# 16.2 Line Integrals

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## 16.2 Line Integrals

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### Line Integrals

In this section we define an integral that is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C.

Such integrals are called line integrals, although “curve integrals” would be better terminology.

They were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism.

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### Line Integrals

We start with a plane curve C given by the parametric equations

x = x(t) y = y(t) a ≤ t ≤ b

or, equivalently, by the vector equation r(t) = x(t) i + y(t) j, and we assume that C is a smooth curve. [This means that r′ is continuous and r′(t) ≠ 0.]

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### Line Integrals

If we divide the parameter interval [a, b] into n subintervals [ti –1, ti] of equal width and we let xi = x(ti), and yi = y(ti), then the corresponding points Pi(xi, yi) divide C into n subarcs

with lengths ∆s1, ∆s2, . . . , ∆sn. (See Figure 1.)

Figure 1

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### Line Integrals

We choose any point in the ith subarc. (This corresponds to a point in [ti –1, ti].)

Now if f is any function of two variables whose domain includes the curve C, we evaluate f at the point

multiply by the length ∆si of the subarc, and form the sum

which is similar to a Riemann sum.

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### Line Integrals

Then we take the limit of these sums and make the following definition by analogy with a single integral.

We have found that the length of C is

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### Line Integrals

A similar type of argument can be used to show that if f is a continuous function, then the limit in Definition 2 always

exists and the following formula can be used to evaluate the line integral:

The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b.

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### Line Integrals

If s(t) is the length of C between r(a) and r(t), then

So the way to remember Formula 3 is to express

everything in terms of the parameter t: Use the parametric equations to express x and y in terms of t and write ds as

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### Line Integrals

In the special case where C is the line segment that joins (a, 0) to (b, 0), using x as the parameter, we can write the parametric equations of C as follows: x = x, y = 0, a ≤ x ≤ b.

Formula 3 then becomes

and so the line integral reduces to an ordinary single integral in this case.

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### Line Integrals

Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area.

In fact, if f(x, y) ≥ 0,

### ∫

C f(x, y) ds represents the area of one side of the “fence” or “curtain” in Figure 2, whose base is C and whose height above the point (x, y) is f(x, y).

Figure 2

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Evaluate

### ∫

C (2 + x2y) ds, where C is the upper half of the unit circle x2 + y2 = 1.

Solution:

In order to use Formula 3, we first need parametric equations to represent C.

Recall that the unit circle can be parametrized by means of the equations

x = cos t y = sin t

and the upper half of the circle is described by the parameter

interval 0 ≤ t ≤ π. (See Figure 3.)

Figure 3

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### Example 1 – Solution

Therefore Formula 3 gives

cont’d

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### Line Integrals

Suppose now that C is a piecewise-smooth curve; that is,

C is a union of a finite number of smooth curves C1, C2, …., Cn, where, as illustrated in Figure 4, the initial

point of Ci + 1 is the terminal point of Ci.

Figure 4

A piecewise-smooth curve

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### Line Integrals

Then we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C:

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### Line Integrals

Any physical interpretation of a line integral

### ∫

C f(x, y) ds depends on the physical interpretation of the function f.

Suppose that ρ(x, y) represents the linear density at a point (x, y) of a thin wire shaped like a curve C.

Then the mass of the part of the wire from Pi – 1 to Pi in

Figure1 is approximately and so the total

mass of the wire is approximately

Figure 1

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### Line Integrals

By taking more and more points on the curve, we obtain the mass m of the wire as the limiting value of these approximations:

[For example, if f(x, y) = 2 + x2y represents the density of a semicircular wire, then the integral in Example 1 would

represent the mass of the wire.]

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### Line Integrals

The center of mass of the wire with density function ρ is located at the point , where

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### Line Integrals

Two other line integrals are obtained by replacing ∆si by either ∆xi = xi – xi–1 or ∆yi = yi – yi –1 in Definition 2.

They are called the line integrals of f along C with respect to x and y:

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### Line Integrals

When we want to distinguish the original line integral

### ∫

C f(x, y) ds from those in Equations 5 and 6, we call it the line integral with respect to arc length.

The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t: x = x(t), y = y(t), dx = x′(t) dt, dy = y′(t) dt.

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### Line Integrals

It frequently happens that line integrals with respect to x and y occur together.

When this happens, it’s customary to abbreviate by writing

C P(x, y) dx +

C Q(x, y) dy =

### ∫

C P(x, y) dx + Q(x, y) dy

When we are setting up a line integral, sometimes the most difficult thing is to think of a parametric representation for a curve whose geometric description is given.

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### Line Integrals

In particular, we often need to parametrize a line segment, so it’s useful to remember that a vector representation of the line segment that starts at r0 and ends at r1 is given by

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### Line Integrals

In general, a given parametrization x = x(t), y = y(t),

a ≤ t ≤ b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the parameter t. (See Figure 8, where the initial point A

corresponds to the parameter value a and the terminal point B corresponds to t = b.)

Figure 8

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### Line Integrals

If –C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to

terminal point A in Figure 8), then we have

### ∫

–C f(x, y) dx = –

C f(x, y) dx

### ∫

–C f(x, y) dy = –

### ∫

C f(x, y) dy

But if we integrate with respect to arc length, the value of the line integral does not change when we reverse the orientation of the curve:

–C f(x, y) ds =

### ∫

C f(x, y) ds

This is because ∆si is always positive, whereas ∆xi and ∆yi change sign when we reverse the orientation of C.

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### Line Integrals in Space

We now suppose that C is a smooth space curve given by the parametric equations

x = x(t) y = y(t) z = z(t) a ≤ t ≤ b or by a vector equation r(t) = x(t) i + y(t) j + z(t) k.

If f is a function of three variables that is continuous on

some region containing C, then we define the line integral of f along C (with respect to arc length) in a manner similar to that for plane curves:

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### Line Integrals in Space

We evaluate it using a formula similar to Formula 3:

Observe that the integrals in both Formulas 3 and 9 can be written in the more compact vector notation

For the special case f(x, y, z) = 1, we get

where L is the length of the curve C.

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### Line Integrals in Space

Line integrals along C with respect to x, y, and z can also be defined. For example,

Therefore, as with line integrals in the plane, we evaluate integrals of the form

by expressing everything (x, y, z, dx, dy, dz) in terms of the parameter t.

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Evaluate

### ∫

C y sin z ds, where C is the circular helix given by the equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.

(See Figure 9.)

Figure 9

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Formula 9 gives

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### Line Integrals of Vector Fields

We know that the work done by a variable force f(x) in moving a particle from a to b along the x-axis is

Then we have found that the work done by a constant force F in moving an object from a point P to another point Q in space is W = F D, where D = PQ is the displacement

vector.

Now suppose that F = P i + Q j + R k is a continuous force field on . (A force field on could be regarded as a

special case where R = 0 and P and Q depend only on x and y.)

We wish to compute the work done by this force in moving a particle along a smooth curve C.

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### Line Integrals of Vector Fields

We divide C into subarcs Pi –1Pi with lengths ∆si by dividing the parameter interval [a, b] into subintervals of equal width.

(See Figure 1 for the two-dimensional case or Figure 11 for the three-dimensional case.)

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### Line Integrals of Vector Fields

Choose a point on the ith subarc corresponding to the parameter value

If ∆si is small, then as the particle moves from Pi –1 to Pi

along the curve, it proceeds approximately in the direction of the unit tangent vector at

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### Line Integrals of Vector Fields

Thus the work done by the force F in moving the particle from Pi – 1 to Pi is approximately

and the total work done in moving the particle along C is approximately

where T(x, y, z) is the unit tangent vector at the point (x, y, z) on C.

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### Line Integrals of Vector Fields

Intuitively, we see that these approximations ought to become better as n becomes larger.

Therefore we define the work W done by the force field F as the limit of the Riemann sums in (11), namely,

Equation 12 says that work is the line integral with respect to arc length of the tangential component of the force.

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### Line Integrals of Vector Fields

If the curve C is given by the vector equation r(t) = x(t) i + y(t) j + z(t) k,

then T(t) = r′(t)/|r′(t)|, so using Equation 9 we can rewrite Equation 12 in the form

This integral is often abbreviated as

### ∫

C F dr and occurs in other areas of physics as well.

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### Line Integrals of Vector Fields

Therefore we make the following definition for the line integral of any continuous vector field.

When using Definition 13, remember that F(r(t)) is just an abbreviation for the vector field F(x(t), y(t), z(t)), so we

evaluate F(r(t)) simply by putting x = x(t), y = y(t), and z = z(t) in the expression for F(x, y, z).

Notice also that we can formally write dr = r′(t) dt.

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### Example 7

Find the work done by the force field F(x, y) = x2 i – xy j in moving a particle along the quarter-circle

r(t) = cos t i + sin t j, 0 ≤ t ≤ π/2.

Solution:

Since x = cos t and y = sin t, we have

F(r(t)) = cos2t i – cos t sin t j and r′(t) = –sin t i + cos t j

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### Example 7 – Solution

Therefore the work done is

cont’d

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### Line Integrals of Vector Fields

Finally, we note the connection between line integrals of vector fields and line integrals of scalar fields. Suppose the vector field F on is given in component form by the

equation F = P i + Q j + R k.

We use Definition 13 to compute its line integral along C:

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### Line Integrals of Vector Fields

But this last integral is precisely the line integral in (10).

Therefore we have

For example, the integral

### ∫

C y dx + z dy + x dz could be expressed as

### ∫

C F dr where

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

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