dt = −π.

Figure 26.30 confirms the sign of the integral. The direction of motion is “against”the direction of the vector field. Thus a particle in a force field would lose energyalong this trajectory.


Examples 26.2.3 and 26.2.4 show that while the line integral does not dependon the parametrization of the curve, it does depend on which curve we take fromone point to another. The vector field and the paths are given in Figures 26.29 and26.30. We see that traversing the lower half of the circle requires work against thefield, while on the upper half of the circle we travel with the field. There are vectorfields for which the line integral only depends on the endpoints of the curve. Suchvector fields are called conservative vector fields. They are important in physics.We will investigate them extensively in Section 26.3.3.

Line integrals are a bit more work when we have to find the parametrization ofthe path.

Example 26.2.5 Integrate the vector fieldEF(x, y) =

(−yx

)over the circle of ra-

dius2, centered at(1, 0) in the positive direction. (We will revisit this integral inExample 28.6.2.)

The standard parametrization for a circle of radius 2 traversed in the positive

direction and centered at the origin isEk(t) =

(2 cos(t)2 sin(t)

), 0 ≤ t ≤ 2π . To center

this circle at the point(1, 0) we add the position vector of the center to obtain the

parametrizationEc(t) =

(10

)+

(2 cos(t)2 sin(t)

)=

(1 + 2 cos(t)

2 sin(t)

), 0 ≤ t ≤ 2π .


With x = 1 + 2 cos(t) and y = 2 sin(t) plugged into EF(x, y) =

(−yx

)we

obtain EF (Ec(t)) =

(−2 sin(t)

1 + 2 cos(t)

)and we can set up and compute the line integral.

∫C

(−yx

)dEc =

∫ 2π

0

(−2 sin(t)

1 + 2 cos(t)

)(−2 sin(t)2 cos(t)

)=

∫ 2π

04 sin2(t) + 2 cos(t) + 4 cos2(t)dt

=

∫ 2π

04 + 2 cos(t)dt = 8π


Figure 26.31 shows the vector field and the circle over which we integrate. Notethat Example 26.2.5 shows that there is a spin in the vector field. This is becausewe can go around a closed circle and the net line integral is positive. This meanswe have gained energy by completing the loop. Section 26.3.3 shows that this isnot possible in a gradient field. Section 28.3 shows how this phenomenon is relatedto the “curl” of a vector field.

Line integrals in three dimensional coordinates can be set up equally easily.

Example 26.2.6Compute the line integral ofEF(x, y, z) =

−x(√

x2+y2+z2)3

−y(√x2+y2+z2

)3

−z(√x2+y2+z2

)3

along

the pathEc(t) =

ttt

with 1 ≤ t ≤ 2.

∫C

EF · dEc =

∫ 2

1

−t(√

t2+t2+t2)3

−t(√t2+t2+t2

)3

−t(√t2+t2+t2

)3

·

111

dt =

∫ 2

13

−t(√3t2)3 dt

= −

∫ 2

13−

12

1

t2dt = 3−

121

t

∣∣∣∣21

= −1

2√

3≈ −0.2887

Finally, when all else fails (in two, three or any other dimensions) we can al-ways set up the integral and compute it with a computer algebra system.

Example 26.2.7Compute the line integral ofEF(x, y) =

−x(√

x2+y2+z2)3

−y(√x2+y2+z2

)3

−z(√x2+y2+z2

)3

along

the pathEc(t) =

2 − sin(

π2 t)

2t−1

1 + cos2(

3π2 t) with 1 ≤ t ≤ 2. (We will revisit this line

integral in Example 26.3.3.)

Figure 26.32: The setup to compute the integralin Example 26.2.7.

If we let the components ofEF be EF(x, y, z) =

P(x, y, z)Q(x, y, z)R(x, y, z)

, then we can

set it up the line integral directly in a computer algebra system as shown in Figure26.32. For complicated line integrals like this one, this is highly appropriate. Theintegral is

Note that we have the same start and end point as in Example 26.2.6 and thatunlike for Examples 26.2.3 and 26.2.4 we appear get the same result for the twointegrals this time. In Example 26.3.3 we will see that the integrals in Examples26.2.6 and 26.2.7 are indeed equal.


Line integrals over paths with corners can be pieced together from severaldifferentiable segments. Because this is rather common, the notation “piecewisesmooth” is usually associated with the curves along which we can compute lineintegrals.

Definition 26.2.8 A curve C is calledpiecewise smoothif and only if there isa parametrizationEc(t), a ≤ t ≤ b of C and numbers a= t0 < t1 < · · · <

tn such thatEc is differentiable with a continuous derivative on each interval[ti −1, ti ]. The line integral of a continuous vector fieldEF over a piecewisesmooth curve is defined to be∫

C

EF · dEc :=∫ b

a

EF (Ec(t)) · Ec′(t) dt

where the fact that the derivative may not exist at ti does not matter. (Theintegral is usually pieced together.)

Just as before, the integral does not depend on the specific parametrization weuse.

Example 26.2.9Compute the line integral of the vector fieldEF(x) =

(ex

xy2

)along

the boundary of the triangle with vertices(0, 0), (1, 0) and(1, 1). Move along theboundary in the mathematically positive direction.

y

xvv

v(1, 1)

(1, 0)(0, 0) -

6

Ech(t) =

(t0

)

Ecv(t) =

(1t

)Ecd(t) =

(1 − t1 − t

)

-

6

Figure 26.33: The triangle in Example 26.2.9with the boundary traversed in the positive di-rection.

The triangle is pictured in Figure 26.33. The boundary naturally decomposesinto a horizontal partCh, a vertical partCv and a diagonal partCd. The parametriza-

tion of the horizontal partCd is Ech =

(t0

), 0 ≤ t ≤ 1. The corresponding integral

isCompute EF(Ech(t)) and Ech'(t)here. ∫

Ch

EF · dEch =

∫ 1

0

(et

0

)·

(10

)dt =

∫ 1

0et dt = et

∣∣10 = e− 1.

For the vertical partCv, the parametrization isEcv =

(1t

), 0 ≤ t ≤ 1. Note

that it does not matter that we use the same parameter interval. This is a separateintegral. The line integral overCv is∫

Cv

EF · dEcv =

∫ 1

0

(e1

t2

)·

(01

)dt =

∫ 1

0t2 dt =

1

3Explain why we could have alsoparametrized Cd withEcd=

( tt), where t goes

\backwards" from 1 to 0.

Finally, the diagonal partCd is parametrized byEcd =

(1 − t1 − t

), 0 ≤ t ≤ 1. For

the line integral we obtain∫Cd

EF · dEcd =

∫ 1

0

(e1−t

(1 − t)3

)·

(−1−1

)dt =

∫ 1

0−e1−t

− (1 − t)3 dt

= e1−t+

1

4(1 − t)4

∣∣∣∣10

= e0+

1

4· 0 −

[e1

+1

4· 1

]=

3

4− e

The overall line integral over the boundary curveB of the triangle is∫B

EF · dEb =

∫Ch

EF · dEch +

∫Cv

EF · dEcv +

∫Cd

EF · dEcd

= (e− 1) +

(1

3

)+

(3

4− e

)=

1

12


Another common notation for line integrals of vector fields is exhibited in Ex-ercise 7

26.2.1 The Scalar Line Integral

The motivation for the line integral of a vector field is the desire to measure theeffects of a field. Because of the fundamental nature of the concept of work, theline integral of a vector field is usually what people mean when they talk aboutline integrals. Yet there is another, scalar, line integral, which arises in certainapplications.

Looking back at the introduction to this section, we see that the curveEc(t) waspartitioned into small curve elements1Ec ≈ Ec′(t)1t . The length of each of thesecurve elements is‖1Ec‖ ≈

∥∥Ec′(t)∥∥1t . If we had a wire that is shaped like the

trajectory ofEc(t) and which has linear densityρ, then we could approximate themass of the wire by the sum

∑ρ(Ec(t∗

k

))‖1Ec‖ ≈

∑ρ(Ec(t∗

k

)) ∥∥Ec′(t∗

k

)∥∥1t. Thissum becomes an integral if we let the1t go to zero. This integral is the scalar lineintegral.

Definition 26.2.10 If the function f is continuous on the curve C withparametrizationEc(t) and the parametrizationEc is differentiable, then we definethescalar line integral of the function f alongC as∫

Cf dc :=

∫ b

af (Ec(t))

∥∥Ec′(t)∥∥ dt

One application of the scalar line integral is the computation of masses andcenters of mass of linearly distributed densities like wires. This application is in-vestigated in detail in Exercise 10.

Scalar line integrals are also connected to vector fields through the law of Biotand Savart. The law of Biot and Savart states how to compute the magnetic fieldgenerated by a currentI through a differential wire segmentdEc. Currents throughisolated differential wire segments are purely conceptual entities. Where would thecurrent come from and where would it go if the wire was not a loop? Yet, just likemass points, they have been useful and the law of Biot and Savart has been testedtime and again and it has always given correct results.

6I

dEc

ZZZZZZZZZZZZ~

Er

uP

6

θ

Figure 26.34: The law of Biot and Savart allowsthe computation of the magnetic field of a cur-rent.

Theorem 26.2.11(The law of Biot and Savart.) Suppose a current I goesthrough a differential segment dEc of a wire (cf. Figure 26.34). Then the mag-netic field generated by this current at a point P is

d EB =µ0I dEc × Er

4πr 3,

whereEr is the vector that points from the location of the wire segment to thepoint P. For the magnitude of the magnetic field we obtain

d B =µ0I sinθ

4πr 2dc,

where r is the distance between the differential wire segment and the point andθ is the angle between a vector parallel to the current and the vectorEr .


Because the direction of a magnetic field is usually determined using symmetryand the right hand rule forEI ×Er , Biot and Savart’s law is normally used to computethe magnitude of the field at the point in question.

Example 26.2.12Consider a current I in a straight (infinitely long) wire W thatgoes along the z-axis. Prove that at a distance R to the wire, the magnitude of the

magnetic field that surrounds the current is B=I

2πε0c2R=

I µ0

2π R. (This proves

the claim of Example 26.1.8 by using the law of Biot and Savart.)As a first step, we coordinatize the situation. We align the wire with thez-axis

and we place the point at(R, 0, 0) (cf. Figure 26.35). This is no loss of generality,because the geometry as described is preserved. At the same time, the choice of aconvenient coordinate system simplifies the resulting computation.

It is now easy to parametrize thez-axis asEc(t) =

00z

. We will need to choose

−∞ < z < ∞ as our parameter domain, because we are considering an infinitelylong wire. This is no problem, because we know how to solve improper integrals.

-

z

x

y

6the currentI

goes up thez-axis

w(R, 0, 0)

dEc =

(001

)dz

Er = Er (z)

6

?

‖Er ‖ =

√R2 + z2

θ

Figure 26.35: A differential wire element asneeded in Example 26.2.12.

To set up the integral, consider a differential elementdEc of the wire, located at

(0, 0, z). For the differential length element we obtain the factor∥∥Ec′(z)

∥∥ =

∥∥∥∥∥∥0

01

∥∥∥∥∥∥ = 1.

The distance from a differential wire element at(0, 0, z) to the pointP is

r =

∥∥∥∥∥∥ R

00

−

00z

∥∥∥∥∥∥ =

√R2 + z2

and the sine of the angle betweendEc andEr is sin(θ) =R

r=

R√

R2 + z2. Overall

we obtain the following integral.

∫W

µ0I sinθ

4πr 2ds =

∫∞

−∞

µ0I R√R2+z2

4π(R2 + z2

) dz =µ0I R

4π

∫∞

−∞

1(R2 + z2

) 32

dz

=µ0I R

4π

z

R2√

R2 + z2

∣∣∣∣z→∞

z→−∞

=µ0I R

4π

[1

R2−

(−

1

R2

) ]=

µ0I R

4π

2

R2=

µ0I

2π R

Integral formula 44 from theintegral table in Section A.1on page 376.∫ 1

(a2+x2)32dx

= xa2√a2+x2 The result is exactly the formula claimed in Example 26.1.8

The connection between the scalar line integral and the line integral for vectorfields is formulated below.

Proposition 26.2.13 If EF is continuous on C andEc is smooth, then∫C

EF dEc =

∫C

EF · ET dc.

Proof. Left to the reader as Exercise 12.


26.2.2 The Idea Behind “Differentials”

Our discussion of the law of Biot and Savart involved a lot of talk about“differential” elementsdEc. Scientists and engineers routinely use differen-tial elements instead of small quantities like1Ec for which we need to letthe length go to zero at some point in time.The notational advantage is obvious. By replacing small differences withdifferentials and sums with integral signs we can directly set up integralswithout any detours through limits of Riemann sums. Readers who do notlike the mystery behind differentials are welcome to re-state the law of Biotand Savart in terms of short wire elements1Ec. (This is a good exercise.)The resulting integrals will be the same.Historically, limits were introduced as a rigorous foundation to calculusabout 200 years after the inception of calculus by Leibnitz and Newton.Leibnitz’ formulation always included the idea of differential quantities.These differentials were to be elements of a size “smaller than any positivereal number but not zero”. It is easy to show that such elements cannot bereal numbers themselves. This made many mathematicians uneasy aboutdifferentials. If they are not real numbers, what are they? Yet the scientificmodels built upon these differentials are correct and extremely powerful.After over a hundred years of searching for a formalism that justified usingdifferentials, limits came about as a formal substitute.Differentials were considered a convenient shorthand until the 1960s whenAbraham Robinson constructed a number system that contains the differ-entials that Leibnitz postulated. This founded a new branch of mathematicscalled “nonstandard analysis”. For us this means that differentials are no-tationally convenient as well as mathematically sound. Hence we shouldnot have any qualms about using them in applications.In this text we give references to Riemann sums as necessary and we vi-sualize the involved quantities as differentials. In this fashion the reader isencouraged to be fluent in both ways of formulating the work with smallquantities. We need to be able to work in both formalisms because model-ing usually is done with differentials, but numerical implementation of themodels requires the use of small finite quantities.

Exercises

1. Consider the vector fieldEF in Figure 26.36.

(a) For which of the paths is the line integral ofEF along thepath fromA to B largest?

(b) For which path is the line integral ofEF along the pathfrom A to B smallest?

(c) What is your estimate for the value of the line integral ofEF along path 2? Explain why this is your estimate.

(d) Consider path 1. What is the relationship between theline integral of EF from A to B and the line integral ofEFfrom B to A? Is your answer a general observation ordoes it only apply here?



2. For this exercise consider the vector fieldEF in Figure 26.37.

(a) Find a path fromA to B such that the integral ofEF alongthe path is positive.

(b) Find a path fromA to B such that the integralEF alongthe path is negative.

(c) Can you find a path fromA to B such that the integralEFalong the path is 0?

(d) Given an arbitrary real numberm, can you find a pathfrom A to B such that the integralEF along the path islarger thanm? How would you construct such a path?


3. Evaluate the vector field on the given path or curve, that is, com-pute EF (Ec(t)) or EF (Es(u, v)) as appropriate.

(a) EF(x, y, z) =

( 1z

y + 2yz

)on the surfaceEs(u, v) =

( uv

u2+ v2

)

4. Line integrals of vector fields.

(a) Integrate the vector fieldEF =

(xy

)along the circle

Ec(t) =

(1 + cos(t)1 + sin(t)

), 0 ≤ t ≤ 2π

(b) Integrate the vector fieldEF =

(x2

y2

)along the circle

Ec(t) =

(1 + cos(t)1 + sin(t)

), 0 ≤ t ≤ 2π

(c) Integrate the vector fieldEF(x, y) =

(cos(x)

−xy

)over the

curveEc(t) =

(t2

t3

)0 ≤ t ≤ 1.

(d) Integrate the vector fieldEF =

(−2yxy

)along the follow-

ing paths.

i. From(0, 0) to (4, 2) by going alongx = y2

ii. Along the vertical line segment from(4, 2) to (4, −3)

(e) Integrate the vector fieldEF = Ei + (x + y) Ej along thefollowing path:

i. First, alongy = x from (0, 0) to (1, 1),

ii. Second, alongx2+ y2

= 2 from (1, 1) to (0,√

2)

in the first quadrant.

(f) Integrate the vector fieldEF =

( x2

yxz

)along the line seg-

mentEc(t) =

( 1 + t)3t − 22 − 4t

), 0 ≤ t ≤ 1

(g) Integrate the vector fieldEF(x, y) =

(yx

)along the bound-

ary of the intersection of the diskx2+ y2

≤ 9 with thefirst quadrant.

(h) Integrate the vector fieldEF(x, y) =

(−yx

)around the

circle of radius 1 around the origin. Traverse the circle inthe mathematically positive direction.

(i) Integrate the vector field

( yx2+ z3

z2

2yz

)along the posi-

tively oriented intersection of the cylinderx2+ y2

= 1and the planez = x + 3. (Use of a computer algebrasystem is acceptable. Also cf. Example 28.3.5.)

5. Line integrals of force fields.

(a) The force field EF =

(x + y

2x

)acts on an object as it

moves in the plane. Calculate the work done byEF as theobject moves from(0, 0) to (6, 2) by going first along thecurvex = y2 from (0, 0) to (4, 2) and then along the linesegment from(4, 2) to (6, 2).

(b) The force fieldEF =

(−y2x

)acts on an object as it moves

in the xy-plane. Calculate the work done byEF as theobject moves from(0, 0) to (2, 7) along the followingpath:

i. First, alongy = x2 from (0, 0) to (2, 4).

ii. Second, along the vertical line segment from(2, 4)

to (2, 7).

(c) The force field EF =

(x − yx + y

)acts on an object as it

moves in the plane. Calculate the work done byEF asthe object moves from(1, 0) to (3, 1) along the follow-ing path:

i. First, alongx2+ y2

= 1 in the first quadrant from(1, 0) to (0, 1)

ii. Second, along the line segment from(0, 1) to (3, 1).

(d) The force field EF = y2Ei + xyEj acts on an object as itmoves in the plane. Suppose the object is moved from(0, 0) to (3, 6) by first going alongy = x2 from (0, 0)

to (2, 4) and then alongy = 2x from (2, 4) to (3, 6).Calculate the work done byEF .

(e) The force fieldEF = yEi + (xy + 1) Ej acts on an object asit moves in the plane. Calculate the work done byEF foreach of the following motions.

i. The object moves from(0, 0) to (4, 2) along thecurvex = y2.


ii. The object moves from(0, 0) to (4, 2) by going firstalong they-axis from(0, 0) to (0, 2) and then alongthe line segment from(0, 2) to (4, 2).

6. (Illustrating Proposition 26.2.2.) Consider the vector field

EF(x, y) =

(sin(πx)

y

).

(a) Integrate the vector field along the pathC with parametriza-

tion Ec(t) =

(t2

t

)with 0 ≤ t ≤ 2.

(b) Integrate the vector field along the same pathC, this time

with parametrizationEc(t) =

(4( t

2

)22t

)with 0 ≤ t ≤ 1.

(c) Compare your results.

7. Another notation for line integrals of vector fields. Sincex′(t) dt =

dx, y′(t) dt = dy andz′(t) dt = dz, a line integral can be re-written as∫

CEF dEc =

∫ b

a

( P(x(t), y(t), z(t))Q(x(t), y(t), z(t))R(x(t), y(t), z(t))

)·

( x′(t)y′(t)z′(t)

)dt

=

∫ b

a

( P(x(t), y(t), z(t))Q(x(t), y(t), z(t))R(x(t), y(t), z(t))

)·

(dxdydz

)

=

∫ b

aP(x(t), y(t), z(t)) dx

+Q(x(t), y(t), z(t)) dy

+R(x(t), y(t), z(t)) dz.

Two dimensional and other dimensional line integrals can be re-written similarly. As long as we remember thatdx = x′(t) dt,dy = y′(t) dt anddz = z′(t) dt, there is no problem computingintegrals formulated in this fashion.

(a) Evaluate the line integral∫

Cxy dx+ x dywhereC con-

sists of the line segment from(2, −4) to (2, 0) followedby the arc of the circlex2

+ y2= 4 from (2, 0) to (0, 2)

which is in the first quadrant.

(b) Evaluate the line integral∫C

(1 + xy)dx +

(1

1 + x2

)dy

for each of the following curves from(0, 0) to (2, 4).

(i) C is the parabolay = x2 from (0, 0) to (2, 4).

ii. C consists of the line segment from(0, 0) to (2, 0)

followed by the line segment from(2, 0) to (2, 4).

iii. C consists of the line segment from(0, 0) to (0, 4)

followed by the line segment from(0, 4) to (2, 4).

8. Connections to other sections.

(a) If the content of Section 23.1.5 is familiar, solve Exercise6c in that section.

9. Compute the given scalar line integral.

(a) Compute the line integral off (x, y) = y over the upperhalf of the circlex2

+ y2= 1.

(b) Compute the line integral off (x, y) = x over the trian-gle with vertices(0, ±1) and(1, 0).

10. For thin, linear objects, such as wires, density often is not recordedas mass per volume, but as mass per length instead. Considersuch a wire bent into the shape given by the curveC and assumethe curve is parametrized byEc(t). The overall mass of the wireis, as noted in the introduction of Section 26.2.1, the scalar line

integralm =

∫C

ρ dc. In this exercise we will compute some

masses and centers of mass.

(a) Compute the mass of a wire shaped like the given curvethat has the given density. Use a CAS if you think the in-tegral is too tedious to solve by hand, but always attemptan exact, “by hand” solution first.

i. ρ(x, y) = 2+x+y, Ec(t) =

(cos(t)sin(t)

), 0 ≤ t ≤ 2π

ii. ρ(x, y, z) = 1, Ec(t) =

( cos(t)sin(t)

t

), 0 ≤ t ≤ 10π

(b) (This exercise requires familiarity with Section 23.1.5.)Prove the following.

Let ρ be a density function defined along the curveCparametrized byEc(t). Prove that the center of mass of thislinearly distributed mass has coordinates

xC =

∫C

xρ ds

yC =

∫C

yρ ds

zC =

∫C

zρ ds

(c) Compute the center of mass of each of the given wire.Use a CAS if you think an integral is too tedious to solveby hand, but always attempt an exact, “by hand” solutionfirst.

i. The wire in Exercise 10(a)i

ii. The wire in Exercise 10(a)ii

11. Use the law of Biot and Savart or other appropriate arguments tocompute the magnitude of the given magnetic field at the givenpoint.

(a) The magnetic field at the origin generated by a currentIthat runs through the circlex2

+ y2= 1 in thexy-plane

in the mathematically positive direction.

(b) The magnetic field at the origin generated by a currentI that runs through the circlex2

+ y2= 1 in the plane

z = z0 in the mathematically positive direction.

(c) The magnetic field at the point(R, 0, 0) generated by acurrentI that runs up thez-axis fromz = a to z = b.

(d) The magnetic field at the point(R, 0, 0) generated by acurrent I that runs up thez-axis and another current ofequal strengthI that runs up the axisx = 2R, y = 0.

12. Prove Proposition 26.2.13.


26.3 The Fundamental Theorem for Line Integrals

In Examples 26.2.6 and 26.2.7 we obtained the same result for two line integrals ofa vector field. The curves had the same starting and ending points, but they werenot equal from start to finish. In Examples 26.2.3 and 26.2.4 we also consideredtwo line integrals of a vector field along different curves with the same starting andending points. Yet the integrals in Examples 26.2.3 and 26.2.4 were different. Theinequality was easy to explain. In Example 26.2.3 we went “with the field”, whichallows us to draw energy from the field. In Example 26.2.4 we traveled “againstthe field”, which requires energy. In this section we shall investigate the fields forwhich the line integral only depends on the end points of the curve, not on the curveitself. That is, we will investigate fields for which the behavior of Examples 26.2.6and 26.2.7 always occurs and the behavior of Examples 26.2.3 and 26.2.4 neveroccurs.

Section 26.3.3 shows that gradient fields are exactly the fields for which the lineintegral does not depend on the path taken. This is a nice geometric characterizationof gradient fields, which also has significant conceptual consequences (cf. Section26.3.4).

26.3.1 Deriving the Fundamental Theorem for Line Integrals

The formal tool that gives us access to this characterization of gradient fields is theFundamental Theorem for Line Integrals. This theorem says that in a gradient field,the potential difference between two points gives the amount of energy needed tomove from one point to the other.

Theorem 26.3.1 (The Fundamental Theorem for Line Integrals.) Let C bepiecewise smooth and parametrized byEc(t), a ≤ t ≤ b. If f is differentiableand E∇ f is continuous on C, then∫

C

E∇ f · dEc = f (Ec(b)) − f (Ec(a)).

Proof. We will assume thatEc(t) is differentiable. The result for piecewisesmooth curves is obtained by repeatedly applying the result for the differentiablepieces of the curve (cf. Exercise 7).

The proof for differentiable curvesEc(t) is an application of the multivariablechain rule in conjunction with the Fundamental Theorem of Calculus.∫

C

E∇ f · dEc =

∫ b

a

E∇ f (Ec(t)) · Ec′(t) dt

=

∫ b

a

∂ f∂x1

(Ec(t))∂ f∂x2

(Ec(t))...

∂ f∂xn

(Ec(t))

·

dx1dt (t)

dx2dt (t)

...dxndt (t)

dt

=

∫ b

a

n∑i =1

∂ f

∂xi(Ec(t))

dxi

dt(t)dt

Chain Rule!

=

∫ b

a

d

dtf (Ec(t) dt

26.3. The Fundamental Theorem for Line Integrals 269

Fundamental Theorem of Calculus

= f (Ec(b)) − f (Ec(a))

26.3.2 Using the Fundamental Theorem for Line Integrals

The Fundamental Theorem for Line Integrals significantly shortens computationsof line integrals for gradient fields.

Example 26.3.2Compute the line integral ofEF(x, y) =

(2x + π cos(πx)y

sin(πx) +1

y+2

)along

the curve parametrized byEc(t) =

(t2

cos(π t)

)with 0 ≤ t ≤ 1.

At first glance, this task looks nearly impossible. The resulting integral willindeed be very complicated. However, via∫

2x + π cos(πx)y ∂x = x2+ sin(πx)y + cy∫

sin(πx) +1

y + 2∂y = sin(πx)y + ln(y + 2) + cx

Verify that E∇f=EF.we conclude thatf (x, y) = x2

+ sin(πx)y + ln(y + 2) is such thatE∇ f = EF .

With Ec(0) =

(01

)andEc(1) =

(1

−1

)we obtain

∫C

EF · dEc = f (1, −1) − f (0, 1) = x2+ sin(πx)y + ln(y + 2)

∣∣(1,−1)

(0,1)

= 1 + 0 · (−1) + ln(1) − [0 + 0 · 1 + ln(3)] = 1 − ln(3)

The Fundamental Theorem for Line Integrals works in any dimension. In par-ticular, it works for three dimensional fields as encountered in Examples 26.2.6 and26.2.7.

Example 26.3.3 (Example 26.2.7 revisited.) Compute the line integral of the vec-

tor field EF(x, y) =

−x(√

x2+y2+z2)3

−y(√x2+y2+z2

)3

−z(√x2+y2+z2

)3

along the pathEc(t) =

2 − sin(

π2 t)

2t−1

1 + cos2(

3π2 t)

with 1 ≤ t ≤ 2.

By Example 25.1.6 we have thatEF = E∇1√

x2 + y2 + z2. Therefore, withEc(1) =

111

andEc(2) =

222

we have

∫C

EF · dEc =1√

x2 + y2 + z2

∣∣∣∣∣(2,2,2)

(1,1,1)

=1

√12

−1

√3

= −1

2√

3,


which is the same result as in Example 26.2.6. The Fundamental Theoremfor Line Integrals explains why this is so. Any line integral of the vector fieldEFdepends only on the starting and ending point of the curve. Whether the path issimple as in Example 26.2.6 or complicated as in Example 26.2.7 does not changethe result.

26.3.3 A Geometric Characterization of Gradient Fields

The Fundamental Theorem for Line Integrals has some immediate general con-sequences for line integrals of gradient fields. For example, because closed curveshave the same start and end point, line integrals of gradient fields over closed curvesare zero. Since line integrals over closed curves are also important in other situa-tions (cf. Section 28.3) they are often denoted with a special symbol.

Definition 26.3.4 A curve K parametrized byEc(t) with a ≤ t ≤ b is calledclosedif and only ifEc(a) = Ec(b). Let EF be a vector field. Then the line integral

of EF over the closed curve K is also denoted by∮

K

EF · dEc :=∫

K

EF · dEc. That

is, the integral with the circle is a regular line integral. The circle simply pointsout that the integral is over a closed curve.

Aside from closed curves, the Fundamental Theorem for Line Integrals alsotells us something about integrals over curves with the same start and end points.The line integral of a gradient field only depends on the start and end point of thecurve.

u

u

p

q

� C

C connectsp andq

Figure 26.38: In a pathwise connected regionD,any two pointsp andq can be connected by acurve that stays within the regionD. (Even ifthe curve has to twist a bit.) Definition 26.3.5 Let D be a region in the domain of the vector fieldEF. The

line integral∫

C

EF · dEc is path-independent in D if and only if for any two

curves C1 and C2 in D with the same initial and terminal points we have∫C1

EF · dEc =

∫C2

EF · dEc

As we are talking about curves, it is reasonable to restrict our attention to re-gions in which any two points can be connected by a curve.

u

u

p

q

~

Any connection fromp to q

must leave the region.

Figure 26.39: In a disconnected regionD, thereare p and q that cannot be connected by acurve that stays within the regionD. Discon-nected regions are made up of several uncon-nected pieces.

Definition 26.3.6 A region D is called(pathwise) connected(cf. Figure26.38) if and only if for all pairs p, q of points in D we have that there isa continuous functionEc on[0, 1] such that

1. Ec(0) is the position vector of p, and

2. Ec(1) is the position vector of q, and

3. For all t in [0, 1] the endpoint of the position vectorEc(t) is in D.

Regions that are not connected are also calleddisconnected(cf. Figure26.39).

Theorem 26.3.7 now sums up the informal observations made above. More-over, it shows that our observations are equivalent to the property of being a gradi-


ent field.

Theorem 26.3.7Let EF be a vector field that is defined on the connected regionD. The the following are equivalent.

1. There is a function f on D such thatEF = E∇ f ,

2. The line integral∮

K

EF · dEc is zero for every closed curve K in D,

3. The line integral∫

C

EF · dEc is path-independent in D.

Proof. We shall prove “1⇒2⇒3⇒1”, which will establish the theorem.For “1⇒2” let K be a closed curve parametrized byEc(t) with a ≤ t ≤ b. Then

Ec(b) = Ec(a) and by the Fundamental Theorem for Line Integrals we obtain∮K

EF · dEc =

∫K

E∇ f · dEc = f (Ec(b)) − f (Ec(a)) = 0.u

uterminal point

initial point

>

�Ec1(t)

Ec2(t)=�

Figure 26.40: Converting between two curvesC1, C2 with the same initial and terminal pointsand a closed curveK in the part “2⇒3” of theproof of Theorem 26.3.7. The dashed arrowsgive the traversal of the joined curveK from ini-tial point to terminal point and back to the initialpoint.

For “2⇒3” note (cf. Figure 26.40) that two curvesC1 andC2 with the samestart- and endpoints can be turned into a closed curveK by starting at the commonstart point, traversingC1 in the forward direction to the common endpoint and thentraversingC2 in the reverse direction back to the common start point. WithC1 andC2 being two arbitrary curves inD with the same start and end points, we thusobtain the following. ∮

K

EF · dEc = 0∫C1

EF · dEc −

∫C2

EF · dEc = 0∫C1

EF · dEc =

∫C2

EF · dEc

SinceC1 andC2 were two arbitrary curves inD with the same start and end

points we have shown that the line integral∫

C

EF · dEc is path-independent inD.

For “3⇒1” we must define a potential functionf for the vector fieldEF . Be-

cause the line integral∫

C

EF · dEc is path-independent inD, we can identify the value

of each line integral ofEF in D from the endpoints of the path. Thus in the rest of

this proof for any pointsp, q in D we set∫ q

p

EF · dEc :=∫

C

EF · dEc for any curveC

which starts at the pointp, ends at the pointq and stays inD.To abbreviate notation, we shall prove the result for the case thatD is a two-

dimensional region andEF(x, y) =

(P(x, y)

Q(x, y)

)is a two-dimensional vector field.

The proof is the same in higher dimensions, but notation becomes a little morecumbersome.

Choosep to be an arbitrary point inD and define f (x, y) :=∫ (x,y)

p

EF · dEc

for (x, y) in D. Since the line integral is path-independent, this defines a func-

tion on D. To show thatE∇ f = EF we need to show∂ f

∂x(x, y) = P(x, y) and


∂ f

∂y(x, y) = Q(x, y). Since the proofs for both equations are similar, we will only

prove the first equation. As is so often the case, the proof goes back to the definitionof partial derivatives via difference quotients.

u

u u

p

(x, y) (x + h, y)

�

O

-

�M

Figure 26.41: The paths used in the proof of“3⇒1” in Theorem 26.3.7. Instead of goingdown the left path and up the right one (separatearrows), we can also go straight across.

∂ f

∂x(x, y) = lim

h→0

f (x + h, y) − f (x, y)

h

= limh→0

1

h

(∫ (x+h,y)

p

EF · dEc −

∫ (x,y)

p

EF · dEc

)As Figure 26.41 shows, the differenceis the integral over the joined path from(x, y) to (x + h, y) that goes throughp.

= limh→0

1

h

(∫(x,y)→p→(x+h,y)

EF · dEc

)As Figure 26.41 shows, by assumption3, the integral over the path throughp isthe same as the integral over the straighthorizontal line segment between the end-points.

= limh→0

1

h

∫ (x+h,y)

(x,y)

EF · dEc

The easiest way to parametrize apath from (x, y) to (x + h, y) is

Ec(t) =

(x + t

y

)with 0 ≤ t ≤ h.

= limh→0

1

h

∫ h

0

EF(x + t, y) ·

(10

)dt

= limh→0

1

h

∫ h

0P(x + t, y) dt

Use the Derivative Form of the Funda-mental Theorem of Calculus

= P(x, y)

Example 26.3.8Theorem 26.3.7 makes it easy to graphically recognize vectorfields that arenotconservative. For example, the field in Figure 26.42 is not conser-vative, because the integral along the indicated circle will not be zero. If we traversethe circle in the mathematically positive direction, the integral will be positive, be-cause we move “with the field” throughout the traversal. If we traverse the circlein the mathematically negative direction, the integral will be negative, because wemove “against the field” throughout the traversal.


It should be noted that it is not possible to prove graphically that a vector fieldis conservative. This is because some disturbances in the field may be too smallto be correctly judged graphically, but might still be sufficient to introduce someintegrals over closed curves with small, nonzero values.

Remark 26.3.9 There is also a computational condition that is equivalent to beinga gradient field. This condition is proved in Theorem 28.4.2 for three dimensional


vector fields. The special case of two dimensional vector fields is an easy conse-quence, given in Corollary 28.4.5.

26.3.4 The Impossibility of a Perpetual Motion Machine

Condition 2 of Theorem 26.3.7 shows that in a gradient field energy is neithergained nor lost when we traverse a closed path. This conservation of energy gaverise to the following notation.

Definition 26.3.10 Gradient fields are also often calledconservativevectorfields.

If the potential function isequal at the starting pointand endpoint of a curve, thenthe line integral is zero.

The level surfaces of the potential function of a conservative vector field definethe regions of equal potential energy as related to the field.

Definition 26.3.11 Let EF = E∇ f be a conservative vector field. The levelsurfaces of f are also called theequipotential surfacesof the vector fieldEF.

By Proposition 25.3.12 the conservative vector fieldEF is perpendicular to itsequipotential surfaces.

In centuries past, ill-fated attempts have been made to construct what is calleda “perpetual motion machine”. This is a machine that, once started, either neverstops, or that possibly even produces some energy to be used elsewhere. Of course,by the principle of conservation of energy and by the steady increase of entropy,this is impossible. Yet that has not stopped the attempts!

A classical attempt is the “idea” of lifting up a mass and then finding a trickyway to lower it that releases more energy than was needed to lift up the mass. Thefact that the gravitational field is a gradient field shows why such a machine willnot produce energy. If we fix the highest and lowest points on the path of the mass,condition 2 of Theorem 26.3.7 guarantees that the amount of energy needed to goup to the highest point is always equal to the amount of energy released on the waydown, no matter how tricky the motion is. Throw in the fact that there will alwaysbe some friction and whatever moves the mass must eventually come to a halt. Weconclude (once more) that there will never be perpetual motion machines.

The impossibility of a perpetual motion machine does not mean however, thatit is impossible to use a conservative vector field to convert energy or that all vectorfields are conservative. Let’s analyze these two phenomena.

It is possible to use a conservative vector field to convert energy. In a hy-droelectric plant the gravitational field pulls water across turbines, which convertthe kinetic energy of the water into electrical energy. The electrical energy that we“gain” for use is offset by the water’s corresponding “loss” of kinetic and ultimatelypotential energy.

Not all vector fields are conservative. Examples 26.2.3 and 26.2.4 show that the

magnetic field of a wire, which is a scaled version ofEF =

(−yx

)is not conserva-

tive. This is not a problem, because the magnetic field does not provide forces that

would pull an object around a circle. Vortices in fluids look similar toEF =

(−yx

)and they do speed up an object that thenseeminglygoes around the same circleover and over again. However, this is an imprecise observation.


In a vortex in a fluid, the vectors are not exactly tangential to circles. Theywill either point a little bit outwards or inwards. Thus it is not the same fluidthat goes around the vortex forever and any energy “gain” by a particle inthe vortex is once more offset by a “loss” of energy in the fluid.

Exercises

1. Determine which of the vector fields below are conservative.

• For the non-conservative fields determine a closed pathalong which the line integral is not zero.

• For the conservative fields find a potential function.

(a) EF(x, y) =

(−yx

)(b) EF(x, y) =

(2x + y

x

)2. For each vector field below, if possible, prove that the vector

field is not conservative by sketching a closed curveC with∮EF · dEc 6= 0. If this is not possible, find a potential function.

(You will need Exercise 1 in Section 26.1.)

(a) The vector field in Figure 26.27, part 1)

(b) The vector field in Figure 26.27, part 2)

(c) The vector field in Figure 26.27, part 3)

(d) The vector field in Figure 26.27, part 4)

(e) The vector field in Figure 26.27, part 5)

(f) The vector field in Figure 26.27, part 6)

(g) The vector field in Figure 26.27, part 7)

(h) The vector field in Figure 26.27, part 8)

(i) The vector field in Figure 26.27, part 9)

(j) The vector field in Figure 26.14

3. Compute the line integral of the given vector fieldEF over thegiven curveEc. If appropriate, use the Fundamental Theorem forLine Integrals.

(a) The line integral of the vector fieldEF(x, y) =

(2xy2

2yx2

)along the pathEr (t) =

(et cos(t)et sin(t)

), where 0≤ t ≤ 2π .

(b) The line integral of the vector fieldEF(x, y) =

(−y2 sin(x)

2y cos(x)

)along the pathEr (t) =

(cos(t)sin(t)

)with 0 ≤ t ≤ 2π .

(c) The line integral of the vector fieldEF(x, y) =

(y2

x2

)along the circle of radius 1 about the origin.

(d) The line integral of the vector fieldEF(x, y) =

(x3

y3

)over the line segment from(2, 4) to (−2, 0).

4. Conservative force fields

(a) Show that the force fieldEF =

2x +1

y

3y2−

x

y2

is conser-

vative in the regiony > 0 by finding a potential functionfor it. Now calculate the work done byEF as it acts on anobject which moves in the upper half plane from(0, 3) to(6, 2).

(b) Show that the force fieldEF =

( 2xy +2x(

x2+y2)

x2+

2y(x2+y2

) + 1

)is

conservative in the region(x, y) 6= (0, 0) by finding apotential function for it. Now use this potential functionto calculate the work done byEF as it acts on an objectwhich moves from(2, 1) to (3, 4) along any path whichdoes not pass through(0, 0).

(c) Show that the force fieldEF =

(−

(4+y2

)(1+x)2

2y +2y

1+x

)is conserva-

tive in the regionx > −1 by finding a potential functionfor it.

(d) Show that the force fieldEF = (2x + ln y)Ei +

(1 +

x

y

)Ej

is conservative in the regiony > 0 by finding a potentialfunction for it. Now use this potential function to cal-culate the work done byEF as it acts on an object whichmoves in the upper half plane from(−1, 1) to (3, 2).

(e) Show that the force fieldEF =

(1

y+ 2x

)Ei +

(3y2

−x

y2

)Ej

is conservative in the regiony 6= 0 by finding a potentialfunction for it. Now use this potential function to cal-culate the work done byEF as it acts on an object whichmoves from(x, y) = (2, 1) to (x, y) = (1, 3) along anycurve in the upper half plane.

(f) Show that the force field

EF =

(2 +

2x

1 + y

)Ei +

(2y −

x2

(1 + y)2

)Ej is conserva-

tive in the regiony > −1 by finding a potential functionfor it. Now calculate the work done byEF as it acts onan object which moves from(1, 0) to (9, 3) along anypiecewise smooth curve in the regiony > −1.

5. State the precise definition of a disconnected region by negatingthe definition of a connected region.

6. Analyzing Theorem 26.3.7.

(a) State what it means tonotbe a gradient field by negatingpart 2 of Theorem 26.3.7.

(b) State what it means tonotbe a gradient field by negatingpart 3 of Theorem 26.3.7.

7. Prove the Fundamental Theorem for Line Integrals for a piece-wise smooth curveEc(t), a ≤ t ≤ b.

Hint. Let a = t0 < t1 < · · · < tn−1 < tn = b be a partitionof the parameter interval such thatEc(t) is differentiable on eachinterval [tn−1, ti ]. Apply the Fundamental Theorem for LineIntegrals on each interval separately and sum the results.

Vector Fields and Line Integrals field...242 26. Vector Fields and Line Integrals Introduction...

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