5.2 Line Integrals

5.2. LINE INTEGRALS

265

5.2 Line Integrals

5.2.1 Introduction

Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

1.

De...nite integral.

Given

a

continuous

real-valued

function

f,

Rb

a

f

(x) dx

represents the area below the graph of f , between x = a and x = b,

assuming that f (x) 0 between x = a and x = b.

2. The de...nite integral can also be used to compute the length of a curve. If a curve C is given by its position vector !r (t) = hx (t) ; y (t)i in 2-D or !r (t) = hx (t) ; y (t) ; z (t)i in 3-D for a t b, then the length L of the

curve C is given by

L = Z b j!r 0 (u)j du

a

The arc length function was de...ned to be

s (t) = Z t j!r 0 (u)j du

a

so that, using the fundamental theorem of Calculus, we have

ds = j!r 0 (t)j dt

ZZ 3. Double integrals. Given a real-valued function f of two variables, f (x; y) dA

D

represents the volume of the solid above the region D and below the graph of z = f (x; y).

4. The double integral can also be used to ...nd the area of a region by the

formula

ZZ

area of D = dA

D

In this section, we study an integral similar to the one in example 1, except that instead of integrating over an interval, we integrate along a curve.

5.2.2 Line Integrals Along Plane Curves

Let us consider the following problem: Suppose that we have a plane curve C given by its position vector

!r (t) = hx (t) ; y (t)i for a t b:

(5.1)

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CHAPTER 5. VECTOR CALCULUS

Let us assume C is a smooth curve (!r 0 is continuous and !r 0 (t) 6= !0 ). Suppose further that we have a continuous function z = f (x; y), we will assume for now f (x; y) 0. Consider the surface S given by hx (t) ; y (t) ; zi. This surface will intersect the graph of z = f (x; y) in a curve C0. We wish to ...nd the surface area of S between the curves C and C0. To help you visualize S, think of a curtain hanging. Except that the curtain in not hanging along a straight rail, but a curved one. Furthermore, the rail is not necessarily horizontal, it has whatever shape f has. This is shown in ...gure 5.2.2. Imagine we want to ...nd the area of the curtain.We will ...rst approximate the area using a technique similar to the one used when de...ning the de...nite integral. We will outline the steps.

1. Split [a; b] into n subintervals [ti 1; ti] of equal length. Let xi = x (ti), yi = y (ti).

2. The corresponding points Pi (xi; yi) divide C into n subarcs of length s1; s2; :::; sn.

3. In each subarc, pick a point Pi (xi ; yi ) (this corresponds to a point ti in [ti 1; ti].

4. Draw the rectangle with base [Pi 1; Pi] and height f (xi ; yi ). The area of this rectangle is f (xi ; yi ) si:

Xn 5. The area of S can be approximated by f (xi ; yi ) si.

i=1

6. The larger n is, the better the approximation.

5.2. LINE INTEGRALS

267

This allows us to de...ne:

De...nition 388 With the notation above, the area of S, denoted A (S) is de...ned to be

Xn

A (S) = lim

n!1

f (xi ; yi ) si

i=1

De...nition 389 If f is any continuous function (not just a positive one), de...ned on a smooth curve C given in equation 5.1, then the line integral of f along C is de...ned by

Z

Xn

C

f

(x;

y)

ds

=

lim

n!1

i=1

f

(xi

;

yi

)

si

(5.2)

if this limit exists.

You will note that we are integrating with respect to arc length. Remem-

bering

that

ds dt

=

j!r 0 (t)j,

it

follows

that

ds

=

j!r 0 (t)j dt

and

therefore,

the

line

integral can be evaluated as follows:

Theorem 390 If f is any continuous function (not just a positive one), de...ned on a smooth curve C given in equation 5.1, then the line integral of f along C can be computed by the following formula

Z f (x; y) ds = Z b f (x (t) ; y (t)) j!r 0 (t)j dt

C

a

Zb

s dx 2

dy 2

= f (x (t) ; y (t))

+

dt

a

dt

dt

(5.3)

Remark 391 We used a and b for the limits of integration because they are the limits of the variable t.

Remark 392 Note that the line integral is with respect to arc length. However, to compute it, we use the parametrization of the curve, whatever it is. We rewrite everything in terms of the parameter used for the curve.

R

Example 393 Evaluate circle x2 + y2 = 1.

C

2 + x2y

ds where C is the upper half of the unit

First, we must write C in parametric form. The upper half of the unit circle is

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CHAPTER 5. VECTOR CALCULUS

Figure 5.5: Piecewise smooth curve

x (t) = cos y, y (t) = sin t, 0 t . Then

Z

Zb

f (x; y) ds =

f (x (t) ; y (t)) j!r 0 (t)j dt

C

Za

p

=

2 + cos2 t sin t sin2 t + cos2 tdt

Z0

=

2 + cos2 t sin t dt

Z0

Z

=

2dt + cos2 t sin tdt

0

0

cos3 t = 2t

30

11 = 2+ +

33

1 = 2 +3

R Remark 394 In the above theorem, the given formula to ...nd C f (x; y) ds requires that C be a smooth curve. However, It is still possible to compute a

line integral when the curve C is not a smooth curve, as long as it is piecewise

smooth, that is made of smooth pieces, as the one shown in ...gure 5.5. In this

case

Z

Z

Z

Z

Z

f (x; y) ds = f (x; y) ds + f (x; y) ds + f (x; y) ds + f (x; y) ds

C

C1

C2

C3

C4

5.2. LINE INTEGRALS

269

R Example 395 Evaluate C 2xds where C = C1 [ C2, C1 being the arc of the parabola y = x2 between (0; 0) and (1; 1) and C2 being the vertical line from

(1; 1) to (1; 2).

Z

Z

Z

2xds = 2xds + 2xds

C

C1

C2

We evaluate each integral separately.

R 1. To evaluate C1 2xds;we need to parametrize C1. y = x2 can be parame-

trized by x = t, y = t2, 0 t 1. Thus

Z

Z1 p

2xds =

2t 1 + 4t2dt

C1

0p

55 1

=

6

R 2. To evaluate C2 2xds;we need to parametrize C2. A vertical line between

the given points can be parametrized by x = 1, y = t, 1 t 2. Thus

Z

Z2 p

2xds =

2 1dt

C2

1

=2

3. Therefore

Z

p

55 1

2xds =

+2

C

6

Two other integrals can be obtained using a similar technique. When we form the sum, we can use xi = xi = xi 1 or yi = yi yi 1 instead of si. The integrals we obtain are

Z

Xn

f (x; y) dx

C

=

lim

n!1

f (xi ; yi )

i=1

xi

Z

Xn

f (x; y) dy

C

=

lim

n!1

f (xi ; yi )

i=1

yi

(5.4) (5.5)

These are called line integrals of f along C with respect to x and y. If x = x (t), then dx = x0 (t) dt. Similarly, dy = y0 (t) dt. So, these integrals can be computed

as follows:

Z

Zb

f (x; y) dx =

f (x (t) ; y (t)) x0 (t) dt

ZC

Zab

f (x; y) dy =

f (x (t) ; y (t)) y0 (t) dt

C

a

(5.6)

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