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**16.2** Line Integrals

### 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.

3 3

### 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.]**

### Line Integrals

*If we divide the parameter interval [a, b] into n subintervals*
*[t*_{i –1}*, t*_{i}*] of equal width and we let x*_{i}*= x(t*_{i}*), and y*_{i}*= y(t** _{i}*), then

*the corresponding points P*

_{i}*(x*

_{i}*, y*

_{i}*) divide C into n subarcs*

with lengths ∆s_{1}, ∆s_{2}, . . . , ∆s* _{n}*. (See Figure 1.)

**Figure 1**

5 5

### Line Integrals

*We choose any point in the ith subarc. (This *
*corresponds to a point in [t*_{i –1}*, t** _{i}*].)

*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 ∆s* _{i}* of the subarc, and form the sum

which is similar to a Riemann sum.

### 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*

7 7

### 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.*

### 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*

9 9

### 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.

### 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**

11 11

### Example 1

Evaluate

### ∫

_{C }*(2 + x*

^{2}

*y) ds, where C is the upper half of the*

*unit circle x*

^{2}

*+ y*

^{2}= 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**

*Example 1 – Solution*

Therefore Formula 3 gives

cont’d

13 13

### Line Integrals

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

*C is a union of a finite number of smooth curves *
*C*_{1}*, C*_{2}*, …., C*_{n}*, where, as illustrated in Figure 4, the initial *

*point of C*_{i + 1 }*is the terminal point of C** _{i}*.

**Figure 4**

A piecewise-smooth curve

### 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:*

15 15

### 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 P*_{i – 1}*to P** _{i }*in

Figure1 is approximately and so the total

mass of the wire is approximately

**Figure 1**

### 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 + x*^{2}*y represents the density of a *
semicircular wire, then the integral in Example 1 would

represent the mass of the wire.]

17 17

### Line Integrals

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

### Line Integrals

Two other line integrals are obtained by replacing ∆s* _{i}* by
either ∆x

_{i }*= x*

_{i}*– x*

*or ∆y*

_{i–1 }

_{i }*= y*

_{i}*– y*

*in Definition 2.*

_{i –1 }**They are called the line integrals of f along C with ****respect to x and y:**

19 19

### 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.*

### 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 r**_{0} **and ends at r**_{1} is given by

### 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 ∆s* _{i }*is always positive, whereas ∆x

*and ∆y*

_{i }

_{i }*change sign when we reverse the orientation of C.*

### Line Integrals in Space

25 25

### 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:

### 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.*

### Example 5

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 ≤ 2*π.

(See Figure 9.)

**Figure 9**

29 29

*Example 5 – Solution*

Formula 9 gives

### Line Integrals of Vector Fields

31 31

### 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. *

### Line Integrals of Vector Fields

*We divide C into subarcs P*_{i –1}*P** _{i}* with lengths ∆s

*by dividing*

_{i}*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 ∆s_{i}*is small, then as the particle moves from P*_{i –1}*to P*_{i}

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

### Line Integrals of Vector Fields

**Thus the work done by the force F in moving the particle **
*from P*_{i – 1}*to P** _{i}* 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.*

35 35

### 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.*

### 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}^{}

*other areas of physics as well.*

**dr and occurs in**37 37

### 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.**

### Example 7

**Find the work done by the force field F(x, y) = x**^{2 }* 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)) = cos**^{2}**t 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

### 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**