INTRODUCTION

The notion of the definite integral ; that is, integration of a function defined over an interval, can be generalized to integration of a function defined along a curve. To this end we need to introduce some terminology about curves.

Terminology

Suppose C is a curve parameterized by x = f(t), y = g(t), a ≤ t ≤ b, and A and B are the points (f(a), g(a)) and (f(b), g(b)), respectively. We say that

(i) C is a smooth curve if f′ and g′ are continuous on the closed interval [a, b] and not simultaneously zero on the open interval (a, b).

(ii) C is piecewise smooth if it consists of a finite number of smooth curves C₁, C₂, ..., C_n joined end to end—that is, C = C₁, ∪ C₂, ∪ ... ∪ C_n.

(iii) C is a closed curve if A = B.

(iv) C is a simple closed curve if A = B and the curve does not cross itself.

(v) If C is not a closed curve, then the positive direction on C is the direction corresponding to increasing values of t.

FIGURE 9.8.1 illustrates each type of curve defined in (i)–(iv).

This same terminology carries over in a natural manner to curves in space. For example, a curve C defined by x = f(t), y = g(t), z = h(t), a ≤ t ≤ b, is smooth if f′, g′, and h′ are continuous on [a, b] and not simultaneously zero on (a, b).

Definite Integral

Before defining integration along a curve, let us review the five steps leading to the definition of the definite integral.

1. Let y = f(x) be defined on a closed interval [a, b].
2. Partition the interval [a, b] into n subintervals [x_k−1, x_k] of lengths Δx_k = x_k – x_k−1. Let P denote the partition shown in FIGURE 9.8.2(a).
3. Let P be the length of the longest subinterval. The number P is called the norm of the partition P.
4. Choose a sample point in each subinterval. See Figure 9.8.2(b).
5. Form the sum

The definite integral of a function of a single variable is given by the limit of a sum:

Line Integrals in the Plane

The following analogous five steps lead to the definitions of three line integrals^* in the plane.

1. Let z = G(x, y) be defined in some region that contains the smooth curve C defined by x = f(t), y = g(t), a ≤ t ≤ b.
2. Divide C into n subarcs of lengths Δs_k according to the partition a = t₀ < t₁ < t₂ < ... < t_n = b of [a, b]. Let the projection of each subarc onto the x- and y-axes have lengths Δx_k and Δy_k, respectively.
3. Let P be the norm of the partition or the length of the longest subarc.
4. Choose a sample point (, ) on each subarc. See FIGURE 9.8.3.
5. Form the sums
   

   

It can be proved that if G(x, y) is continuous on C, then the integrals defined in (i), (ii), and (iii) exist. We shall assume continuity of G as a matter of course.

Method of Evaluation—Curve Defined Parametrically

The line integrals in Definition 9.8.1 can be evaluated in two ways, depending on whether the curve C is defined parametrically or by an explicit function. In either case the basic idea is to convert the line integral to a definite integral in a single variable. If C is a smooth curve parameterized by x = f(t), y = g(t), a ≤ t ≤ b, then we simply replace x and y in the integral by the functions f(t) and g(t), and the appropriate differential dx, dy, or ds by f′(t) dt, g′(t) dt, or dt. The expression ds = dt is called the differential of arc length. The integration is carried out with respect to the variable t in the usual manner:

(1)

(2)

. (3)

EXAMPLE 1 Evaluation of Line Integrals

Evaluate (a) ∫_C xy² dx, (b) ∫_C xy² dy, and (c) ∫_C xy² ds on the quarter-circle C defined by x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ π/2. See FIGURE 9.8.4.

SOLUTION

(a) From (1),

(b) From (2),

(c) From (3),

≡

Method of Evaluation—Curve Defined by an Explicit Function

If the curve C is defined by an explicit function y = f(x), a ≤ x ≤ b, we can use x as a parameter. With dy = f′(x) dx and ds = dx, the foregoing line integrals become, in turn,

(4)

(5)

(6)

A line integral along a piecewise-smooth curve C is defined as the sum of the integrals over the various smooth curves whose union comprises C. For example, if C is composed of smooth curves C₁ and C₂, then

Notation

In many applications, line integrals appear as a sum

It is common practice to write this sum as one integral without parentheses as

(7)

A line integral along a closed curve C is very often denoted by

EXAMPLE 2 Curve Defined by an Explicit Function

Evaluate ∫_C xy dx + x² dy, where C is given by y = x³, –1 ≤ x ≤ 2.

SOLUTION

The curve C is illustrated in FIGURE 9.8.5 and is defined by the explicit function y = x³. Hence we can use x as the parameter. Using dy = 3x² dx, we get

EXAMPLE 3 Curve Defined Parametrically

EXAMPLE 4 Closed Curve

Evaluate y² dx – x² dy on the closed curve C that is shown in FIGURE 9.8.6(a).

SOLUTION

Since C is piecewise smooth, we express the integral as a sum of integrals. Symbolically, we write

where C₁, C₂, and C₃ are the curves shown in Figure 9.8.6(b). On C₁, we use x as a parameter. Since y = 0, dy = 0; therefore,

On C₂, we use y as a parameter. From x = 2, dx = 0, we have

Finally, on C₃, we again use x as a parameter. From y = x², we get dy = 2x dx and so

Hence,

≡

It is important to be aware that a line integral is independent of the parameterization of the curve C provided C is given the same orientation by all sets of parametric equations defining the curve. See Problem 37 in Exercises 9.8. Also, recall for definite integrals that . Line integrals possess a similar property. Suppose, as shown in FIGURE 9.8.7, that –C denotes the curve having the opposite orientation of C. Then it can be shown that

or equivalently,

(8)

For example, in part (a) of Example 1, ∫_−C xy² dx = 64.

Line Integrals in Space

Line integrals of a function G of three variables, ∫_C G(x, y, z) dx, ∫_C G(x, y, z) dy, and ∫_C G(x, y, z) ds, are defined in a manner analogous to Definition 9.8.1. However, to that list we add a fourth line integral along a space curve C with respect to z:

. (9)

Method of Evaluation

If C is a smooth curve in 3-space defined by the parametric equations x = f(t), y = g(t), z = h(t), a ≤ t ≤ b, then the integral in (9) can be evaluated by using

.

The integrals ∫_C G(x, y, z) dx and ∫_C G(x, y, z) dy are evaluated in a similar fashion. The line integral with respect to arc length is

.

As in (7), in 3-space we are often concerned with line integrals in the form of a sum:

EXAMPLE 5 Line Integral on a Curve in 3-Space

Evaluate ∫_C y dx + x dy + z dz, where C is the helix x = 2 cos t, y = 2 sin t, z = t, 0 ≤ t ≤ 2π.

SOLUTION

Substituting the expressions for x, y, and z along with dx = –2 sin t dt,dy = 2 cos t dt, dz = dt, we get

We can use the concept of a vector function of several variables to write a general line integral in a compact fashion. For example, suppose the vector-valued function F(x, y) = P(x, y)i + Q(x, y)j is defined along a curve C: x = f(t), y = g(t), a ≤ t ≤ b, and suppose r(t) = f(t)i + g(t)j is the position vector of points on C. Then the derivative of r(t),

,

prompts us to define dr = dt = dxi + dyj. Since F(x, y) · dr = P(x, y) dx + Q(x, y) dy we can write

(10)

Similarly, for a line integral on a space curve,

(11)

where F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k and dr = dxi + dyj + dzk.

Work

In Section 7.3 we saw that the work W done by a constant force F that causes a straight-line displacement d of an object is W = F · d. In beginning courses in calculus or physics it is then shown that the work done in moving an object from x = a to x = b by a force F(x), which varies in magnitude but not in direction, is given by the definite integral . In general, a force field F(x, y) = P(x, y)i + Q(x, y)j acting at each point on a smooth curve C: x = f(t), y = g(t), a ≤ t ≤ b, varies in both magnitude and direction. See FIGURE 9.8.8(a). If A and B are the points (f(a), g(a)) and (f(b), g(b)), respectively, we ask: What is the work done by F as its point of application moves along C from A to B ? To answer this question, suppose C is divided into n subarcs of lengths Δs_k. On each subarc F is a constant force. If, as shown in Figure 9.8.8(b), the length of the vector Δr_k = (x_k – x_k−1)i + (y_k – y_k−1)j = Δx_k i + Δy_k j is an approximation to the length of the kth subarc, then the approximate work done by F over the subarc is

By summing these elements of work and passing to the limit, we naturally define the work done by F along C as the line integral

. (12)

Of course, (12) extends to force fields acting at points on a space curve. In this case, work ∫_C F · dr is defined as in (11).

Now, since

we let dr = T ds, where T = dr/ds is a unit tangent to C. Hence,

(13)

In other words:

EXAMPLE 6 Work Done by a Force

Find the work done by (a) F = xi + yj and (b) F = i + j along the curve C traced by r(t) = cos ti + sin tj from t = 0 to t = π.

SOLUTION

(a) The vector function r(t) gives the parametric equations x = cos t, y = sin t, 0 ≤ t ≤ π, which we recognize as a half-circle. As seen in FIGURE 9.8.9, the force field F is perpendicular to C at every point. Since the tangential components of F are zero, the work done along C is zero. To see this we use (12):

(b) In FIGURE 9.8.10 the vectors in red are the projections of F on the unit tangent vectors. The work done by F is

The units of work depend on the units of F and on the units of distance. ≡

Circulation

A line integral of a vector field F around a simple closed curve C is said to be the circulation of F around C; that is,

In particular, if F is the velocity field of a fluid, then the circulation is a measure of the amount by which the fluid tends to turn the curve C by rotating, or circulating, around it. For example, if F is perpendicular to T for every (x, y) on C, then ∫_C F · T ds = 0, and the curve does not move at all. On the other hand, ∫_C F · T ds > 0 and ∫_C F · T ds < 0 mean that the fluid tends to rotate C in the counterclockwise and clockwise directions, respectively. See FIGURE 9.8.11.

REMARKS

In the case of two variables, the line integral with respect to arc length ∫_C G(x, y) ds can be interpreted in a geometric manner when G(x, y) ≥ 0 on C. In Definition 9.8.1 the symbol Δs_k represents the length of the kth subarc on the curve C. But from Figure 9.8.3 accompanying that definition, we have the approximation Δs_k = . With this interpretation of Δs_k we see from FIGURE 9.8.12(a) that the product G Δs_k is the area of a vertical rectangle of height G and width Δs_k. The integral ∫_C G(x, y) ds then represents the area of one side of a “fence” or “curtain” extending from the curve C in the xy-plane up to the graph of G(x, y) that corresponds to points (x, y) on C. See Figure 9.8.12(b).

9.8 Exercises Answers to selected odd-numbered problems begin on page ANS-25.

In Problems 1–4, evaluate ∫_C G(x, y) dx, ∫_C G(x, y) dy, and ∫_C G(x, y) ds on the indicated curve C.

1. G(x, y) = 2xy; x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ π/4
2. G(x, y) = x³ + 2xy² + 2x; x = 2t, y = t², 0 ≤ t ≤ 1
3. G(x, y) = 3x² + 6y²; y = 2x + 1, –1 ≤ x ≤ 0
4. G(x, y) = x²/y³; 2y = 3x^2/3, 1 ≤ x ≤ 8

In Problems 5 and 6, evaluate ∫_C G(x, y, z) dx, ∫_C G(x, y, z) dy, ∫_C G(x, y, z) dz, and ∫_C G(x, y, z) ds on the indicated curve C.

5. G(x, y, z) = z; x = cos t, y = sin t, z = t, 0 ≤ t ≤ π/2
6. G(x, y, z) = 4xyz; x = t³, y = t², z = 2t, 0 ≤ t ≤ 1

In Problems 7–10, evaluate ∫_C (2x + y) dx + xy dy on the given curve C between (–1, 2) and (2, 5).

7. y = x + 3
8. y = x² + 1

9. 

10. 

In Problems 11–14, evaluate ∫_C y dx + x dy on the given curve C between (0, 0) and (1, 1).

11. y = x²
12. y = x
13. C consists of the line segments from (0, 0) to (0, 1) and from (0, 1) to (1, 1).
14. C consists of the line segments from (0, 0) to (1, 0) and from (1, 0) to (1, 1).
15. Evaluate ∫_C (6x² + 2y²) dx + 4xy dy, where C is given by x = , y = t, 4 ≤ t ≤ 9.
16. Evaluate ∫_C –y² dx + xy dy, where C is given by x = 2t, y = t³, 0 ≤ t ≤ 2.
17. Evaluate ∫_C 2x³y dx + (3x + y) dy, where C is given by x = y² from (1, –1) to (1, 1).
18. Evaluate ∫_C 4x dx + 2y dy, where C is given by x = y³ + 1 from (0, –1) to (9, 2).

In Problems 19 and 20, evaluate ∮_C (x² + y²) dx – 2xy dy on the given closed curve C.

In Problems 21 and 22, evaluate ∮_C x²y³ dx – xy² dy on the given closed curve C.

21. 

22. 

23. Evaluate ∮_C (x² – y²) ds, where C is given by
    

    x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ 2π.

24. Evaluate ∫_−C y dx – x dy, where C is given by
    

    x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ π.

In Problems 25–28, evaluate ∫_C y dx + z dy + x dz on the given curve C between (0, 0, 0) and (6, 8, 5).

25. C consists of the line segments from (0, 0, 0) to (2, 3, 4) and from (2, 3, 4) to (6, 8, 5).
26. x = 3t, y = t³, z = t², 0 ≤ t ≤ 2

27. 

28. 

In Problems 29 and 30, evaluate ∫_C F · dr.

29. F(x, y) = y³i – x²yj; r(t) = e^−2ti + e^tj, 0 ≤ t ≤ ln 2
30. F(x, y, z) = e^xi + xe^xyj + xye^xyzk; r(t) = ti + t²j + t³k, 0 ≤ t ≤ 1
31. Find the work done by the force F(x, y) = yi + xj acting along y = ln x from (1, 0) to (e, 1).
32. Find the work done by the force F(x, y) = 2xyi + 4y²j acting along the piecewise-smooth curve consisting of the line segments from (–2, 2) to (0, 0) and from (0, 0) to (2, 3).
33. Find the work done by the force F(x, y) = (x + 2y)i + (6y – 2x)j acting counterclockwise once around the triangle with vertices (1, 1), (3, 1), and (3, 2).
34. Find the work done by the force F(x, y, z) = yzi + xzj + xyk acting along the curve given by r(t) = t³i + t²j + tk from t = 1 to t = 3.
35. Find the work done by a constant force F(x, y) = ai + bj acting counterclockwise once around the circle x² + y² = 9.
36. In an inverse square force field F = cr/r³, where c is a constant and r = xi + yj + zk,^* find the work done in moving a particle along the line from (1, 1, 1) to (3, 3, 3).
37. Verify that the line integral ∫_C y² dx + xy dy has the same value on C for each of the following parameterizations:
    
38. Consider the three curves between (0, 0) and (2, 4):
    
    

    Show that ∫_C1 xy ds = ∫_C3 xy ds, but ∫_C1 xy ds ≠ ∫_C2 xy ds. Explain.

39. Assume a smooth curve C is described by the vector function r(t) for a ≤ t ≤ b. Let acceleration, velocity, and speed be given by a = dv/dt,v = dr/dt, and v = v, respectively. Using Newton’s second law F = ma, show that, in the absence of friction, the work done by F in moving a particle of constant mass m from point A at t = a to point B at t = b is the same as the change in kinetic energy:
    

    

    [Hint: Consider .]

40. If ρ(x, y) is the density of a wire (mass per unit length), then m = ∫_C ρ(x, y) ds is the mass of the wire. Find the mass of a wire having the shape of the semicircle x = 1 + cos t, y = sin t, 0 ≤ t ≤ π, if the density at a point P is directly proportional to distance from the y-axis.
41. The coordinates of the center of mass of a wire with variable density are given by = M_y/m, = M_x/m, where
    

    

    and

    

    Find the center of mass of the wire in Problem 40.

42. A force field F(x, y) acts at each point on the curve C, which is the union of C₁, C₂, and C₃ shown in red in FIGURE 9.8.21. is measured in pounds and distance is measured in feet using the scale given in the figure. Use the representative vectors shown to approximate the work done by F along C. [Hint: Use W = ∫_C F · T ds.]
    

    

^*An unfortunate choice of names. Curve integrals would be more appropriate.

^*Note that the magnitude of F is inversely proportional to r².