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Calculus Volume 3

6.3 Conservative Vector Fields

Calculus Volume 36.3 Conservative Vector Fields
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 6.3.1. Describe simple and closed curves; define connected and simply connected regions.
  • 6.3.2. Explain how to find a potential function for a conservative vector field.
  • 6.3.3. Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.
  • 6.3.4. Explain how to test a vector field to determine whether it is conservative.

In this section, we continue the study of conservative vector fields. We examine the Fundamental Theorem for Line Integrals, which is a useful generalization of the Fundamental Theorem of Calculus to line integrals of conservative vector fields. We also discover show how to test whether a given vector field is conservative, and determine how to build a potential function for a vector field known to be conservative.

Curves and Regions

Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here.

We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve (Figure 6.25).

Definition

Curve C is a closed curve if there is a parameterization r(t),atbr(t),atb of C such that the parameterization traverses the curve exactly once and r(a)=r(b).r(a)=r(b). Curve C is a simple curve if C does not cross itself. That is, C is simple if there exists a parameterization r(t),atbr(t),atb of C such that r is one-to-one over (a,b).(a,b). It is possible for r(a)=r(b),r(a)=r(b), meaning that the simple curve is also closed.

An image showing eight curves and their types. The first curve is neither simple nor closed; it has two endpoints and crosses itself twice. The second curve is simple but not closed; it does not cross itself and has two endpoints. The third curve is closed but is not simple; it crosses itself a few times. The fourth is a simple closed curve; it does not cross itself and has no endpoints. The fifth is a simple, not closed curve; it does not cross itself, but it has endpoints. The sixth is a simple, closed curve; it does not cross itself and has no endpoints. The seventh is closed but not a simple curve; it crosses itself but has no endpoints. The last is not simple and not closed; it crosses itself and has endpoints.
Figure 6.25 Types of curves that are simple or not simple and closed or not closed.

Example 6.27

Determining Whether a Curve Is Simple and Closed

Is the curve with parameterization r(t)=cost,sin(2t)2,0t2πr(t)=cost,sin(2t)2,0t2π a simple closed curve?

Solution

Note that r(0)=1,0=r(2π);r(0)=1,0=r(2π); therefore, the curve is closed. The curve is not simple, however. To see this, note that r(π2)=0,0=r(3π2),r(π2)=0,0=r(3π2), and therefore the curve crosses itself at the origin (Figure 6.26).

A diagram in the (x,y) coordinate plane that shows a closed but not simple curve. It looks like a horizontal figure eight with the crossing point at the origin.
Figure 6.26 A curve that is closed but not simple.
Checkpoint 6.24

Is the curve given by parameterization r(t)=2cost,3sint,0t6π,r(t)=2cost,3sint,0t6π, a simple closed curve?

Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region’s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.

Definition

A region D is a connected region if, for any two points P1P1 and P2,P2, there is a path from P1P1 to P2P2 with a trace contained entirely inside D. A region D is a simply connected region if D is connected for any simple closed curve C that lies inside D, and curve C can be shrunk continuously to a point while staying entirely inside D. In two dimensions, a region is simply connected if it is connected and has no holes.

All simply connected regions are connected, but not all connected regions are simply connected (Figure 6.27).

A diagram showing simply connected, connected, and not connected regions. The simply connected regions have no holes. The connected regions may have holes, but a path can still be found between any two points in the region. The not connected region has some points that cannot be connected by a path in the region. Here, this is illustrated by showing two circular shapes that are defined as part of region D1 but are separated by white space.
Figure 6.27 Not all connected regions are simply connected. (a) Simply connected regions have no holes. (b) Connected regions that are not simply connected may have holes but you can still find a path in the region between any two points. (c) A region that is not connected has some points that cannot be connected by a path in the region.
Checkpoint 6.25

Is the region in the below image connected? Is the region simply connected?

A shaded circle with an open space in the shape of a circle inside it but very close to the boundary.

Fundamental Theorem for Line Integrals

Now that we understand some basic curves and regions, let’s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function ff has an antiderivative F, then the integral of ff from a to b depends only on the values of F at a and at b—that is,

abf(x)dx=F(b)F(a).abf(x)dx=F(b)F(a).

If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.

Example 6.28

Evaluating a Line Integral and the Antiderivatives of the Endpoints

Let F(x,y)=2x,4y.F(x,y)=2x,4y. Calculate CFdr,CFdr, where C is the line segment from (0,0) to (2,2)(Figure 6.28).

Solution

We use Equation 6.9 to calculate CFdr.CFdr. Curve C can be parameterized by r(t)=2t,2t,0t1.r(t)=2t,2t,0t1. Then, F(r(t))=4t,8tF(r(t))=4t,8t and r(t)=2,2,r(t)=2,2, which implies that

CF·dr=014t,8t·2,2dt=01(8t+16t)dt=0124tdt=[12t2]01=12.CF·dr=014t,8t·2,2dt=01(8t+16t)dt=0124tdt=[12t2]01=12.
A vector field in two dimensions. The arrows are longer the further away from the origin they are. They stretch out from the origin, forming a rectangular pattern. A line segment is drawn from P_0 at (0,0) to P_1 at (2,2).
Figure 6.28 The value of line integral CFdrCFdr depends only on the value of the potential function of F at the endpoints of the curve.

Notice that F=f,F=f, where f(x,y)=x2+2y2.f(x,y)=x2+2y2. If we think of the gradient as a derivative, then ff is an “antiderivative” of F. In the case of single-variable integrals, the integral of derivative g(x)g(x) is g(b)g(a),g(b)g(a), where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral F to be f(P1)f(P0),f(P1)f(P0), where P1P1 is the endpoint of the curve of integration and P0P0 is the start point. Notice that this is the case for this example:

CFdr=Cfdr=12CFdr=Cfdr=12

and

f(2,2)f(0,0)=4+80=12.f(2,2)f(0,0)=4+80=12.

In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.

The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the Fundamental Theorem for Line Integrals.

Theorem 6.7

The Fundamental Theorem for Line Integrals

Let C be a piecewise smooth curve with parameterization r(t),atb.r(t),atb. Let ff be a function of two or three variables with first-order partial derivatives that exist and are continuous on C. Then,

Cfdr=f(r(b))f(r(a)).Cfdr=f(r(b))f(r(a)).
6.12

Proof

By Equation 6.9,

Cfdr=abf(r(t))r(t)dt.Cfdr=abf(r(t))r(t)dt.

By the chain rule,

ddt(f(r(t))=f(r(t))r(t).ddt(f(r(t))=f(r(t))r(t).

Therefore, by the Fundamental Theorem of Calculus,

Cfdr=abf(r(t))r(t)dt=abddt(f(r(t))dt=[f(r(t))]t=at=b=f(r(b))f(r(a)).Cfdr=abf(r(t))r(t)dt=abddt(f(r(t))dt=[f(r(t))]t=at=b=f(r(b))f(r(a)).

We know that if F is a conservative vector field, there are potential functions ff such that f=F.f=F. Therefore CF·dr=Cf·dr=f(r(b))f(r(a)).CF·dr=Cf·dr=f(r(b))f(r(a)). In other words, just as with the Fundamental Theorem of Calculus, computing the line integral CF·dr,CF·dr, where F is conservative, is a two-step process: (1) find a potential function (“antiderivative”) ff for F and (2) compute the value of ff at the endpoints of C and calculate their difference f(r(b))f(r(a)).f(r(b))f(r(a)). Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.

Example 6.29

Applying the Fundamental Theorem

Calculate integral CFdr,CFdr, where F(x,y,z)=2xlny,x2y+z2,2yzF(x,y,z)=2xlny,x2y+z2,2yz and C is a curve with parameterization r(t)=t2,t,t,1ter(t)=t2,t,t,1te

  1. without using the Fundamental Theorem of Line Integrals and
  2. using the Fundamental Theorem of Line Integrals.

Solution

  1. First, let’s calculate the integral without the Fundamental Theorem for Line Integrals and instead use Equation 6.9:
    CFdr=1eF(r(t))r(t)dt=1e2t2lnt,t4t+t2,2t22t,1,1dt=1e(4t3lnt+t3+3t2)dt=1e4t3lntdt+1e(t3+3t2)dt=1e4t3lntdt+[t44+t3]1e=r1et3lntdt+e44+e354.CFdr=1eF(r(t))r(t)dt=1e2t2lnt,t4t+t2,2t22t,1,1dt=1e(4t3lnt+t3+3t2)dt=1e4t3lntdt+1e(t3+3t2)dt=1e4t3lntdt+[t44+t3]1e=r1et3lntdt+e44+e354.

    Integral 1et3lntdt1et3lntdt requires integration by parts. Let u=lntu=lnt and dv=t3.dv=t3. Then u=lnt,dv=t3u=lnt,dv=t3
    and
    du=1tdt,v=t44.du=1tdt,v=t44.

    Therefore,
    1et3lntdt=[t44lnt]1e141et3dt=e441r(e4414).1et3lntdt=[t44lnt]1e141et3dt=e441r(e4414).

    Thus,
    CFdr=41et3lntdt+e44+e354=4(e4414(e4414))+e44+e354=e4e44+14+e44+e354=e4+e31.CFdr=41et3lntdt+e44+e354=4(e4414(e4414))+e44+e354=e4e44+14+e44+e354=e4+e31.
  2. Given that f(x,y,z)=x2lny+yz2f(x,y,z)=x2lny+yz2 is a potential function for F, let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that
    CFdr=Cfdr=f(r(e))f(r(1))=f(e2,e,e)f(1,1,1)=e4+e31.CFdr=Cfdr=f(r(e))f(r(1))=f(e2,e,e)f(1,1,1)=e4+e31.

    This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.

Example 6.29 illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.

Checkpoint 6.26

Given that f(x,y)=(x1)2y+(y+1)2xf(x,y)=(x1)2y+(y+1)2x is a potential function for F=2xy2y+(y+1)2,(x1)2+2yx+2x,F=2xy2y+(y+1)2,(x1)2+2yx+2x, calculate integral CF·dr,CF·dr, where C is the lower half of the unit circle oriented counterclockwise.

A vector field in two dimensions. The arrows near the origin are the shortest, and the arrows in the upper right and lower left corners of quadrants 1 and 3 are the shortest. The arrows go up and to the left in quadrants 1 and 3. In quadrant 2, the arrows stretch up and to the right for values greater than x=-1. The closer the arrows are to y=1, the more horizontal they become. For values less than x=-1, the arrows point up and form a curve to the left. The closer the arrows are to y=1, the more horizontal they become. Above y=1, it looks like the arrows are shifting from vertical, going down to horizontal. In quadrant 4, the arrows go up and to the right fairly regularly, but they tend to be curving to the right the larger the x value becomes. For y values less than -1, the arrows shift from pointing up to pointing down, following x=1. The lower half of the unit circle with center at the origin is drawn in quadrants 3 and 4.

The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if F is conservative and C is a closed curve, then the circulation of F along C is zero—that is, CF·dr=0.CF·dr=0. To see why this is true, let ff be a potential function for F. Since C is a closed curve, the terminal point r(b) of C is the same as the initial point r(a) of C—that is, r(a)=r(b).r(a)=r(b). Therefore, by the Fundamental Theorem for Line Integrals,

CF·dr=Cf·dr=f(r(b))f(r(a))=f(r(b))f(r(b))=0.CF·dr=Cf·dr=f(r(b))f(r(a))=f(r(b))f(r(b))=0.

Recall that the reason a conservative vector field F is called “conservative” is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field F in integral CF·drCF·dr as a gravitational field, then the equation CF·dr=0CF·dr=0 follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.

The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path—meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.

Definition

Let F be a vector field with domain D. The vector field F is independent of path (or path independent) if C1F·dr=C2F·drC1F·dr=C2F·dr for any paths C1C1 and C2C2 in D with the same initial and terminal points.

The second consequence is stated formally in the following theorem.

Theorem 6.8

Path Independence of Conservative Fields

If F is a conservative vector field, then F is independent of path.

Proof

Let D denote the domain of F and let C1C1 and C2C2 be two paths in D with the same initial and terminal points (Figure 6.29). Call the initial point P1P1 and the terminal point P2.P2. Since F is conservative, there is a potential function ff for F. By the Fundamental Theorem for Line Integrals,

C1F·dr=f(P2)f(P1)=C2F·dr.C1F·dr=f(P2)f(P1)=C2F·dr.

Therefore, C1F·dr=C2F·drC1F·dr=C2F·dr and F is independent of path.

A vector field in two dimensions. The arrows are shorter the closer to the x axis and line x=1.5 they become. The arrows point up, converging around x=1.5 in quadrant 1. That line is approached from the left and from the right. Below, in quadrant 4, the arrows in the rough interval [1,2.5] curve out, away from the given line x=1.5, but do turn back in and converge to x=1.5 above the x axis. Outside of that interval, the arrows go to the left and right horizontally for x values less than 1 and greater than 2.5, respectively. A line is drawn from P_1 at the origin to P_2 at (3,.75) and labeled C_2. C_1 is a simple curve that connects the given endpoints above C_2, C_3 is a simple curve that connects the given endpoints below C_2.
Figure 6.29 The vector field is conservative, and therefore independent of path.

To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.

We have shown that if F is conservative, then F is independent of path. It turns out that if the domain of F is open and connected, then the converse is also true. That is, if F is independent of path and the domain of F is open and connected, then F is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.

Theorem 6.9

The Path Independence Test for Conservative Fields

If F is a continuous vector field that is independent of path and the domain D of F is open and connected, then F is conservative.

Proof

We prove the theorem for vector fields in 2.2. The proof for vector fields in 33 is similar. To show that F=P,QF=P,Q is conservative, we must find a potential function ff for F. To that end, let X be a fixed point in D. For any point (x,y)(x,y) in D, let C be a path from X to (x,y).(x,y). Define ff(x,y)(x,y) by f(x,y)=CF·dr.f(x,y)=CF·dr. (Note that this definition of ff makes sense only because F is independent of path. If F was not independent of path, then it might be possible to find another path CC from X to (x,y)(x,y) such that CF·drCF·dr,CF·drCF·dr, and in such a case ff(x,y)(x,y) would not be a function.) We want to show that ff has the property f=F.f=F.

Since domain D is open, it is possible to find a disk centered at (x,y)(x,y) such that the disk is contained entirely inside D. Let (a,y)(a,y) with a<xa<x be a point in that disk. Let C be a path from X to (x,y)(x,y) that consists of two pieces: C1C1 and C2.C2. The first piece, C1,C1, is any path from C to (a,y)(a,y) that stays inside D; C2C2 is the horizontal line segment from (a,y)(a,y) to (x,y)(x,y) (Figure 6.30). Then

f(x,y)=C1F·dr+C2F·dr.f(x,y)=C1F·dr+C2F·dr.

The first integral does not depend on x, so

fx=xC2Fdr.fx=xC2Fdr.

If we parameterize C2C2 by r(t)=t,y,atx,r(t)=t,y,atx, then

fx=xC2Fdr=xaxF(r(t))r(t)dt=xaxF(r(t))ddt(t,y)dt=xaxF(r(t))1,0dt=xaxP(t,y)dt.fx=xC2Fdr=xaxF(r(t))r(t)dt=xaxF(r(t))ddt(t,y)dt=xaxF(r(t))1,0dt=xaxP(t,y)dt.

By the Fundamental Theorem of Calculus (part 1),

fx=xaxP(t,y)dt=P(x,y).fx=xaxP(t,y)dt=P(x,y).
A diagram of a region D in the rough shape of a backwards C. It is a simply connected region formed by a closed curve. Another curve C_1 is drawn inside D from point X to (a,y). C_2 is a horizontal line segment drawn from (a,y) to (x,y). Arrowheads point to (a,y) on C_1 and to (x,y) on C_2.
Figure 6.30 Here, C1C1 is any path from C to (a,y)(a,y) that stays inside D, and C2C2 is the horizontal line segment from (a,y)(a,y) to (x,y).(x,y).

A similar argument using a vertical line segment rather than a horizontal line segment shows that fy=Q(x,y).fy=Q(x,y).

Therefore f=Ff=F and F is conservative.

We have spent a lot of time discussing and proving Path Independence of Conservative Fields and The Path Independence Test for Conservative Fields, but we can summarize them simply: a vector field F on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.

Example 6.30

Showing That a Vector Field Is Not Conservative

Use path independence to show that vector field F(x,y)=x2y,y+5F(x,y)=x2y,y+5 is not conservative.

Solution

We can indicate that F is not conservative by showing that F is not path independent. We do so by giving two different paths, C1C1 and C2,C2, that both start at (0,0)(0,0) and end at (1,1),(1,1), and yet C1FdrC2Fdr.C1FdrC2Fdr.

Let C1C1 be the curve with parameterization r1(t)=t,t,0t1r1(t)=t,t,0t1 and let C2C2 be the curve with parameterization r2(t)=t,t2,0t1r2(t)=t,t2,0t1 (Figure 6.31). Then

C1F·dr=01F(r1(t))·r1(t)dt=01t3,t+5·1,1dt=01(t3+t+5)dt=[t44+t22+5t]01=234C1F·dr=01F(r1(t))·r1(t)dt=01t3,t+5·1,1dt=01(t3+t+5)dt=[t44+t22+5t]01=234

and

C2F·dr=01F(r2(t))·r2(t)dt=01t4,t2+5·1,2tdt=01(t4+2t3+10t)dt=[t55+t42+5t2]01=5710.C2F·dr=01F(r2(t))·r2(t)dt=01t4,t2+5·1,2tdt=01(t4+2t3+10t)dt=[t55+t42+5t2]01=5710.

Since C1FdrC2Fdr,C1FdrC2Fdr, the value of a line integral of F depends on the path between two given points. Therefore, F is not independent of path, and F is not conservative.

A vector field drawn in two dimensions. The arrows are roughly the same length. They point directly up but tend to shift to the right in the upper right portion of quadrant 1. Curves C_1 and C_2 connect the origin to point (1,1). They are both simple curves, and their arrowheads point to (1,1).
Figure 6.31 Curves C1C1 and C2C2 are both oriented from left to right.

Checkpoint 6.27

Show that F(x,y)=xy,x2y2F(x,y)=xy,x2y2 is not path independent by considering the line segment from (0,0)(0,0) to (0,2)(0,2) and the piece of the graph of y=x22y=x22 that goes from (0,0)(0,0) to (0,2).(0,2).

Conservative Vector Fields and Potential Functions

As we have learned, the Fundamental Theorem for Line Integrals says that if F is conservative, then calculating CF·drCF·dr has two steps: first, find a potential function ff for F and, second, calculate f(P1)f(P0),f(P1)f(P0), where P1P1 is the endpoint of C and P0P0 is the starting point. To use this theorem for a conservative field F, we must be able to find a potential function ff for F. Therefore, we must answer the following question: Given a conservative vector field F, how do we find a function ff such that f=F?f=F? Before giving a general method for finding a potential function, let’s motivate the method with an example.

Example 6.31

Finding a Potential Function

Find a potential function for F(x,y)=2xy3,3x2y2+cos(y),F(x,y)=2xy3,3x2y2+cos(y), thereby showing that F is conservative.

Solution

Suppose that f(x,y)f(x,y) is a potential function for F. Then, f=F,f=F, and therefore

fx=2xy3andfy=3x2y2+cosy.fx=2xy3andfy=3x2y2+cosy.

Integrating the equation fx=2xy3fx=2xy3 with respect to x yields the equation

f(x,y)=x2y3+h(y).f(x,y)=x2y3+h(y).

Notice that since we are integrating a two-variable function with respect to x, we must add a constant of integration that is a constant with respect to x, but may still be a function of y. The equation f(x,y)=x2y3+h(y)f(x,y)=x2y3+h(y) can be confirmed by taking the partial derivative with respect to x:

fx=x(x2y3)+x(h(y))=2xy3+0=2xy3.fx=x(x2y3)+x(h(y))=2xy3+0=2xy3.

Since ff is a potential function for F,

fy=3x2y2+cos(y),fy=3x2y2+cos(y),

and therefore

3x2y2+h(y)=3x2y2+cos(y).3x2y2+h(y)=3x2y2+cos(y).

This implies that h(y)=cosy,h(y)=cosy, so h(y)=siny+C.h(y)=siny+C. Therefore, any function of the form f(x,y)=x2y3+sin(y)+Cf(x,y)=x2y3+sin(y)+C is a potential function. Taking, in particular, C=0C=0 gives the potential function f(x,y)=x2y3+sin(y).f(x,y)=x2y3+sin(y).

To verify that ff is a potential function, note that f=2xy3,3x2y2+cosy=F.f=2xy3,3x2y2+cosy=F.

Checkpoint 6.28

Find a potential function for F(x,y)=exy3+y,3exy2+x.F(x,y)=exy3+y,3exy2+x.

The logic of the previous example extends to finding the potential function for any conservative vector field in 2.2. Thus, we have the following problem-solving strategy for finding potential functions:

Problem-Solving Stragegy: Finding a Potential Function for a Conservative Vector Field F(x,y)=P(x,y),Q(x,y)F(x,y)=P(x,y),Q(x,y)
  1. Integrate P with respect to x. This results in a function of the form g(x,y)+h(y),g(x,y)+h(y), where h(y)h(y) is unknown.
  2. Take the partial derivative of g(x,y)+h(y)g(x,y)+h(y) with respect to y, which results in the function gy(x,y)+h(y).gy(x,y)+h(y).
  3. Use the equation gy(x,y)+h(y)=Q(x,y)gy(x,y)+h(y)=Q(x,y) to find h(y).h(y).
  4. Integrate h(y)h(y) to find h(y).h(y).
  5. Any function of the form f(x,y)=g(x,y)+h(y)+C,f(x,y)=g(x,y)+h(y)+C, where C is a constant, is a potential function for F.

We can adapt this strategy to find potential functions for vector fields in 3,3, as shown in the next example.

Example 6.32

Finding a Potential Function in 33

Find a potential function for F(x,y)=2xy,x2+2yz3,3y2z2+2z,F(x,y)=2xy,x2+2yz3,3y2z2+2z, thereby showing that F is conservative.

Solution

Suppose that ff is a potential function. Then, f=Ff=F and therefore fx=2xy.fx=2xy. Integrating this equation with respect to x yields the equation f(x,y,z)=x2y+g(y,z)f(x,y,z)=x2y+g(y,z) for some function g. Notice that, in this case, the constant of integration with respect to x is a function of y and z.

Since ff is a potential function,

x2+2yz3=fy=x2+gy.x2+2yz3=fy=x2+gy.

Therefore,

gy=2yz3.gy=2yz3.

Integrating this function with respect to y yields

g(y,z)=y2z3+h(z)g(y,z)=y2z3+h(z)

for some function h(z)h(z) of z alone. (Notice that, because we know that g is a function of only y and z, we do not need to write g(y,z)=y2z3+h(x,z).)g(y,z)=y2z3+h(x,z).) Therefore,

f(x,y,z)=x2y+g(y,z)=x2y+y2z3+h(z).f(x,y,z)=x2y+g(y,z)=x2y+y2z3+h(z).

To find ff, we now must only find h. Since ff is a potential function,

3y2z2+2z=gz=3y2z2+h(z).3y2z2+2z=gz=3y2z2+h(z).

This implies that h(z)=2z,h(z)=2z, so h(z)=z2+C.h(z)=z2+C. Letting C=0C=0 gives the potential function

f(x,y,z)=x2y+y2z3+z2.f(x,y,z)=x2y+y2z3+z2.

To verify that ff is a potential function, note that f=2xy,x2+2yz3,3y2z2+2z=F.f=2xy,x2+2yz3,3y2z2+2z=F.

Checkpoint 6.29

Find a potential function for F(x,y,z)=12x2,cosycosz,1sinysinz.F(x,y,z)=12x2,cosycosz,1sinysinz.

We can apply the process of finding a potential function to a gravitational force. Recall that, if an object has unit mass and is located at the origin, then the gravitational force in 22 that the object exerts on another object of unit mass at the point (x,y)(x,y) is given by vector field

F(x,y)=Gx(x2+y2)3/2,y(x2+y2)3/2,F(x,y)=Gx(x2+y2)3/2,y(x2+y2)3/2,

where G is the universal gravitational constant. In the next example, we build a potential function for F, thus confirming what we already know: that gravity is conservative.

Example 6.33

Finding a Potential Function

Find a potential function ff for F(x,y)=Gx(x2+y2)3/2,y(x2+y2)3/2.F(x,y)=Gx(x2+y2)3/2,y(x2+y2)3/2.

Solution

Suppose that ff is a potential function. Then, f=Ff=F and therefore

fx=Gx(x2+y2)3/2.fx=Gx(x2+y2)3/2.

To integrate this function with respect to x, we can use u-substitution. If u=x2+y2,u=x2+y2, then du2=xdx,du2=xdx, so

Gx(x2+y2)3/2dx=G2u3/2du=Gu+h(y)=Gx2+y2+h(y)Gx(x2+y2)3/2dx=G2u3/2du=Gu+h(y)=Gx2+y2+h(y)

for some function h(y).h(y). Therefore,

f(x,y)=Gx2+y2+h(y).f(x,y)=Gx2+y2+h(y).

Since ff is a potential function for F,

fy=Gy(x2+y2)3/2.fy=Gy(x2+y2)3/2.

Since f(x,y)=Gx2+y2+h(y),f(x,y)=Gx2+y2+h(y), fyfy also equals Gy(x2+y2)3/2+h(y).Gy(x2+y2)3/2+h(y).

Therefore,

Gy(x2+y2)3/2+h(y)=Gy(x2+y2)3/2,Gy(x2+y2)3/2+h(y)=Gy(x2+y2)3/2,

which implies that h(y)=0.h(y)=0. Thus, we can take h(y)h(y) to be any constant; in particular, we can let h(y)=0.h(y)=0. The function

f(x,y)=Gx2+y2f(x,y)=Gx2+y2

is a potential function for the gravitational field F. To confirm that ff is a potential function, note that

f=12G(x2+y2)3/2(2x),12G(x2+y2)3/2(2y)=Gx(x2+y2)3/2,Gy(x2+y2)3/2=F.f=12G(x2+y2)3/2(2x),12G(x2+y2)3/2(2y)=Gx(x2+y2)3/2,Gy(x2+y2)3/2=F.
Checkpoint 6.30

Find a potential function ff for the three-dimensional gravitational force F(x,y,z)=Gx(x2+y2+z2)3/2,Gy(x2+y2+z2)3/2,Gz(x2+y2+z2)3/2.F(x,y,z)=Gx(x2+y2+z2)3/2,Gy(x2+y2+z2)3/2,Gz(x2+y2+z2)3/2.

Testing a Vector Field

Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if F is conservative, then F has the cross-partial property (see The Cross-Partial Property of Conservative Vector Fields). That is, if F=P,Q,RF=P,Q,R is conservative, then Py=Qx,Pz=Rx,Py=Qx,Pz=Rx, and Qz=Ry.Qz=Ry. So, if F has the cross-partial property, then is F conservative? If the domain of F is open and simply connected, then the answer is yes.

Theorem 6.10

The Cross-Partial Test for Conservative Fields

If F=P,Q,RF=P,Q,R is a vector field on an open, simply connected region D and Py=Qx,Pz=Rx,Py=Qx,Pz=Rx, and Qz=RyQz=Ry throughout D, then F is conservative.

Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.

Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:

Theorem 6.11

Cross-Partial Property of Conservative Fields

Let F=P,Q,RF=P,Q,R be a vector field on an open, simply connected region D. Then Py=Qx,Pz=Rx,Py=Qx,Pz=Rx, and Qz=RyQz=Ry throughout D if and only if F is conservative.

The version of this theorem in 22 is also true. If F=P,QF=P,Q is a vector field on an open, simply connected domain in 2,2, then F is conservative if and only if Py=Qx.Py=Qx.

Example 6.34

Determining Whether a Vector Field Is Conservative

Determine whether vector field F(x,y,z)=xy2z,x2yz,z2F(x,y,z)=xy2z,x2yz,z2 is conservative.

Solution

Note that the domain of F is all of 22 and 33 is simply connected. Therefore, we can use Cross-Partial Property of Conservative Fields to determine whether F is conservative. Let

P(x,y,z)=xy2z,Q(x,y,z)=x2yz,andR(x,y,z)=z2.P(x,y,z)=xy2z,Q(x,y,z)=x2yz,andR(x,y,z)=z2.

Since Qz=x2yQz=x2y and Ry=0,Ry=0, the vector field is not conservative.

Example 6.35

Determining Whether a Vector Field Is Conservative

Determine vector field F(x,y)=xln(y),x22yF(x,y)=xln(y),x22y is conservative.

Solution

Note that the domain of F is the part of 22 in which y>0.y>0. Thus, the domain of F is part of a plane above the x-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use Cross-Partial Property of Conservative Fields to determine whether F is conservative. Let

P(x,y)=xln(y)andQ(x,y)=x22y.P(x,y)=xln(y)andQ(x,y)=x22y.

Then Py=xy=QxPy=xy=Qx and thus F is conservative.

Checkpoint 6.31

Determine whether F(x,y)=sinxcosy,cosxsinyF(x,y)=sinxcosy,cosxsiny is conservative.

When using Cross-Partial Property of Conservative Fields, it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of Cross-Partial Property of Conservative Fields, the theorem can be applied only if the domain of the vector field is simply connected.

To see what can go wrong when misapplying the theorem, consider the vector field from Example 6.30:

F(x,y)=yx2+y2i+xx2+y2j.F(x,y)=yx2+y2i+xx2+y2j.

This vector field satisfies the cross-partial property, since

y(yx2+y2)=(x2+y2)y(2y)(x2+y2)2=x2y2(x2+y2)2y(yx2+y2)=(x2+y2)y(2y)(x2+y2)2=x2y2(x2+y2)2

and

x(xx2+y2)=(x2+y2)+x(2x)(x2+y2)2=x2y2(x2+y2)2.x(xx2+y2)=(x2+y2)+x(2x)(x2+y2)2=x2y2(x2+y2)2.

Since F satisfies the cross-partial property, we might be tempted to conclude that F is conservative. However, F is not conservative. To see this, let

r(t)=cost,sint,0tπr(t)=cost,sint,0tπ

be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this C1)C1) and let

s(t)=cost,sint,0tπs(t)=cost,sint,0tπ

be a parameterization of the lower half of a unit circle oriented clockwise (denote this C2).C2). Notice that C1C1 and C2C2 have the same starting point and endpoint. Since sin2t+cos2t=1,sin2t+cos2t=1,

F(r(t))r(t)=sin(t),cos(t)sin(t),cos(t)=−1F(r(t))r(t)=sin(t),cos(t)sin(t),cos(t)=−1

and

F(s(t))·s(t)=sint,cost·sint,cost=sin2t+cos2t=1.F(s(t))·s(t)=sint,cost·sint,cost=sin2t+cos2t=1.

Therefore,

C1F·dr=0π−1dt=πandC2F·dr=0π1dt=π.C1F·dr=0π−1dt=πandC2F·dr=0π1dt=π.

Thus, C1C1 and C2C2 have the same starting point and endpoint, but C1F·drC2F·dr.C1F·drC2F·dr. Therefore, F is not independent of path and F is not conservative.

To summarize: F satisfies the cross-partial property and yet F is not conservative. What went wrong? Does this contradict Cross-Partial Property of Conservative Fields? The issue is that the domain of F is all of 22 except for the origin. In other words, the domain of F has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, Cross-Partial Property of Conservative Fields does not apply to F.

We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form CF·dr,CF·dr, then our first question should be: Is F conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can’t help us and we have to use other methods, such as using Equation 6.9.

Example 6.36

Using the Fundamental Theorem for Line Integrals

Calculate line integral CF·dr,CF·dr, where F(x,y,z)=2xeyz+exz,x2eyz,x2ey+exF(x,y,z)=2xeyz+exz,x2eyz,x2ey+ex and C is any smooth curve that goes from the origin to (1,1,1).(1,1,1).

Solution

Before trying to compute the integral, we need to determine whether F is conservative and whether the domain of F is simply connected. The domain of F is all of 3,3, which is connected and has no holes. Therefore, the domain of F is simply connected. Let

P(x,y,z)=2xeyz+exz,Q(x,y,z)=x2eyz,andR(x,y,z)=x2ey+exP(x,y,z)=2xeyz+exz,Q(x,y,z)=x2eyz,andR(x,y,z)=x2ey+ex

so that F=P,Q,R.F=P,Q,R. Since the domain of F is simply connected, we can check the cross partials to determine whether F is conservative. Note that

Py=2xeyz=QxPz=2xey+ex=RxQz=x2ey=Ry.Py=2xeyz=QxPz=2xey+ex=RxQz=x2ey=Ry.

Therefore, F is conservative.

To evaluate CF·drCF·dr using the Fundamental Theorem for Line Integrals, we need to find a potential function ff for F. Let ff be a potential function for F. Then, f=F,f=F, and therefore fx=2xeyz+exz.fx=2xeyz+exz. Integrating this equation with respect to x gives f(x,y,z)=x2eyz+exz+h(y,z)f(x,y,z)=x2eyz+exz+h(y,z) for some function h. Differentiating this equation with respect to y gives x2eyz+hy=Q=x2eyz,x2eyz+hy=Q=x2eyz, which implies that hy=0.hy=0. Therefore, h is a function of z only, and f(x,y,z)=x2eyz+exz+h(z).f(x,y,z)=x2eyz+exz+h(z). To find h, note that fz=x2ey+ex+h(z)=R=x2ey+ex.fz=x2ey+ex+h(z)=R=x2ey+ex. Therefore, h(z)=0h(z)=0 and we can take h(z)=0.h(z)=0. A potential function for F is f(x,y,z)=x2eyz+exz.f(x,y,z)=x2eyz+exz.

Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,

CF·dr=Cf·dr=f(1,1,1)f(0,0,0)=2e.CF·dr=Cf·dr=f(1,1,1)f(0,0,0)=2e.

Analysis

Notice that if we hadn’t recognized that F is conservative, we would have had to parameterize C and use Equation 6.9. Since curve C is unknown, using the Fundamental Theorem for Line Integrals is much simpler.

Checkpoint 6.32

Calculate integral CF·dr,CF·dr, where F(x,y)=sinxsiny,5cosxcosyF(x,y)=sinxsiny,5cosxcosy and C is a semicircle with starting point (0,π)(0,π) and endpoint (0,π).(0,π).

Example 6.37

Work Done on a Particle

Let F(x,y)=2xy2,2x2yF(x,y)=2xy2,2x2y be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the x-axis or the y-axis. Furthermore, the particle’s motion can be modeled with a smooth parameterization. Show that F does positive work on the particle.

Solution

We show that F does positive work on the particle by showing that F is conservative and then by using the Fundamental Theorem for Line Integrals.

To show that F is conservative, suppose f(x,y)f(x,y) were a potential function for F. Then, f=F=2xy2,2x2yf=F=2xy2,2x2y and therefore fx=2xy2fx=2xy2 and fy=2x2y.fy=2x2y. Equation fx=2xy2fx=2xy2 implies that f(x,y)=x2y2+h(y).f(x,y)=x2y2+h(y). Deriving both sides with respect to y yields fy=2x2y+h(y).fy=2x2y+h(y). Therefore, h(y)=0h(y)=0 and we can take h(y)=0.h(y)=0.

If f(x,y)=x2y2,f(x,y)=x2y2, then note that f=2xy2,2x2y=F,f=2xy2,2x2y=F, and therefore ff is a potential function for F.

Let (a,b)(a,b) be the point at which the particle stops is motion, and let C denote the curve that models the particle’s motion. The work done by F on the particle is CF·dr.CF·dr. By the Fundamental Theorem for Line Integrals,

CF·dr=Cf·dr=f(a,b)f(0,0)=a2b2.CF·dr=Cf·dr=f(a,b)f(0,0)=a2b2.

Since a0a0 and b0,b0, by assumption, a2b2>0.a2b2>0. Therefore, CF·dr>0,CF·dr>0, and F does positive work on the particle.

Analysis

Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use Equation 6.9. Since the path of motion C can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.

Checkpoint 6.33

Let F(x,y)=4x3y4,4x4y3,F(x,y)=4x3y4,4x4y3, and suppose that a particle moves from point (4,4)(4,4) to (1,1)(1,1) along any smooth curve. Is the work done by F on the particle positive, negative, or zero?

Section 6.3 Exercises

99.

True or False? If vector field F is conservative on the open and connected region D, then line integrals of F are path independent on D, regardless of the shape of D.

100.

True or False? Function r(t)=a+t(ba),r(t)=a+t(ba), where 0t1,0t1, parameterizes the straight-line segment from atob.atob.

101.

True or False? Vector field F(x,y,z)=(ysinz)i+(xsinz)j+(xycosz)kF(x,y,z)=(ysinz)i+(xsinz)j+(xycosz)k is conservative.

102.

True or False? Vector field F(x,y,z)=yi+(x+z)jykF(x,y,z)=yi+(x+z)jyk is conservative.

103.

Justify the Fundamental Theorem of Line Integrals for CF·drCF·dr in the case when F(x,y)=(2x+2y)i+(2x+2y)jF(x,y)=(2x+2y)i+(2x+2y)j and C is a portion of the positively oriented circle x2+y2=25x2+y2=25 from (5, 0) to (3, 4).

104.

[T] Find CF·dr,,]CF·dr,,] where F(x,y)=(yexy+cosx)i+(xexy+1y2+1)jF(x,y)=(yexy+cosx)i+(xexy+1y2+1)j and C is a portion of curve y=sinxy=sinx from x=0x=0 to x=π2.x=π2.

105.

[T] Evaluate line integral CF·dr,CF·dr, where F(x,y)=(exsinyy)i+(excosyx2)j,F(x,y)=(exsinyy)i+(excosyx2)j, and C is the path given by r(t)=[t3sinπt2]i[π2cos(πt2+π2)]jr(t)=[t3sinπt2]i[π2cos(πt2+π2)]j for 0t1.0t1.

A vector field in three dimensions. The arrows are roughly the same length and all point up into the z-plane. A curve is drawn seemingly parallel to the (x,y)-plane. In the (x,y)-plane, it would look like a decreasing concave down curve in quadrant 1.

For the following exercises, determine whether the vector field is conservative and, if it is, find the potential function.

106.

F(x,y)=2xy3i+3y2x2jF(x,y)=2xy3i+3y2x2j

107.

F(x,y)=(y+exsiny)i+[(x+2)excosy]jF(x,y)=(y+exsiny)i+[(x+2)excosy]j

108.

F(x,y)=(e2xsiny)i+[e2xcosy]jF(x,y)=(e2xsiny)i+[e2xcosy]j

109.

F(x,y)=(6x+5y)i+(5x+4y)jF(x,y)=(6x+5y)i+(5x+4y)j

110.

F(x,y)=[2xcos(y)ycos(x)]i+[x2sin(y)sin(x)]jF(x,y)=[2xcos(y)ycos(x)]i+[x2sin(y)sin(x)]j

111.

F(x,y)=[yex+sin(y)]i+[ex+xcos(y)]jF(x,y)=[yex+sin(y)]i+[ex+xcos(y)]j

For the following exercises, evaluate the line integrals using the Fundamental Theorem of Line Integrals.

112.

C(yi+xj)·dr,C(yi+xj)·dr, where C is any path from (0, 0) to (2, 4)

113.

C(2ydx+2xdy),C(2ydx+2xdy), where C is the line segment from (0, 0) to (4, 4)

114.

[T] C[arctanyxxyx2+y2]dx+[x2x2+y2+ey(1y)]dy,C[arctanyxxyx2+y2]dx+[x2x2+y2+ey(1y)]dy, where C is any smooth curve from (1, 1) to (−1,2)(−1,2)

115.

Find the conservative vector field for the potential function

f(x,y)=5x2+3xy+10y2.f(x,y)=5x2+3xy+10y2.

For the following exercises, determine whether the vector field is conservative and, if so, find a potential function.

116.

F(x,y)=(12xy)i+6(x2+y2)jF(x,y)=(12xy)i+6(x2+y2)j

117.

F(x,y)=(excosy)i+6(exsiny)jF(x,y)=(excosy)i+6(exsiny)j

118.

F(x,y)=(2xyex2y)i+6(x2ex2y)jF(x,y)=(2xyex2y)i+6(x2ex2y)j

119.

F(x,y,z)=(yez)i+(xez)j+(xyez)kF(x,y,z)=(yez)i+(xez)j+(xyez)k

120.

F(x,y,z)=(siny)i(xcosy)j+kF(x,y,z)=(siny)i(xcosy)j+k

121.

F(x,y,z)=(1y)i+(xy2)j+(2z1)kF(x,y,z)=(1y)i+(xy2)j+(2z1)k

122.

F(x,y,z)=3z2icosyj+2xzkF(x,y,z)=3z2icosyj+2xzk

123.

F(x,y,z)=(2xy)i+(x2+2yz)j+y2kF(x,y,z)=(2xy)i+(x2+2yz)j+y2k

For the following exercises, determine whether the given vector field is conservative and find a potential function.

124.

F(x,y)=(excosy)i+6(exsiny)jF(x,y)=(excosy)i+6(exsiny)j

125.

F(x,y)=(2xyex2y)i+6(x2ex2y)jF(x,y)=(2xyex2y)i+6(x2ex2y)j

For the following exercises, evaluate the integral using the Fundamental Theorem of Line Integrals.

126.

Evaluate Cf·dr,Cf·dr, where f(x,y,z)=cos(πx)+sin(πy)xyzf(x,y,z)=cos(πx)+sin(πy)xyz and C is any path that starts at (1,12,2)(1,12,2) and ends at (2,1,−1).(2,1,−1).

127.

[T] Evaluate Cf·dr,Cf·dr, where f(x,y)=xy+exf(x,y)=xy+ex and C is a straight line from (0,0)(0,0) to (2,1).(2,1).

128.

[T] Evaluate Cf·dr,Cf·dr, where f(x,y)=x2yxf(x,y)=x2yx and C is any path in a plane from (1, 2) to (3, 2).

129.

Evaluate Cf·dr,Cf·dr, where f(x,y,z)=xyz2yzf(x,y,z)=xyz2yz and C has initial point (1, 2) and terminal point (3, 5).

For the following exercises, let F(x,y)=2xy2i+(2yx2+2y)jF(x,y)=2xy2i+(2yx2+2y)j and G(x,y)=(y+x)i+(yx)j,G(x,y)=(y+x)i+(yx)j, and let C1 be the curve consisting of the circle of radius 2, centered at the origin and oriented counterclockwise, and C2 be the curve consisting of a line segment from (0, 0) to (1, 1) followed by a line segment from (1, 1) to (3, 1).

A vector fields in two dimensions is shown. It has short arrows close to the origin. Longer arrows are in the upper right corner of quadrant 1 and somewhat in the bottom right of quadrant 4, upper left of quadrant 2, and lower left of quadrant 3. The arrows all point away from the origin at about 90-degrees in their respective quadrants. A unit circle with center at the origin is drawn as C_1. Curve C_2 connects the origin, (1,1), and (3,1) with arrowheads pointing in that order. A vector field has the same curves C_1 and C_2. However, the arrows are different. Here, the arrows spiral out from the origin in a clockwise manner. The further away they are from the origin, the longer they become. They are largely horizontal in quadrants 1 and 3 and largely vertical in quadrants 2 and 4.
130.

Calculate the line integral of F over C1.

131.

Calculate the line integral of G over C1.

132.

Calculate the line integral of F over C2.

133.

Calculate the line integral of G over C2.

134.

[T] Let F(x,y,z)=x2i+zsin(yz)j+ysin(yz)k.F(x,y,z)=x2i+zsin(yz)j+ysin(yz)k. Calculate CF·dr,CF·dr, where C is a path from A=(0,0,1)A=(0,0,1) to B=(3,1,2).B=(3,1,2).

135.

[T] Find line integral CF·drCF·dr of vector field F(x,y,z)=3x2zi+z2j+(x3+2yz)kF(x,y,z)=3x2zi+z2j+(x3+2yz)k along curve C parameterized by r(t)=(lntln2)i+t3/2j+tcos(πt),1t4.r(t)=(lntln2)i+t3/2j+tcos(πt),1t4.

For the following exercises, show that the following vector fields are conservative by using a computer. Calculate CF·drCF·dr for the given curve.

136.

F=(xy2+3x2y)i+(x+y)x2j;F=(xy2+3x2y)i+(x+y)x2j; C is the curve consisting of line segments from (1,1)(1,1) to (0,2)(0,2) to (3,0).(3,0).

137.

F=2xy2+1i2y(x2+1)(y2+1)2j;F=2xy2+1i2y(x2+1)(y2+1)2j; C is parameterized by x=t31,y=t6t,0t1.x=t31,y=t6t,0t1.

138.

[T] F=[cos(xy2)xy2sin(xy2)]i2x2ysin(xy2)j;F=[cos(xy2)xy2sin(xy2)]i2x2ysin(xy2)j; C is curve (et,et+1),−1t0.(et,et+1),−1t0.

139.

The mass of Earth is approximately 6×1027g6×1027g and that of the Sun is 330,000 times as much. The gravitational constant is 6.7×10−8cm3/s2·g.6.7×10−8cm3/s2·g. The distance of Earth from the Sun is about 1.5×1012cm.1.5×1012cm. Compute, approximately, the work necessary to increase the distance of Earth from the Sun by 1cm.1cm.

140.

[T] Let F=(x,y,z)=(exsiny)i+(excosy)j+z2k.F=(x,y,z)=(exsiny)i+(excosy)j+z2k. Evaluate the integral CF·ds,CF·ds, where c(t)=(t,t3,et),0t1.c(t)=(t,t3,et),0t1.

141.

[T] Let c:[1,2]2c:[1,2]2 be given by x=et1,y=sin(πt).x=et1,y=sin(πt). Use a computer to compute the integral CF·ds=C2xcosydxx2sinydy,CF·ds=C2xcosydxx2sinydy, where F=(2xcosy)i(x2siny)j.F=(2xcosy)i(x2siny)j.

142.

[T] Use a computer algebra system to find the mass of a wire that lies along curve r(t)=(t21)j+2tk,0t1,r(t)=(t21)j+2tk,0t1, if the density is 32t.32t.

143.

Find the circulation and flux of field F=yi+xjF=yi+xj around and across the closed semicircular path that consists of semicircular arch r1(t)=(acost)i+(asint)j,0tπ,r1(t)=(acost)i+(asint)j,0tπ, followed by line segment r2(t)=ti,ata.r2(t)=ti,ata.

A vector field in two dimensions. The arrows are shorter the closer they are to the origin. They surround the origin in a counterclockwise radial pattern. The upper half of a circle with radius 2 and center at the origin is drawn. (-2,0) and (2,0) are labeled as –a and a, respectively, and the curve is labeled r_1.
144.

Compute Ccosxcosydxsinxsinydy,Ccosxcosydxsinxsinydy, where c(t)=(t,t2),0t1.c(t)=(t,t2),0t1.

145.

Complete the proof of The Path Independence Test for Conservative Fields by showing that fy=Q(x,y).fy=Q(x,y).

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