Thomas Calculus !6th Chapter

8/9/2019 Thomas Calculus !6th Chapter

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OVERVIEWIn this chapter we extend the theory of integration to curves and surfaces inspace. The resulting theory of line and surface integrals gives powerful mathematical toolsfor science and engineering. Line integrals are used to find the work done by a force inmoving an object along a path, and to find the mass of a curved wire with variable density.Surface integrals are used to find the rate of flow of a fluid across a surface. We presentthe fundamental theorems of vector integral calculus, and discuss their mathematical con-sequences and physical applications. In the final analysis, the key theorems are shown asgeneralized interpretations of the Fundamental Theorem of Calculus.

16.1 Line IntegralsTo calculate the total mass of a wire lying along a curve in space, or to f ind the work done

by a variable force acting along such a curve, we need a more general notion of integralthan was defined in Chapter 5. We need to integrate over a curve C rather than over an in-terval [ a , b]. These more general integrals are called line integrals (although path integralsmight be more descriptive). We make our definitions for space curves, with curves in the

xy-plane being the special case with z -coordinate identically zero.Suppose that ( x, y, z ) is a real-valued function we wish to integrate over the curve C

lying within the domain of and parametrized by .The values of along the curve are given by the composite function ( g (t ), h(t ), k (t )). Weare going to integrate this composite with respect to arc length from to To be-gin, we first partition the curve C into a finite number n of subarcs (Figure 16.1). The typ-ical subarc has length In each subarc we choose a point and form the sum

which is similar to a Riemann sum. Depending on how we partition the curve C and pick

in the k th subarc, we may get different values for S n. If is continuous and thefunctions g , h, and k have continuous first derivatives, then these sums approach a limit asn increases and the lengths approach zero. This limit gives the following definition,similar to that for a single integral. In the def inition, we assume that the partition satisfies

as n : q . sk : 0

sk

s xk , yk , z k d

S n = ank = 1

s xk , yk , z k d sk ,

s xk , yk , z k d sk .

t = b.t = a

r st d = g st di + hst d j + k st dk , a t b

919

16I NTEGRATION INV ECTOR F IELDS

DEFINITION If is defined on a curve C given parametrically bythen the line integral of over C is

(1)

provided this limit exists.

LC s x, y, z d ds = limn : q an

k = 1s xk , yk , z k d sk ,

g (t )i + h(t ) j + k (t )k , a t b,r (t ) =

z

y

x

r ( t )

t 5 b

t 5 a( x k , yk , zk )

D sk

FIGURE 16.1 The curve r (t ) partitioned into small arcs from to Thelength of a typical subarc is sk .

t = b.t = a


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If the curve C is smooth for (so is continuous and never 0) and thefunction f is continuous on C , then the limit in Equation (1) can be shown to exist. We canthen apply the Fundamental Theorem of Calculus to differentiate the arc length equation,

to express ds in Equation (1) as and evaluate the integral of over C as

(2)

Notice that the integral on the right side of Equation (2) is just an ordinary (single) defi-nite integral, as defined in Chapter 5, where we are integrating with respect to the parame-ter t . The formula evaluates the line integral on the left side correctly no matter what

parametrization is used, as long as the parametrization is smooth. Note that the parameter t defines a direction along the path. The starting point on C is the position and move-ment along the path is in the direction of increasing t (see Figure 16.1).

r sa d

LC s x, y, z d ds = Lb

a s g st d, hst d, k st ddvst d dt .

ds = vst d dt

sst d = Lt

a vst d d t ,

v = d r>dt a t b920 Chapter 16: Integration in Vector Fields

Eq. (3) of Section 13.3with t 0 = a

dsdt

= v = A adxdt b2 + adydt b2 + adz dt b2

How to Evaluate a Line IntegralTo integrate a continuous function ( x, y, z ) over a curve C :

1. Find a smooth parametrization of C ,

.

2. Evaluate the integral as

LC s x, y, z d ds =

Lb

a s g st d, hst d, k st ddvst d dt .

r st d = g st di + hst d j + k st dk , a t b

FIGURE 16.2 The integration path inExample 1.

If has the constant value 1, then the integral of over C gives the length of C fromto in Figure 16.1.

EXAMPLE 1 Integrate over the line segment C joining theorigin to the point (1, 1, 1) (Figure 16.2).

Solution We choose the simplest parametrization we can think of:

The components have continuous first derivatives and

is never 0, so the parametrization is smooth. The integral of over C is

= 2 3

L1

0 s2t - 3t 2d dt = 2 3

Ct 2 - t 3

D01 = 0.

= L1

0 st - 3t 2 + t d2 3 dt

LC s x, y, z d ds = L1

0 st , t , t dA2 3B dt

2 12 + 12 + 12 = 2 3i + j + k =vst d =

r st d = t i + t j + t k , 0 t 1.

s x, y, z d = x - 3 y 2 + z

t = bt = a

z

x

C

(1, 1, 0)

(1, 1, 1)

y

Eq. (2)


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AdditivityLine integrals have the useful property that if a piecewise smooth curve C is made by join-

ing a finite number of smooth curves end to end (Section 13.1), then the in-tegral of a function over C is the sum of the integrals over the curves that make it up:

(3)

EXAMPLE 2 Figure 16.3 shows another path from the origin to (1, 1, 1), the union of line segments and Integrate over

Solution We choose the simplest parametrizations for and we can find, calculatingthe lengths of the velocity vectors as we go along:

With these parametrizations we f ind that

Notice three things about the integrations in Examples 1 and 2. First, as soon as thecomponents of the appropriate curve were substituted into the formula for , the integration

became a standard integration with respect to t . Second, the integral of over wasobtained by integrating over each section of the path and adding the results. Third, the in-tegrals of over C and had different values.C 1 C 2

C 1 C 2

= 2 2 ct 22 - t 3d01 + ct 22 - 2t d01 = - 2 22 - 32 . = L

1

0 st - 3t 2 + 0d2 2 dt + L

1

0 s1 - 3 + t ds1d dt

= L1

0 st , t , 0 d2 2 dt + L

1

0 s1, 1, t ds1d dt

LC 1 C 2 s x, y, z d ds = LC 1 s x, y, z d ds + LC 2 s x, y, z d ds

C 2: r st d = i + j + t k , 0 t 1; v = 2 02 + 02 + 12 = 1. C 1: r st d = t i + t j , 0 t 1; v = 2 12 + 12 = 2 2

C 2C 1

C 1 C 2.s x, y, z d = x - 3 y 2 + z C 2.C 1

LC ds = LC 1 ds + LC 2 ds + + LC n ds.C 1, C 2 , , C n

16.1 Line Integrals 921

The value of the line integral along a path joining two points can change if youchange the path between them.

FIGURE 16.3 The path of integration inExample 2. Eq. (3)

Eq. (2)

We investigate this third observation in Section 16.3.

Mass and Moment CalculationsWe treat coil springs and wires as masses distributed along smooth curves in space. Thedistribution is described by a continuous density function representing mass per unit length. When a curve C is parametrized bythen x, y, and z are functions of the parameter t , the density is the functionand the arc length differential is given by

ds =

A adxdt b

2

+

ady

dt b2

+

adz dt b

2

dt .

z st dd, yst d,d s xst d,a t b,r st d = xst di + yst d j + z st dk ,

d s x, y, z d

z

x

(0, 0, 0)

(1, 1, 0)

(1, 1, 1)

C 1

C 2 y


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(See Section 13.3.) The springs or wires mass, center of mass, and moments are then cal-culated with the formulas in Table 16.1, with the integrations in terms of the parameter t over the interval [ a , b]. For example, the formula for mass becomes

These formulas also apply to thin rods, and their derivations are similar to those in Section6.6. Notice how alike the formulas are to those in Tables 15.1 and 15.2 for double and triple integrals. The double integrals for planar regions, and the triple integrals for solids,

become line integrals for coil springs, wires, and thin rods.

M = Lb

a d s xst d, yst d, z st dd A adxdt b2 + adydt b2 + adz dt b2 dt .

922 Chapter 16: Integration in Vector Fields

TABLE 16.1 Mass and moment formulas for coil springs, wires, and thin rods lyingalong a smooth curve C in space

is the density at ( x, y, z )

First moments about the coordinate planes:

Coordinates of the center of mass:

Moments of inertia about axes and other lines:

r s x, y, z d = distance from the point s x, y, z d to line L I L = LC r 2 d ds I x = LC s y 2 + z 2d d ds,

I y = LC s x 2 + z 2d d ds, I z = LC s x 2 + y 2d d ds,

x = M yz > M , y = M xz > M , z = M xy > M M yz = LC x d ds,

M xz = LC y d ds,

M xy = LC z d ds

d = d ( x, y, z )Mass: M = LC d ds

Notice that the element of mass dm is equal to in the table rather than as inTable 15.1, and that the integrals are taken over the curve C.

EXAMPLE 3 A slender metal arch, denser at the bottom than top, lies along the semi-circle in the yz -plane (Figure 16.4). Find the center of the archs massif the density at the point ( x, y, z ) on the arch is

Solution We know that and because the arch lies in the yz -plane with itsmass distributed symmetrically about the z -axis. To find we parametrize the circle as

For this parametrization,

so ds = v dt = dt .

vst d = B adxdt b2 + adydt b2 + adz dt b2 = 2 s0d2 + s- sin t d2 + scos t d2 = 1,

r st d = scos t d j + ssin t dk , 0 t p .

z , y = 0 x = 0

d s x, y, z d = 2 - z . y

2+ z

2= 1, z 0,

d dV d ds

z

y x

1

1

c.m.

y2 z2 1, z 0

1

FIGURE 16.4 Example 3 shows how tofind the center of mass of a circular arch of variable density.


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The formulas in Table 16.1 then give

With to the nearest hundredth, the center of mass is (0, 0, 0.57).

Line Integrals in the PlaneThere is an interesting geometric interpretation for line integrals in the plane. If C is asmooth curve in the xy-plane parametrized by we gener-ate a cylindrical surface by moving a straight line along C orthogonal to the plane, holdingthe line parallel to the z -axis, as in Section 12.6. If is a nonnegative continuousfunction over a region in the plane containing the curve C , then the graph of is a surfacethat lies above the plane. The cylinder cuts through this surface, forming a curve on it thatlies above the curve C and follows its winding nature. The part of the cylindrical surfacethat lies beneath the surface curve and above the xy-plane is like a winding wall or fence standing on the curve C and orthogonal to the plane. At any point ( x, y) along thecurve, the height of the wall is We show the wall in Figure 16.5, where the top of the wall is the curve lying on the surface (We do not display the surface

formed by the graph of in the f igure, only the curve on it that is cut out by the cylinder.)From the definition

where as we see that the line integral is the area of the wallshown in the figure.

1C dsn : q , sk : 0LC ds = limn : q a

n

k = 1 s xk , yk d sk ,

z = s x, yd.s x, yd.

z = s x, yd

r st d = xst di + yst d j , a t b,

z

z = M xy M

= 8 - p

2# 1

2p - 2 = 8 -

p

4p - 4 L 0.57.

= Lp

0s2 sin t - sin2 t d dt = 8

- p2

M xy = LC z d ds = LC z s2 - z d ds = Lp

0ssin t ds2 - sin t d dt

M = LC d

ds = LC s2 - z d ds = Lp

0 s2 - sin t d dt = 2p

- 2

16.1 Line Integrals 923

FIGURE 16.5 The line integralgives the area of the portion of thecylindrical surface or wall beneath z = s x, yd 0.

1C ds

Exercises 16.1

Graphs of Vector EquationsMatch the vector equations in Exercises 18 with the graphs (a)(h)given here.

a. b.

y

z

x

2

1

y

z

x

1

1

c. d.

y

z

x

2

2

(2, 2, 2)

y

z

x

1 1

z

y

x

t a

t b

( x , y)

height f ( x , y)

Plane curve C sk


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e. f.

g. h.

1.

2.

3.

4.

5.6.

7.

8.

Evaluating Line Integrals over Space Curves9. Evaluate where C is the straight-line segment

from (0, 1, 0) to (1, 0, 0).

10. Evaluate where C is the straight-line seg-ment from (0, 1, 1) to (1, 0, 1).

11. Evaluate along the curve

12. Evaluate along the curve

13. Find the line integral of over the straight-line segment from (1, 2, 3) to

14. Find the line integral of over the curve

15. Integrate over the path from (0, 0, 0)to (1, 1, 1) (see accompanying figure) given by

C 2: r st d = i + j + t k , 0 t 1 C 1: r st d = t i + t 2 j , 0 t 1

s x, y, z d = x + 1 y - z 2r st d = t i + t j + t k , 1 t q .

s x, y, z d = 2 3>s x2 + y 2 + z 2ds0, - 1, 1 d.

s x, y, z d = x + y + z

s4 sin t d j + 3t k , - 2p t 2p .r st d = s4 cos t di +1C 2 x2 + y 2 ds

t j + s2 - 2t dk , 0 t 1.r st d = 2t i +1C s xy + y + z d ds

x = t , y = s1 - t d, z = 1,1C s x - y + z - 2d ds

x = t , y = s1 - t d, z = 0,1C s x + yd ds

r st d = s2 cos t di + s2 sin t dk , 0 t p

r st d = st 2 - 1d j + 2t k , - 1 t 1

r st d = t j + s2 - 2t dk , 0 t 1r st d = t i + t j + t k , 0 t 2

r st d = t i, - 1 t 1

r st d = s2 cos t di + s2 sin t d j , 0 t 2p

r st d = i + j + t k , - 1 t 1

r st d = t i + s1 - t d j , 0 t 1

y

z

x

2

2

2

y

z

x

2

2

y

z

x

2

2

1

y

z

x

11

(1, 1, 1)

(1, 1, 1)

16. Integrate over the path from (0, 0, 0)to (1, 1, 1) (see accompanying f igure) given by

17. Integrate over the path

18. Integrate over the circle

Line Integrals over Plane Curves19. Evaluate where C is

a. the straight-line segment from (0, 0) to (4, 2).

b. the parabolic curve from (0, 0) to (2, 4).20. Evaluate where C is

a. the straight-line segment from (0, 0) to (1, 4).

b. is the line segment from (0, 0) to (1, 0) and isthe line segment from (1, 0) to (1, 2).

21. Find the line integral of along the curve


23. Evaluate , where C is the curve for


25. Evaluate where C is given in the accompanyingfigure.

x

y

y 5 x 2 y 5 x

(0, 0)

(1, 1)C

1C A x + 2 yB ds1>2 t 1.r st d = t 3i + t 4 j ,

s x, yd = 2 y> x1 t 2.

x = t 2, y = t 3,LC

x2

y 4>3 dsr st d = (cos t )i + (sin t ) j , 0 t 2p .

s x, yd = x - y + 3

- 1 t 2.r st d = 4t i - 3t j ,s x, yd = ye x

2

C 2C 1 C 2; C 1

x = t , y = 4t ,1C 2 x + 2 y ds,

x = t , y = t 2,

x = t , y = t >2,1C x ds,

r st d = sa cos t d j + sa sin t dk , 0 t 2p .

s x, y, z d = - 2 x2 + z 2r st d = t i + t j + t k , 0 6 a t b.

s x, y, z d = s x + y + z d>s x2 + y 2 + z 2d C 3: r st d = t i + j + k , 0 t 1 C 2: r st d = t j + k , 0 t 1 C 1: r st d = t k , 0 t 1

s x, y, z d = x + 1 y - z 2

z

y

x

(a)(1, 1, 0)

(1, 1, 1)(0, 0, 0)

z

y x

(b)

(0, 0, 0)(1, 1, 1)

(0, 0, 1)(0, 1, 1)

C 1

C 1

C 2

C 2

C 3

The paths of integration for Exercises 15 and 16.


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26. Evaluate where C is given in the accompanying

figure.

In Exercises 2730, integrate over the given curve.

27.

28. from (1, 1 2) to(0, 0)

29. in the first quadrant from(2, 0) to (0, 2)

30. in the first quadrant from(0, 2) to

31. Find the area of one side of the winding wall standing orthogo-nally on the curve and beneath the curve onthe surface

32. Find the area of one side of the wall standing orthogonally onthe curve and beneath the curve on thesurface

Masses and Moments33. Mass of a wire Find the mass of a wire that lies along the curve

if the density is

34. Center of mass of a curved wire A wire of densitylies along the curve

Find its center of mass. Then sketch the curveand center of mass together.

35. Mass of wire with variable density Find the mass of a thinwire lying along the curve

if the density is (a) and (b)

36. Center of mass of wire with variable density Find the center of mass of a thin wire lying along the curve

if the density is

37. Moment of inertia of wire hoop A circular wire hoop of con-stant density lies along the circle in the xy-plane.Find the hoops moment of inertia about the z -axis.

38. Inertia of a slender rod A slender rod of constant density liesalong the line segment in thes2 - 2t dk , 0 t 1,r st d = t j +

x 2 + y 2 = a 2d

d = 3 1 5 + t .s2>3dt 3>2k , 0 t 2,r st d = t i + 2t j +

d = 1.d = 3t 0 t 1,r st d = 2 2t i + 2 2t j + s4 - t 2dk ,

2t k , - 1 t 1.r st d = st 2 - 1d j +d s x, y, z d = 15 2 y + 2

d = s3>2dt .r st d = st 2 - 1d j + 2t k , 0 t 1,

s x, yd = 4 + 3 x + 2 y.2 x + 3 y = 6, 0 x 6,

s x, yd = x + 2 y . y = x2, 0 x 2,

s1 2, 1 2ds x, yd = x2 - y, C : x2 + y 2 = 4

s x, yd = x + y, C : x2 + y 2 = 4 >s x, yd = s x + y 2d

>2 1 + x2, C : y = x2

>2

s x, yd = x3> y, C : y = x2>2, 0 x 2

x

y

(0, 0)

(0, 1)

(1, 0)

(1, 1)

LC

1

x2 + y 2 + 1 ds

16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux 925

yz -plane. Find the moments of inertia of the rod about the threecoordinate axes.

39. Two springs of constant density A spring of constant densitylies along the helix

a. Find .

b. Suppose that you have another spring of constant densitythat is twice as long as the spring in part (a) and lies along thehelix for Do you expect for the longer springto be the same as that for the shorter one, or should it bedifferent? Check your prediction by calculating for thelonger spring.

40. Wire of constant density A wire of constant density liesalong the curve

Find

41. The arch in Example 3 Find for the arch in Example 3.

42. Center of mass and moments of inertia for wire with variabledensity Find the center of mass and the moments of inertiaabout the coordinate axes of a thin wire lying along the curve

if the density is .

COMPUTER EXPLORATIONSIn Exercises 4346, use a CAS to perform the following steps to eval-

uate the line integrals.a. Find for the path

b. Express the integrand as a function of the parameter t .

c. Evaluate using Equation (2) in the text.

43.

44.

45.

46.

0 t 2pt 5>2k , s x, y, z d = a1 + 94 z 1>3b

1>4; r st d = scos 2t di + ssin 2t d j +

0 t 2ps x, y, z d = x1 y - 3 z 2 ; r st d = scos 2t di + ssin 2t d j + 5t k ,0 t 2s x, y, z d = 2 1 + x3 + 5 y 3 ; r st d = t i + 13 t

2 j + 1 t k , 0 t 2s x, y, z d = 2 1 + 30 x2 + 10 y ; r st d = t i + t 2 j + 3t 2k ,

1C dss g st d, hst d, k st ddvst d

k st dk .r st d = g st di + hst d j +ds = vst d dt

d = 1>st + 1dr st d = t i + 2

2 23

t 3>2 j + t 22 k ,

0 t 2,

I x

z and I z .

r st d = st cos t di + st sin t d j + A

2 2 2

>3

Bt 3>2k , 0 t 1.d = 1

I z

I z 0 t 4p .

d

I z

r st d = scos t di + ssin t d j + t k , 0 t 2p .d

16.2 Vector Fields and Line Integrals: Work, Circulation, and FluxGravitational and electric forces have both a direction and a magnitude. They are repre-sented by a vector at each point in their domain, producing a vector field. In this sectionwe show how to compute the work done in moving an object through such a field by usinga line integral involving the vector f ield. We also discuss velocity fields, such as the vector


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field representing the velocity of a flowing fluid in its domain. A line integral can be used to find the rate at which the fluid flows along or across a curve within the domain.

Vector FieldsSuppose a region in the plane or in space is occupied by a moving fluid, such as air or wa-ter. The fluid is made up of a large number of particles, and at any instant of time, a parti-cle has a velocity v. At different points of the region at a given (same) time, these veloci-ties can vary. We can think of a velocity vector being attached to each point of the fluid representing the velocity of a particle at that point. Such a fluid flow is an example of avector field. Figure 16.6 shows a velocity vector field obtained from air flowing around anairfoil in a wind tunnel. Figure 16.7 shows a vector field of velocity vectors along thestreamlines of water moving through a contracting channel. Vector fields are also associ-ated with forces such as gravitational attraction (Figure 16.8), and to magnetic fields, elec-tric fields, and also purely mathematical fields.

Generally, a vector field is a function that assigns a vector to each point in its domain.A vector field on a three-dimensional domain in space might have a formula like

The field is continuous if the component functions M , N , and P are continuous; it isdifferentiable if each of the component functions is differentiable. The formula for a f ield of two-dimensional vectors could look like

We encountered another type of vector f ield in Chapter 13. The tangent vectors T and normal vectors N for a curve in space both form vector fields along the curve. Along acurve r (t ) they might have a component formula similar to the velocity f ield expression

If we attach the gradient vector of a scalar function ( x, y, z ) to each point of alevel surface of the function, we obtain a three-dimensional field on the surface. If we at-tach the velocity vector to each point of a flowing fluid, we have a three-dimensional field defined on a region in space. These and other fields are illustrated in Figures 16.916.15.To sketch the fields, we picked a representative selection of domain points and drew the

v(t ) = (t )i + g (t ) j + h(t )k .

Fs x, yd = M s x, ydi + N s x, yd j .

Fs x, y, z d = M s x, y, z di + N s x, y, z d j + P s x, y, z dk .


FIGURE 16.7 Streamlines in acontracting channel. The water speeds upas the channel narrows and the velocityvectors increase in length.

FIGURE 16.6 Velocity vectors of a flowaround an airfoil in a wind tunnel.

y

z

x

FIGURE 16.8 Vectors in agravitational field point toward

the center of mass that gives thesource of the field.

z

x

y

FIGURE 16.9 A surface, like a mesh net or parachute, in a vector field representing water or wind flow velocity vectors. The arrowsshow the direction and their lengths indicate speed.


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f ( x , y, z) 5 c

FIGURE 16.10 The field of gradient vectors on asurface s x, y, z d = c.

y

x

FIGURE 16.13 The flow of fluidin a long cylindrical pipe. The vectors

inside the cylinder thathave their bases in the xy-plane havetheir tips on the paraboloid z = a 2 - r 2.

v = sa 2 - r 2dk

FIGURE 16.11 The radial field of position vectors of points in

the plane. Notice the convention that anarrow is drawn with its tail, not its head, atthe point where F is evaluated.

F = xi + y j

WIND SPEED, M/S

0 2 4 6 8 10 12 14 16+

FIGURE 16.15 NASAs Seasat used radar to take 350,000 wind measurementsover the worlds oceans. The arrows show wind direction; their length and the color contouring indicate speed. Notice the heavy storm south of Greenland.

x

y

FIGURE 16.12 A spin field of rotatingunit vectors

in the plane. The field is not defined at theorigin.

F = s- y i + x j d>s x2 + y 2d1>2

y

x 0

FIGURE 16.14 Thevelocity vectors v(t ) of a

projectiles motion make avector field along thetrajectory.

z

y

x

x 2 y2 a2

z a2 r 2

0

vectors attached to them. The arrows are drawn with their tails, not their heads, attached tothe points where the vector functions are evaluated.

Gradient FieldsThe gradient vector of a differentiable scalar-valued function at a point gives the directionof greatest increase of the function. An important type of vector field is formed by all the


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gradient vectors of the function (see Section 14.5). We define the gradient field of a dif-ferentiable function ( x, y, z ) to be the f ield of gradient vectors

At each point the gradient field gives a vector pointing in the direction of greatestincrease of , with magnitude being the value of the directional derivative in that direction.The gradient field is not always a force field or a velocity field.

EXAMPLE 1 Suppose that the temperature T at each point ( x, y, z ) in a region of spaceis given by

,

and that F ( x, y, z ) is defined to be the gradient of T . Find the vector field F .

Solution The gradient field F is the field At each pointin space, the vector field F gives the direction for which the increase in temperature isgreatest.

Line Integrals of Vector Fields

In Section 16.1 we defined the line integral of a scalar function over a path C .We turn our attention now to the idea of a line integral of a vector field F along the curveC .

Assume that the vector field has contin-

uous components, and that the curve C has a smooth parametrizationAs discussed in Section 16.1, the parametrization defines

a direction (or orientation) along C which we call the forward direction . At each pointalong the path C , the tangent vector is a unit vector tangent to the pathand pointing in this forward direction. (The vector is the velocity vector tangentto C at the point, as discussed in Sections 13.1 and 13.3.) Intuitively, the line integral of thevector field is the line integral of the scalar tangential component of F along C . This tan-gential component is given by the dot product

so we have the following formal definition, where in Equation (1) of Section16.1.

= F #T

F #T = F #d rds ,

v = d r>dt T = d r>ds = v> v

r st dhst d j + k st d k , a t b.r st d = g st d i +

F = M s x, y, z d i + N s x, y, z d j + P s x, y, z d k

s x, y, z d

F = T = - 2 xi - 2 y j - 2 z k .

T = 100 - x2 - y 2 - z 2

s x, y, z d,

= 00 x i + 00 y j +

00 z k .


DEFINITION Let F be a vector field with continuous components defined along a smooth curve C parametrized by Then the line integralof F along C is

LC F #T ds = LC aF #d rds b ds = LC F #d r .r st d, a t b.

We evaluate line integrals of vector fields in a way similar to how we evaluate line in-tegrals of scalar functions (Section 16.1).


11/42

EXAMPLE 2 Evaluate where along the curve C

given by

Solution We have

and

Thus,

Line Integrals With Respect to the xyz CoordinatesIt is sometimes useful to write a line integral of a scalar function with respect to one of thecoordinates, such as This integral is not the same as the arc length line integral

we defined in Section 16.1. To define the new integral for the scalar functionwe specify a vector field over the curve C parametrized by

With this notation we have and Then,

So we define the line integral of M over C with respect to the coordinate x as

In the same way, by defining or we obtain the integralsand Expressing everything in terms of the parameter t , we have the fol-

lowing formulas for these integrals:1C P dz .1C N dy F = P s x, y, z dk ,F = N s x, y, z d j ,

LC M s x, y, z d dx = LC F #d r , where F = M s x, y, z d i.

F #d r = F #d rdt dt = M s x, y, z d g st d dt = M s x, y, z d dx.

dx = g st d dt . x = g st dr st d = g st di + hst d j + k st dk , a t b.

F = M s x, y, z di M s x, y, z d,1C M ds

1C M dx.

= c a32b a25 t 5>2b + 14 t 4d1

0= 17

20 .

= L1

0 a2t 3>2 + t 3 - 12 t 3>2b dt

LC F #d r = L1

0 Fsr st dd#d rdt dt

d rdt

= 2t i + j + 12 2 t

k .

Fsr st dd = 2 t i + t 3 j - t 2k

r st d = t 2i + t j + 2 t k , 0 t 1.

Fs x, y, z d = z i + xy j - y 2k 1C F #d r ,


Evaluating the Line Integral of along

1. Express the vector field F in terms of the parametrized curve C as bysubstituting the components of r into the scalar components of F .

2. Find the derivative (velocity) vector

3. Evaluate the line integral with respect to the parameter to obtain

LC F #d r = Lb

a Fsr st dd#d rdt dt .

t , a t b,d r>dt .

M s x, y, z d, N s x, y, z d, P s x, y, z d x = g st d, y = hst d, z = k st d

Fsr st ddC : r( t ) g ( t )i h( t )j k ( t )k

F Mi N j P k


12/42

It often happens that these line integrals occur in combination, and we abbreviate the nota-tion by writing

EXAMPLE 3 Evaluate the line integral where C is the helix

Solution We express everything in terms of the parameter t , soand Then,

Work Done by a Force over a Curve in SpaceSuppose that the vector field represents aforce throughout a region in space (it might be the force of gravity or an electromagneticforce of some kind) and that

is a smooth curve in the region. The formula for the work done by the force in moving anobject along the curve is motivated by the same kind of reasoning we used in Chapter 6 toderive the formula for the work done by a continuous force of magnitude

F ( x) directed along an interval of the x-axis. For a curve C in space, we define the work done by a continuous force field F to move an object along C from a point A to another

point B as follows.We divide C into n subarcs with lengths starting at A and ending at B. We

choose any point in the subarc and let be the unit tangentvector at the chosen point. The work done to move the object along the subarcis approximated by the tangential component of the force times the arclength

approximating the distance the object moves along the subarc (see Figure 16.16). sk Fs xk , yk , z k d

P k - 1 P k W k Ts xk , yk , z k d P k - 1 P k s xk , yk , z k d

sk , P k - 1 P k

W = 1ba F ( x) dx

r st d = g st di + hst d j + k st dk , a t b,

F = M s x, y, z di + N s x, y, z d j + P s x, y, z dk

= p . = [0 + s0 + 1d + sp - 0d] - [0 + s0 + 1d + s0 - 0d]

= c2 sin t + st sin t + cos t d + at 2 - sin 2t 4 b d2p

0

= L2p

0 [2 cos t + t cos t + sin2 t ] dt

LC - y dx + z dy + 2 x dz = L2p

0 [s - sin t ds- sin t d + t cos t + 2 cos t ] dt

dy = cos t dt , dz = dt .dx = - sin t dt , z = t , y = sin t , x = cos t ,

(sin t ) j + t k , 0 t 2p .r st d = (cos t )i +1C - y dx + z dy + 2 x dz ,

LC M s x, y, z d dx + LC N s x, y, z d dy + LC P s x, y, z d dz = LC M dx + N dy + P dz .


(1)

(2)

(3)LC P s x, y, z d dz = Lb

a P s g st d, hst d, k st dd k st d dt

LC N s x, y, z d dy = Lb

a N s g st d, hst d, k st dd hst d dt

LC M s x, y, z d dx = Lb

a M s g st d, hst d, k st dd g st d dt

Pk 2 1

T k

F k . T k

F k Pk

( x k , yk , zk )

FIGURE 16.16 The work done along thesubarc shown here is approximately

where and T k = Ts xk , yk , z k d.

F k = Fs xk , yk , z k dF k #T k sk ,


13/42

The total work done in moving the object from point A to point B is then approximated bysumming the work done along each of the subarcs, so

For any subdivision of C into n subarcs, and for any choice of the points withineach subarc, as and these sums approach the line integral

.

This is just the line integral of F along C , which is defined to be the total work done.

LC F #T ds sk : 0,n : q

s xk , yk , z k d

W L an

k = 1 W k L a

n

k = 1 Fs xk , yk , z k d#Ts xk , yk , z k d sk .


DEFINITION Let C be a smooth curve parametrized by and F be a continuous force field over a region containing C . Then the work done in

moving an object from the point to the point along C is

(4)W = LC F #T ds = Lb

a Fsr st dd#d rdt dt .

B =

r sbd A =

r sa d

r st d, a t b,T

F

A

B t b

t a

FIGURE 16.17 The work done by a forceF is the line integral of the scalar component over the smooth curvefrom A to B.

F #T

The sign of the number we calculate with this integral depends on the direction in whichthe curve is traversed. If we reverse the direction of motion, then we reverse the directionof T in Figure 16.17 and change the sign of and its integral.

Using the notations we have presented, we can express the work integral in a variety of ways, depending upon what seems most suitable or convenient for a particular discussion.Table 16.2 shows five ways we can write the work integral in Equation (4).

F #T

TABLE 16.2 Different ways to write the work integral for overthe curve

The definition

Vector differential form

Parametric vector evaluation

Parametric scalar evaluation

Scalar differential form= LC M dx + N dy + P dz = L

b

a a M dxdt + N dydt + P dz dt b dt

= Lb

a F #d rdt dt

= LC F #d r W = LC F #T ds

C : r (t ) = g (t )i + h(t ) j + k (t )k , a t bF = Mi + N j + P k

EXAMPLE 4 Find the work done by the force field along the curve from (0, 0, 0) to (1, 1, 1)

(Figure 16.18).

Solution First we evaluate F on the curve

.

0(')'*

= st 2 - t 2di + st 3 - t 4d j + st - t 6dk

F = s y - x2di + s z - y 2d j + s x - z 2dk

r st d:

t i + t 2 j + t 3k , 0 t 1,r st d =s x - z 2dk F = s y - x2di + s z - y 2d j +

Substitute. z = t 3 y = t 2,

x = t ,

y

z

x

(0, 0, 0)

(1, 1, 0)

(1, 1, 1)

r (t ) t i t 2 j t 3k

FIGURE 16.18 The curve in Example 4.


14/42

Then we find d r dt ,

.

Finally, we find and integrate from to

so,

EXAMPLE 5 Find the work done by the force field in moving anobject along the curve C parametrized by

Solution We begin by writing F along C as a function of t ,

Next we compute d r dt ,

We then calculate the dot product,

The work done is the line integral

Flow Integrals and Circulation for Velocity FieldsSuppose that F represents the velocity field of a fluid flowing through a region in space (a

tidal basin or the turbine chamber of a hydroelectric generator, for example). Under thesecircumstances, the integral of along a curve in the region gives the fluids flow alongthe curve.

F #T

Lb

a F(r (t )) #d rdt dt = L

1

0 2t 3 dt = t

4

2 d1

0= 1

2.

F (r (t )) #d rdt = - p sin ( p t ) cos ( p t ) + 2t 3 + p sin ( p t ) cos( p t ) = 2t 3.

d rdt

= - p sin ( p t ) i + 2t j + p cos ( p t ) k .

>Fsr st dd = cos ( p t ) i + t 2 j + sin ( p t ) k .

0 t 1.r (t ) = cos ( p t ) i + t 2 j + sin ( p t ) k ,F = xi + y j + z k

=

c25

t 5 - 26

t 6 + 34

t 4 - 39

t 9

d01

= 2960

.

Work = L1

0 s2t 4 - 2t 5 + 3t 3 - 3t 8d dt

= st 3 - t 4ds2t d + st - t 6ds3t 2d = 2t 4 - 2t 5 + 3t 3 - 3t 8

F #d rdt = [st 3 - t 4d j + st - t 6dk ] #si + 2t j + 3t 2k d

t = 1:t = 0F #d r>dt d r

dt = d

dt st i + t 2 j + t 3k d = i + 2t j + 3t 2k

>932 Chapter 16: Integration in Vector Fields

DEFINITIONS If r (t ) parametrizes a smooth curve C in the domain of a continu-ous velocity field F , the flow along the curve from is

(5)

The integral in this case is called a flow integral . If the curve starts and ends at thesame point, so that , the flow is called the circulation around the curve. A = B

Flow = LC F #T ds. A = r (a) to B = r (b)


15/42

The direction we travel along C matters. If we reverse the direction, then T is replaced by and the sign of the integral changes. We evaluate flow integrals the same way weevaluate work integrals.

EXAMPLE 6 A fluids velocity field is Find the flow along the helix

Solution We evaluate F on the curve,

.

and then find d r dt :

Then we integrate from to

so,

EXAMPLE 7 Find the circulation of the field around the circle(Figure 16.19).

Solution On the circle, and

Then

gives

As Figure 16.19 suggests, a fluid with this velocity f ield is circulating counterclockwisearound the circle.

Flux Across a Simple Plane CurveA curve in the xy-plane is simple if it does not cross itself (Figure 16.20). When a curvestarts and ends at the same point, it is a closed curve or loop . To find the rate at which a fluid is entering or leaving a region enclosed by a smooth simple closed curve C in the xy-plane,

= ct - sin2 t 2 d02p

= 2p .

Circulation = L2p

0 F # d rdt dt = L

2p

0 s1 - sin t cos t d dt

- sin t cos t + sin2 t + cos 2 t F # d rdt =

d rdt

= s- sin t di + scos t d j.

F = s x - ydi + x j = scos t - sin t di + scos t d j ,

ssin t d j , 0 t 2pr st d = scos t di +F = s x - ydi + x j

= ccos 2 t 2 + t sin t d0p >2

= a0 + p2 b - a12 + 0b = p2 - 12 . Flow = L

t = b

t = a F # d rdt dt = L

p >20

s - sin t cos t + t cos t + sin t d dt

= - sin t cos t + t cos t + sin t

F # d rdt = scos t ds- sin t d + st dscos t d + ssin t ds1d

t = p

2 :t = 0F #sd r

>dt d

d rdt

= s- sin t di + scos t d j + k .

>Substitute x = cos t , z = t , y = sin t F = xi + z j + yk = scos t di + t j + ssin t dk

ssin t d j + t k , 0 t p >2.r st d = scos t di +F = xi + z j + yk .

- T


x

y

FIGURE 16.19 The vector field F and curve in Example 7.r (t )

Simple,closed

Not simple,closed

Simple,not closed

Not simple,not closed

FIGURE 16.20 Distinguishing curves thatare simple or closed. Closed curves are alsocalled loops.

(''')'''*

1


16/42

we calculate the line integral over C of the scalar component of the fluids velocityfield in the direction of the curves outward-pointing normal vector. The value of this inte-gral is the flux of F across C . Flux is Latin for flow, but many flux calculations involve nomotion at all. If F were an electric field or a magnetic field, for instance, the integral of

would still be called the flux of the field across C .F #n

F #n ,


DEFINITION If C is a smooth simple closed curve in the domain of a contin-uous vector field in the plane, and if n is the outward-

pointing unit normal vector on C , the flux of F across C is

(6)Flux of F across C = LC F #n ds.F = M s x, ydi + N s x, yd j

T

z

y

x k

C

T

z

y

x k

C

For clockwise motion,k T points outward.

For counterclockwisemotion, T k pointsoutward.

k T

T k

FIGURE 16.21 To f ind an outwardunit normal vector for a smooth simplecurve C in the xy-plane that is traversed

counterclockwise as t increases, we takeFor clockwise motion, we

take n = k * T .n = T * k .

Notice the difference between flux and circulation. The flux of F across C is the lineintegral with respect to arc length of the scalar component of F in the direction of theoutward normal. The circulation of F around C is the line integral with respect to arclength of the scalar component of F in the direction of the unit tangent vector. Fluxis the integral of the normal component of F ; circulation is the integral of the tangentialcomponent of F .

To evaluate the integral for flux in Equation (6), we begin with a smooth parametrization

that traces the curve C exactly once as t increases from a to b. We can find the outward unit normal vector n by crossing the curves unit tangent vector T with the vector k . Butwhich order do we choose, or Which one points outward? It depends onwhich way C is traversed as t increases. If the motion is clockwise, points outward;if the motion is counterclockwise, points outward (Figure 16.21). The usual choiceis the choice that assumes counterclockwise motion. Thus, although the valueof the integral in Equation (6) does not depend on which way C is traversed, the formulaswe are about to derive for computing n and evaluating the integral assume counterclock-wise motion.

In terms of components,

If then

Hence,

We put a directed circle on the last integral as a reminder that the integration around the closed curve C is to be in the counterclockwise direction. To evaluate this integral, weexpress M , dy, N , and dx in terms of the parameter t and integrate from to Wedo not need to know n or ds explicitly to f ind the flux.

t = b.t = a

~

LC F #n ds = LC a M dyds - N dxds b ds = FC

M dy - N dx.

F #n = M s x, yd dyds - N s x, yd dxds .

F = M s x, ydi + N s x, yd j ,

n = T * k = adxds i + dyds jb * k = dyds i - dxds j .

n = T * k ,T * k

k * Tk * T?T * k

x = g st d, y = hst d, a t b,

F #T,F #n ,


17/42

EXAMPLE 8 Find the flux of across the circle in the xy-plane. (The vector f ield and curve were shown previously in Figure 16.19.)

Solution The parametrization traces the circlecounterclockwise exactly once. We can therefore use this parametrization in Equation (7).With

we find

Eq. (7)

The flux of F across the circle is Since the answer is positive, the net flow across thecurve is outward. A net inward flow would have given a negative flux.

p .

= L2p

0 cos 2 t dt = L

2p

0 1 + cos 2t

2 dt = ct 2 + sin 2t 4 d0

2p

= p .

Flux = LC M dy - N dx = L2p

0 scos 2 t - sin t cos t + cos t sin t d dt

N = x = cos t , dx = d scos t d = - sin t dt , M = x - y = cos t - sin t , dy = d ssin t d = cos t dt

r st d = scos t di + ssin t d j , 0 t 2p ,

x2 + y 2 = 1F = s x - ydi + x j


Calculating Flux Across a Smooth Closed Plane Curve

(7)

The integral can be evaluated from any smooth parametrizationthat traces C counterclockwise exactly once.a t b,

x = g st d, y = hst d,

sFlux of F = M i + N j across C d = FC M dy - N dx

Exercises 16.2

Vector FieldsFind the gradient f ields of the functions in Exercises 14.

1.

2.

3.

4.

5. Give a formula for the vector field inthe plane that has the property that F points toward the origin withmagnitude inversely proportional to the square of the distancefrom ( x, y) to the origin. (The field is not defined at (0, 0).)

6. Give a formula for the vector field inthe plane that has the properties that at (0, 0) and that atany other point ( a , b), F is tangent to the circle

and points in the clockwise direction with magnitude

Line Integrals of Vector FieldsIn Exercises 712, find the line integrals of F from (0, 0, 0) to(1, 1, 1) over each of the following paths in the accompanying f igure.

F = 2 a 2 + b 2.a 2 + b 2

x 2 + y 2 =F = 0

F = M s x, ydi + N s x, yd j

F = M s x, ydi + N s x, yd j g s x, y, z d = xy + yz + xz g s x, y, z d = e z - ln s x2 + y 2ds x, y, z d = ln 2 x2 + y 2 + z 2s x, y, z d = s x2 + y 2 + z 2d- 1>2

a. The straight-line path

b. The curved path

c. The path consisting of the line segment from (0, 0, 0)to (1, 1, 0) followed by the segment from (1, 1, 0) to (1, 1, 1)

7. 8.

9. 10.

11.

12.

z

y

x

(0, 0, 0)

(1, 1, 0)

(1, 1, 1)C 1

C 2

C 3

C 4

F = s y + z di + s z + xd j + s x + ydk F = s3 x2 - 3 xdi + 3 z j + k

F = xyi + yz j + xz k F = 1 z i - 2 x j + 1 yk F = [1

>s x2 + 1d] jF = 3 yi + 2 x j + 4 z k

C 3 C 4

r st d = t i + t 2 j + t 4k , 0 t 1C 2:r st d = t i + t j + t k , 0 t 1C 1:


18/42

Line Integrals with Respect to x , y , and z In Exercises 1316, find the line integrals along the given path C .

13. , where C : for

14. , where C : for

15. , where C is given in the accompanying figure.

16. , where C is given in the accompanying figure.

17. Along the curve evaluate eachof the following integrals.

a.

b.

c.

18. Along the curveevaluate each of the following integrals.

a. b. c.

WorkIn Exercises 1922, find the work done by F over the curve in thedirection of increasing t .

19.

20.

21.

22.

r st d = ssin t di + scos t d j + st >6dk , 0 t 2pF = 6 z i + y 2 j + 12 xk r st d = ssin t di + scos t d j + t k , 0 t 2pF = z i + x j + yk r st d = scos t di + ssin t d j + st >6dk , 0 t 2pF = 2 yi + 3 x j + s x + ydk r st d = t i + t 2 j + t k , 0 t 1F = xyi + y j - yz k

LC xyz dz

LC xz dy

LC xz dx

r (t ) = (cos t )i + (sin t ) j - (cos t )k , 0 t p ,

LC ( x + y - z ) dz LC ( x + y - z ) dyLC ( x + y - z ) dx

r (t ) = t i - j + t 2k , 0 t 1,

x

y

(0, 0)

(0, 3) (1, 3)C

y 5 3 x

LC 2 x + y dx

x

y

(0, 0) (3, 0)

(3, 3)

C

LC ( x2 + y 2) dy1 t 2 x = t , y = t 2,LC x y dy

0 t 3 x = t , y = 2t + 1,LC ( x - y) dx


Line Integrals in the Plane23. Evaluate along the curve from

to (2, 4).

24. Evaluate counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1).

25. Evaluate for the vector field along thecurve from (4, 2) to

26. Evaluate for the vector field counter-clockwise along the unit circle from (1, 0) to (0, 1).

Work, Circulation, and Flux in the Plane27. Work Find the work done by the force

over the straight line from (1, 1) to (2, 3).

28. Work Find the work done by the gradient ofcounterclockwise around the circle from (2, 0) toitself.

29. Circulation and flux Find the circulation and flux of the f ields

around and across each of the following curves.

a. The circle

b. The ellipse

30. Flux across a circle Find the flux of the fields

across the circle

In Exercises 3134, find the circulation and flux of the field F around

and across the closed semicircular path that consists of the semicircu-lar arch followed by theline segment

31. 32.

33. 34.

35. Flow integrals Find the flow of the velocity fieldalong each of the following paths from (1, 0)

to in the xy-plane.

a. The upper half of the circle

b. The line segment from (1, 0) to

c. The line segment from (1, 0) to followed by the linesegment from to

36. Flux across a triangle Find the flux of the field F in Exercise35 outward across the triangle with vertices (1, 0) , (0, 1),

37. Find the flow of the velocity field along each of the following paths from (0, 0) to (2, 4).

a. b.

c. Use any path from (0, 0) to (2, 4) different from parts (a)and (b).

x

y

(0, 0)

(2, 4)

2

y 5 x 2

x

y

(0, 0)

(2, 4)

2

y 5 2 x

F = y 2i + 2 xy js - 1, 0 d.

s - 1, 0 ds0, - 1ds0, - 1ds - 1, 0 d

x2 + y 2 = 1s - 1, 0 d

s x2 + y 2d js x + ydi -F =

F = - y2 i + x2 jF = - yi + x jF = x2 i + y 2 jF = xi + y j

r 2st d = t i, - a t a .r 1st d = sa cos t di + sa sin t d j , 0 t p ,

r st d = sa cos t di + sa sin t d j , 0 t 2p .

F 1 = 2 xi - 3 y j

and

F 2 = 2 xi + s x - yd j

r st d = scos t di + s4 sin t d j , 0 t 2pr st d = scos t di + ssin t d j , 0 t 2p

F 1 = xi + y j and F 2 = - yi + x j x2 + y 2 = 4

s x, yd = s x + yd2

F = xyi + s y - xd j

x2 + y 2 = 1F = yi - x j1C F #d r

s1, - 1d . x = y 2F = x2i - y j1C F #T ds

1C s x - yd dx + s x + yd dys - 1, 1 d

y = x21C xy dx + s x + yd dy


19/42

38. Find the circulation of the field around eachof the following closed paths.

a.

b.

c. Use any closed path different from parts (a) and (b).

Vector Fields in the Plane39. Spin field Draw the spin field

(see Figure 16.12) along with its horizontal and vertical compo-nents at a representative assortment of points on the circle

40. Radial field Draw the radial field

(see Figure 16.11) along with its horizontal and vertical compo-nents at a representative assortment of points on the circle

41. A field of tangent vectors

a. Find a field in the xy-plane with the property that at any point G is a vector of

magnitude tangent to the circleand pointing in the counterclockwise direction.

(The field is undefined at (0, 0).)

b. How is G related to the spin field F in Figure 16.12?

42. A field of tangent vectors

a. Find a field in the xy-plane with the property that at any point G is a unit vector tangent to the circle and pointing in theclockwise direction.

b. How is G related to the spin field F in Figure 16.12?

x2 + y 2 = a 2 + b 2sa , bd Z s0, 0 d,

G = P s x, ydi + Qs x, yd j

a 2 + b 2 x2 + y 2 =2 a 2 + b 2

sa , bd Z s0, 0 d,G = P s x, ydi + Qs x, yd j

x2 + y 2 = 1.

F = x i + y j

x2 + y 2 = 4.

F = - y2 x2 + y 2

i + x2 x2 + y 2

j

x

y

x 2 1 y2 5 4

x

y(1, 1)

(1, 1)

(1, 1)

(1, 1)

F = yi + ( x + 2 y) j


43. Unit vectors pointing toward the origin Find a fieldin the xy-plane with the property that at each

point F is a unit vector pointing toward the origin.

(The field is undefined at (0, 0).)44. Two central fields Find a field in

the xy-plane with the property that at each pointF points toward the origin and is (a) the distance from ( x, y)to the origin, (b) inversely proportional to the distance from ( x, y)to the origin. (The field is undefined at (0, 0).)

45. Work and area Suppose that ( t ) is differentiable and positivefor Let C be the pathand Is there any relation between the value of the work integral

and the area of the region bounded by the t -axis, the graph of ,and the lines and Give reasons for your answer.

46. Work done by a radial force with constant magnitude A par-ticle moves along the smooth curve from ( a , (a)) to(b, (b)). The force moving the particle has constant magnitude k and always points away from the origin. Show that the work done

by the force is

Flow Integrals in SpaceIn Exercises 4750, F is the velocity field of a fluid flowing through aregion in space. Find the flow along the given curve in the direction of increasing t .

47.

48.

49.

50.

51. Circulation Find the circulation ofaround the closed path consisting of the following three curves

traversed in the direction of increasing t .

y

z

x

(1, 0, 0) (0, 1, 0)

0, 1,

C 1 C 2

C 3

2

C 3: r st d = t i + s1 - t d j , 0 t 1 C 2: r st d = j + sp >2ds1 - t dk , 0 t 1 C 1: r st d = scos t di + ssin t d j + t k , 0 t p >2

F = 2 xi + 2 z j + 2 yk 0 t 2pr st d = s- 2 cos t di + s2 sin t d j + 2t k ,

F = - yi + x j + 2k r st d = scos t di + ssin t dk , 0 t pF = s x - z di + xk r st d = 3t j + 4t k , 0 t 1F = x2i + yz j + y 2k r st d = t i + t 2 j + k , 0 t 2F = - 4 xyi + 8 y j + 2k

LC F #T ds = k Csb 2 + ssbdd2d1>2 - sa 2 + ssa dd2d1>2D.

y = s xd

t = b?t = a

LC F #d rF = yi.

a t b,r st d = t i + st d j ,a t b.

F s x, yd Z s0, 0 d,

F = M s x, ydi + N s x, yd j

s x, yd Z s0, 0 d, M s x, ydi + N s x, yd j

F =


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52. Zero circulation Let C be the ellipse in which the planemeets the cylinder Show, with-

out evaluating either line integral directly, that the circulation of

the field around C in either direction is zero.53. Flow along a curve The field is the

velocity field of a flow in space. Find the flow from (0, 0, 0) to(1, 1, 1) along the curve of intersection of the cylinder and the plane ( Hint: Use as the parameter.)

54. Flow of a gradient field Find the flow of the field

a. Once around the curve C in Exercise 52, clockwise as viewed from above

b. Along the line segment from (1, 1, 1) to s2, 1, - 1d.

F = s xy2 z 3d:

y

z

x

(1, 1, 1)

y x 2

z x

t = x z = x . y = x2

F = xyi + y j - yz k F = xi + y j + z k

x2 + y 2 = 12.2 x + 3 y - z = 0


COMPUTER EXPLORATIONSIn Exercises 5560, use a CAS to perform the following steps for finding the work done by force F over the given path:

a. Find d r for the pathb. Evaluate the force F along the path.

c. Evaluate

55.

56.

57.

58.

59.

60.

0 t 2ps2 sin 2 t - 1dk , F = s x2 ydi + 13 x

3 j + xyk ; r st d = scos t di + ssin t d j +r st d = ssin t di + scos t d j + ssin 2t dk , - p >2 t p >2F = s2 y + sin xdi + s z 2 + s1>3dcos yd j + x4 k ; 1 t 4F = 2 xyi - y 2 j + ze x k ; r st d = - t i + 1 t j + 3t k , 0 t 2p

r st d = (2 cos t )i + (3 sin t ) j + k , s z + xy cos xyz dk ; F = s y + yz cos xyz di + s x2 + xz cos xyz d j +0 t p

F = 31 + x2

i + 21 + y 2

j ; r st d = scos t di + ssin t d j , 0 t 2pF = xy6 i + 3 xs xy5 + 2d j ; r st d = s2 cos t di + ssin t d j ,

LC F #d r .r st d = g st di + hst d j + k st dk .

16.3Path Independence, Conservative Fields, and Potential Functions

A gravitational field G is a vector field that represents the effect of gravity at a point inspace due to the presence of a massive object. The gravitational force on a body of mass m

placed in the field is given by F mG . Similarly, an electric field E is a vector field inspace that represents the effect of electric forces on a charged particle placed within it. Theforce on a body of charge q placed in the field is given by F qE . In gravitational and elec-tric fields, the amount of work it takes to move a mass or charge from one point to another depends on the initial and final positions of the objectnot on which path is taken betweenthese positions. In this section we study vector fields with this property and the calculationof work integrals associated with them.

Path Independence

If A and B are two points in an open region D in space, the line integral of F along C from A to B for a field F defined on D usually depends on the path C taken, as we saw in Sec-tion 16.1. For some special fields, however, the integrals value is the same for all pathsfrom A to B.

=

=

DEFINITIONS Let F be a vector f ield defined on an open region D in space,and suppose that for any two points A and B in D the line integral alonga path C from A to B in D is the same over all paths from A to B. Then the integral

is path independent in D and the field F is conservative on D .1C F #d r1C F #d r

The word conservative comes from physics, where it refers to fields in which the principleof conservation of energy holds. When a line integral is independent of the path C from


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16.3 Path Independence, Conservative Fields, and Potential Functions 939

point A to point B, we sometimes represent the integral by the symbol rather than the

usual line integral symbol . This substitution helps us remember the path-independence

property.Under differentiability conditions normally met in practice, we will show that a f ield

F is conservative if and only if it is the gradient field of a scalar function that is, if and only if for some . The function then has a special name.F =

1C 1 B A

DEFINITION If F is a vector field defined on D and for some scalar function on D, then is called a potential function for F .

F =

A gravitational potential is a scalar function whose gradient field is a gravitational field,

an electric potential is a scalar function whose gradient field is an electric field, and so on.As we will see, once we have found a potential function for a f ield F , we can evaluate allthe line integrals in the domain of F over any path between A and B by

(1)

If you think of for functions of several variables as being something like the deriv-ative for functions of a single variable, then you see that Equation (1) is the vector cal-culus analogue of the Fundamental Theorem of Calculus formula

Conservative fields have other remarkable properties. For example, saying that F isconservative on D is equivalent to saying that the integral of F around every closed path in

D is zero. Certain conditions on the curves, fields, and domains must be satisfied for Equation (1) to be valid. We discuss these conditions next.

Assumptions on Curves, Vector Fields, and DomainsIn order for the computations and results we derive below to be valid, we must assume cer-tain properties for the curves, surfaces, domains, and vector fields we consider. We givethese assumptions in the statements of theorems, and they also apply to the examples and

exercises unless otherwise stated.The curves we consider are piecewise smooth . Such curves are made up of finitely

many smooth pieces connected end to end, as discussed in Section 13.1. We will treat vec-tor fields F whose components have continuous f irst partial derivatives.

The domains D we consider are open regions in space, so every point in D is the cen-ter of an open ball that lies entirely in D (see Section 13.1). We also assume D to be conn-ected . For an open region, this means that any two points in D can be joined by a smoothcurve that lies in the region. Finally, we assume D is simply connected , which means thatevery loop in D can be contracted to a point in D without ever leaving D. The plane with adisk removed is a two-dimensional region that is not simply connected; a loop in the planethat goes around the disk cannot be contracted to a point without going into the hole left

by the removed disk (see Figure 16.22c). Similarly, if we remove a line from space, the re-maining region D is not simply connected. A curve encircling the line cannot be shrunk toa point while remaining inside D.

Lb

a s xd dx = sbd - sa d.

L B

A F #d r = L

B

A #d r = s Bd - s Ad.


22/42


THEOREM 1Fundamental Theorem of Line Integrals Let C be a smooth curve joining the point A to the point B in the plane or in space and parametrized byr (t ). Let be a differentiable function with a continuous gradient vector Fon a domain D containing C . Then

LC F #d r = ( B) - ( A).

=

Connectivity and simple connectivity are not the same, and neither property impliesthe other. Think of connected regions as being in one piece and simply connected re-gions as not having any loop-catching holes. All of space itself is both connected and simply connected. Figure 16.22 illustrates some of these properties.

Caution Some of the results in this chapter can fail to hold if applied to situations wherethe conditions weve imposed do not hold. In particular, the component test for conservativefields, given later in this section, is not valid on domains that are not simply connected (seeExample 5).

Line Integrals in Conservative FieldsGradient fields F are obtained by differentiating a scalar function . A theorem analogousto the Fundamental Theorem of Calculus gives a way to evaluate the line integrals of gradient fields.

Like the Fundamental Theorem, Theorem 1 gives a way to evaluate line integrals with-out having to take limits of Riemann sums or finding the line integral by the procedureused in Section 16.2. Before proving Theorem 1, we give an example.

EXAMPLE 1 Suppose the force field F is the gradient of the function

Find the work done by F in moving an object along a smooth curve C joining (1, 0, 0) to(0, 0, 2) that does not pass through the origin.

Solution An application of Theorem 1 shows that the work done by F along any smoothcurve C joining the two points and not passing through the origin is

The gravitational force due to a planet, and the electric force associated with acharged particle, can both be modeled by the field F given in Example 1 up to a constantthat depends on the units of measurement.

Proof of Theorem 1 Suppose that A and B are two points in region D and thatis a smooth curve in D joining A to B.C : r st d = g st di + hst d j + k st dk , a t b,

LC F #d r = (0, 0, 2) - (1, 0, 0) = - 14 - ( - 1) = 34.

( x, y, z ) = - 1 x2 + y 2 + z 2

.

=

y

x

(a)

Simply connected

b

Simply connected

z

y

x

y

x

C 1

(c)

Not simply connected

z

y

x

C 2

(d)

Not simply connected

FIGURE 16.22 Four connected regions.In (a) and (b), the regions are simplyconnected. In (c) and (d), the regions arenot simply connected because the curves

and cannot be contracted to a pointinside the regions containing them.

C 2C 1


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We use the abbreviated form for the parametrization of the curve.Along the curve, is a differentiable function of t and

Therefore,

So we see from Theorem 1 that the line integral of a gradient f ield F is straight-forward to compute once we know the function . Many important vector fields arising inapplications are indeed gradient fields. The next result, which follows from Theorem 1,shows that any conservative field is of this type.

=

= s g st d, hst d, k st ddda

b

= s Bd - s Ad.

r (a) = A , r (b) = BLC F #d r = Lt = b

t = a F #d rdt dt = L

b

a d dt

dt

= # adxdt i + dydt j + dz dt k b = #d rdt = F #d rdt . d dt

= 00 x dxdt

+ 00 y dydt

+ 00 z dz dt

r (t ) = xi + y j + z k


Chain Rule in Section 14.4with z = k st d

y = hst d, x = g st d,

Because F =

THEOREM 2Conservative Fields are Gradient Fields Let be a vector field whose components are continuous throughout an open connected region D in space. Then F is conservative if and only if F is a gradient fieldfor a differentiable function .

F = M i + N j + P k

Theorem 2 says that F if and only if for any two points A and B in the region D,the value of line integral is independent of the path C joining A to B in D.

Proof of Theorem 2 If F is a gradient field, then F for a differentiable function ,and Theorem 1 shows that The value of the line integral doesnot depend on C , but only on its endpoints A and B. So the line integral is path independ-ent and F satisfies the definition of a conservative field.

On the other hand, suppose that F is a conservative vector field. We want to find afunction on D satisfying . First, pick a point A in D and set For anyother point B in D define to equal where C is any smooth path in D from Ato B. The value of does not depend on the choice of C , since F is conservative. To showthat we need to demonstrate that , and

Suppose that B has coordinates ( x, y, z ). By definition, the value of the function ata nearby point located at is , where is any path from A to . Wetake a path from A to B formed by first traveling along to arrive atand then traveling along the line segment L from to B (Figure 16.23). When isclose to B, the segment L lies in D and, since the value is independent of the pathfrom A to B,

Differentiating, we have

00 x ( x, y, z ) =

00 x a

LC 0

F #d r +

L L F #d r b.

( x, y, z ) = LC 0 F #d r + L L F #d r .( B)

B0 B0 B0C 0C = C 0 h L B0C 01C 0 F

#d r( x0, y, z ) B0

0

>0 z = P .0

>0 x = M , 0

>0 y = N = F

( B)1C F #d r ,( B) ( A)

= 0. = F

1C F #d r = ( B) - ( A).=

1C F #d r=

z

y

x

B L

A

D

B0

x 0

x

C 0( x 0, y, z)

( x , y, z)

FIGURE 16.23 The functionin the proof of Theorem 2 is computed

by a line integralfrom A to , plus a line integralalong a line segment L parallel to the

x-axis and joining to B located at( x, y, z ). The value of at A is ( A) = 0.

B0

1 L F #d r B01C 0 F #d r = ( B0)

( x, y, z )


24/42

Only the last term on the right depends on x, so

Now parametrize L as Then ,and Substitution gives

by the Fundamental Theorem of Calculus. The partial derivatives and follow similarly, showing that

EXAMPLE 2 Find the work done by the conservative field where

along any smooth curve C joining the point A to B

Solution With we have

A very useful property of line integrals in conservative fields comes into play when the path of integration is a closed curve, or loop. We often use the notation for integrationaround a closed path (discussed with more detail in the next section).

DC

= - 24 + 27 = 3.

= s1ds6ds- 4d - s- 1ds3ds9d

= xyz s1,6, - 4d - xyz s- 1,3,9 d

= s Bd - s Ad

LC F #d r = L B

A #d r

s x, y, z d = xyz ,

s1, 6, - 4d.s - 1, 3, 9 d

( x, y, z ) = xyz ,F = yz i + xz j + xyk = ,

F = .0>0 z = P 0>0 y = N

00 x ( x, y, z ) =

00 x L

x

x0

M (t , y, z ) dt = M ( x, y, z )

1 L F #d r = 1 x x0 M (t , y, z ) dt .F #d r>dt = M d r>dt = i,r (t ) = t i + y j + z k , x0 t x.

0

0 x ( x, y, z ) =

0

0 x

L L F #d r .


and pathindependenceF =

Theorem 1

THEOREM 3Loop Property of Conservative Fields The following statementsare equivalent.

1. around every loop (that is, closed curve C ) in D.

2. The field F is conservative on D.

DC

F #d r = 0

Proof that Part 1 Part 2 We want to show that for any two points A and B in D, theintegral of has the same value over any two paths and from A to B. We reversethe direction on to make a path from B to A (Figure 16.24). Together, andmake a closed loop C , and by assumption,

Thus, the integrals over and give the same value. Note that the definition ofshows that changing the direction along a curve reverses the sign of the line integral.

F #d rC 2C 1

LC 1 F #d r - LC 2 F #d r = LC 1 F #d r + L- C 2 F #d r = LC F #d r = 0.

- C 2C 1- C 2C 2C 2C 1F #d r

Q

A

B

A

B

C 1 C 1

C 2 C 2

FIGURE 16.24 If we have two paths from A to B, one of them can be reversed tomake a loop.


25/42

Proof that Part 2 Part 1 We want to show that the integral of is zero over anyclosed loop C . We pick two points A and B on C and use them to break C into two pieces:

from A to B followed by from B back to A (Figure 16.25). Then

The following diagram summarizes the results of Theorems 2 and 3.

Theorem 2 Theorem 3

Two questions arise:

1. How do we know whether a given vector field F is conservative?

2. If F is in fact conservative, how do we find a potential function (so that )?

Finding Potentials for Conservative Fields

The test for a vector field being conservative involves the equivalence of certain partialderivatives of the f ield components.

F =

F = on D 3 F conservative 3 FC

F #d r = 0on D

over any loop in D

FC

F #d r = LC 1 F #d r + LC 2 F #d r = L B

A F #d r - L

B

A F #d r = 0.

C 2C 1

F #d rQ


Mixed Derivative Theorem,Section 14.3

A

B

A

B

C 2

C 1

C 2

C 1

FIGURE 16.25 If A and B lie on a loop,we can reverse part of the loop to maketwo paths from A to B.

Component Test for Conservative FieldsLet be a field on a connected and simply connected domain whose component functions have continuous first partialderivatives. Then, F is conservative if and only if

(2)0 P 0 y =

0 N 0 z ,

0 M 0 z =

0 P 0 x , and 0 N 0 x = 0 M 0 y .

F = M s x, y, z di + N s x, y, z d j + P s x, y, z dk

Proof that Equations (2) hold if F is conservative There is a potential function suchthat

Hence,

The others in Equations (2) are proved similarly.

The second half of the proof, that Equations (2) imply that F is conservative, is a con-sequence of Stokes Theorem, taken up in Section 16.7, and requires our assumption thatthe domain of F be simply connected.

= 00 z a00 yb = 0 N 0 z .

= 020 z 0 y

0 P 0 y =

00 y a00 z b = 0

20 y 0 z

F = M i + N j + P k = 00 x i +

00 y j +

00 z k .


26/42

Once we know that F is conservative, we usually want to find a potential function for F .This requires solving the equation or

for . We accomplish this by integrating the three equations

as illustrated in the next example.

EXAMPLE 3 Show that is con-servative over its natural domain and find a potential function for it.

Solution The natural domain of F is all of space, which is connected and simply con-nected. We apply the test in Equations (2) to

and calculate

The partial derivatives are continuous, so these equalities tell us that F is conservative, sothere is a function with (Theorem 2).

We find by integrating the equations

(3)

We integrate the first equation with respect to x, holding y and z fixed, to get

We write the constant of integration as a function of y and z because its value may depend on y and z , though not on x. We then calculate from this equation and match it withthe expression for in Equations (3). This gives

so Therefore, g is a function of z alone, and

We now calculate from this equation and match it to the formula for in Equa-tions (3). This gives

so

hs z d = z 2

2 + C .

xy + dhdz = xy + z ,

or dh

dz = z ,

0>0 z 0>0 z s x, y, z d = e x cos y + xyz + hs z d.

0 g >0 y = 0.- e x sin y + xz +

0 g

0 y = xz - e x sin y,

0>0 y0>0 y

s x, y, z d = e x cos y + xyz + g s y, z d.

00 x = e

x cos y + yz ,00 y = xz - e

x sin y,00 z = xy + z .

= F

0 P 0 y = x =

0 N 0 z ,

0 M 0 z = y =

0 P 0 x ,

0 N 0 x = - e

x sin y + z = 0 M 0 y .

M = e x cos y + yz , N = xz - e x sin y, P = xy + z

F = se x cos y + yz di + s xz - e x sin yd j + s xy + z dk

00 x = M ,

00 y = N ,

00 z = P ,

00 x i +

00 y j +

00 z k = M i + N j + P k

= F



27/42

Hence,

We have infinitely many potential functions of F , one for each value of C .

EXAMPLE 4 Show that is not conservative.

Solution We apply the component test in Equations (2) and find immediately that

The two are unequal, so F is not conservative. No further testing is required.

EXAMPLE 5 Show that the vector field

satisfies the equations in the Component Test, but is not conservative over its natural do-main. Explain why this is possible.

Solution We have and If we apply theComponent Test, we find

So it may appear that the field F passes the Component Test. However, the test assumesthat the domain of F is simply connected, which is not the case. Since cannotequal zero, the natural domain is the complement of the z -axis and contains loops that can-not be contracted to a point. One such loop is the unit circle C in the xy-plane. The circle is

parametrized by This loop wraps around the z -axisand cannot be contracted to a point while staying within the complement of the z -axis.

To show that F is not conservative, we compute the line integral around theloop C . First we write the field in terms of the parameter t :

Next we find and then calculate the line integral as

Since the line integral of F around the loop C is not zero, the field F is not conservative, byTheorem 3.

Example 5 shows that the Component Test does not apply when the domain of the field is not simply connected. However, if we change the domain in the example so that it is re-stricted to the ball of radius 1 centered at the point (2, 2, 2), or to any similar ball-shaped re-gion which does not contain a piece of the z -axis, then this new domain D is simply connected.

Now the partial derivative Equations (2), as well as all the assumptions of the ComponentTest, are satisfied. In this new situation, the field F in Example 5 is conservative on D.

FC

F #d r = FC

F #d rdt dt = L2p

0 Asin2 t + cos 2 t B dt = 2p .

d r>dt = ( - sin t )i + (cos t ) j ,

F = - y

x2 + y 2 i + x

x2 + y 2 j =

- sin t sin2 t + cos2 t

i + cos t sin2 t + cos2 t

j = (- sin t )i + (cos t ) j .

DC

F #d rr (t ) = (cos t )i + (sin t ) j , 0 t 2p .

x2 + y 2

0 P 0 y = 0 =

0 N 0 z ,

0 P 0 x = 0 =

0 M 0 z , and

0 M 0 y =

y 2 - x2

( x2 + y 2)2 = 0 N 0 x .

P = 0. M = - y>( x2 + y 2), N = x>( x2 + y 2),

F = - y

x2 + y 2 i + x

x2 + y 2 j + 0k

0 P 0 y =

00 y scos z d = 0,

0 N 0 z =

00 z s - z d = - 1.

F = s2 x - 3di - z j + scos z dk

s x, y, z d = e x cos y + xyz + z 2

2 + C .



28/42

Just as we must be careful with a function when determining if it satisfies a propertythroughout its domain (like continuity or the intermediate value property), so must we also

be careful with a vector field in determining the properties it may or may not have over itsassigned domain.

Exact Differential FormsIt is often convenient to express work and circulation integrals in the differential form

discussed in Section 16.2. Such line integrals are relatively easy to evaluate if is the total differential of a function and C is any path joining the

two points from A to B. For then

is conservative.

Theorem 1

Thus,

just as with differentiable functions of a single variable.

L B

A df = s Bd - s Ad,

= s Bd - s Ad.

= L B

A #d r

LC M dx + N dy + P dz = LC 00 x dx +

00 y dy +

00 z dz

M dx + N dy + P dz

LC M dx + N dy + P dz


DEFINITIONS Any expressionis a differential form . A differential form is exact on a domain D in space if

for some scalar function throughout D.

M dx + N dy + P dz = 0 f 0 x dx +

00 y dy +

00 z dz = d

M s x, y, z d dx + N s x, y, z d dy + P s x, y, z d dz

Component Test for Exactness ofThe differential form is exact on a connected and simplyconnected domain if and only if

This is equivalent to saying that the field is conservative.F = M i + N j + P k

0 P 0 y =

0 N 0 z ,

0 M 0 z =

0 P 0 x , and

0 N 0 x =

0 M 0 y .

M dx + N dy + P dz M dx + N dy + P dz

Notice that if on D, then is the gra-dient field of on D. Conversely, if then the form is exact.The test for the forms being exact is therefore the same as the test for F being conservative.

M dx + N dy + P dz F = ,F = M i + N j + P k M dx + N dy + P dz = d


29/42


Exercises 16.3

Testing for Conservative FieldsWhich fields in Exercises 16 are conservative, and which are not?

1.

2.

3.

4.

5.

6.

Finding Potential FunctionsIn Exercises 712, find a potential function for the field F .

7. F = 2 xi + 3 y j + 4 z k

F = se x cos ydi - se x sin yd j + z k F = s z + ydi + z j + s y + xdk F = - yi + x jF = yi + s x + z d j - yk F = s y sin z di + s x sin z d j + s xy cos z dk F = yz i + xz j + xyk

8.

9.

10.

11.

12.

a y

2 1 - y 2 z 2+ 1

z bk

F = y

1 + x2 y 2 i + a x1 + x2 y 2 + z 2 1 - y 2 z 2b j +

asec 2s x + yd + y y 2 + z 2b j + z y 2 + z 2 k F = sln x + sec 2s x + yddi +

F = s y sin z di + s x sin z d j + s xy cos z dk

F = e y + 2 z si + x j + 2 xk d

F = s y + z di + s x + z d j + s x + ydk

EXAMPLE 6 Show that is exact and evaluate the integral

over any path from (1, 1, 1) to

Solution We let and apply the Test for Exactness:

These equalities tell us that is exact, so

for some function , and the integrals value isWe find up to a constant by integrating the equations

(4)

From the first equation we get

The second equation tells us that

Hence, g is a function of z alone, and

The third of Equations (4) tells us that

Therefore,

The value of the line integral is independent of the path taken from (1, 1, 1) to (2, 3, ,and equals

s2, 3, - 1d - s1, 1, 1 d = 2 + C - s5 + C d = - 3.

- 1)s x, y, z d = xy + 4 z + C .

00 z = 0 +

dhdz

= 4, or hs z d = 4 z + C .s x, y, z d = xy + hs z d.

00 y = x +

0 g 0 y = x, or

0 g 0 y = 0.

s x, y, z d = xy + g s y, z d.

00 x = y,

00 y = x,

00 z = 4.

s2, 3, - 1d - s1, 1, 1 d.

y dx + x dy + 4 dz = d

y dx + x dy + 4 dz

0 P 0 y = 0 =

0 N 0 z ,

0 M 0 z = 0 =

0 P 0 x ,

0 N 0 x = 1 =

0 M 0 y .

M = y, N = x, P = 4

s2, 3, - 1d.Ls2,3, - 1d

s1,1,1 d y dx + x dy + 4 dz

y dx + x dy + 4 dz


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Exact Differential FormsIn Exercises 1317, show that the differential forms in the integralsare exact. Then evaluate the integrals.

13.

14.

15.

16.

17.

Finding Potential Functions to Evaluate Line Integrals

Although they are not defined on all of space the fields associated with Exercises 1822 are simply connected and the Component Testcan be used to show they are conservative. Find a potential functionfor each field and evaluate the integrals as in Example 6.

18.

19.

20.

21.

22.

Applications and Examples23. Revisiting Example 6 Evaluate the integral

from Example 6 by f inding parametric equations for the line seg-ment from (1, 1, 1) to and evaluating the line integralof along the segment. Since F is conservative,the integral is independent of the path.

24. Evaluate

along the line segment C joining (0, 0, 0) to (0, 3, 4).

Independence of path Show that the values of the integrals inExercises 25 and 26 do not depend on the path taken from A to B.

25.

26.

In Exercises 27 and 28, f ind a potential function for F .

27.

28. F = se x ln ydi + ae x

y + sin z b j + s y cos z dk F = 2 x y i + a1 - x2 y 2 b j , {( x, y): y 7 0}

L B

A x dx + y dy + z dz

2 x2 + y 2 + z 2

L B

A z 2 dx + 2 y dy + 2 xz dz

LC x2 dx + yz dy + s y 2

>2d dz

F = yi + x j + 4k s2, 3, - 1d

Ls2,3, - 1d

s1,1,1 d y dx + x dy + 4 dz

Ls2,2,2 d

s- 1, - 1, - 1d 2 x dx

+2 y dy

+2 z dz

x2 + y 2 + z 2

Ls2,2,2 d

s1,1,1 d 1 y dx + a1 z - x y 2b dy -

y

z 2 dz

Ls2,1,1 d

s1,2,1 d s2 x ln y - yz d dx + a x2 y - xz b dy - xy dz

Ls1,2,3 d

s1,1,1 d 3 x2 dx + z

2

y dy + 2 z ln y dz

Ls1,p >2,2d

s0,2,1 d 2 cos y dx + a1 y - 2 x sin yb dy + 1 z dz

R3

,

Ls0,1,1 d

s1,0,0 d sin y cos x dx + cos y sin x dy + dz

Ls3,3,1 d

s0,0,0 d 2 x dx - y 2 dy - 4

1 + z 2 dz

Ls1,2,3 d

s0,0,0 d 2 xy dx + s x2 - z 2d dy - 2 yz dz

Ls3,5,0 d

s1,1,2 d yz dx + xz dy + xy dz

Ls2,3, - 6ds0,0,0 d 2 x dx + 2 y dy + 2 z dz


29. Work along different paths Find the work done byover the following paths from

(1, 0, 0) to (1, 0, 1).

a. The line segmentb. The helix

c. The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabolafrom (0, 0, 0) to (1, 0, 1)

30. Work along different paths Find the work done byover the following

paths from (1, 0, 1) to

a. The line segment

b. The line segment from (1, 0, 1) to the origin followed by theline segment from the origin to

c. The line segment from (1, 0, 1) to (1, 0, 0), followed by the x-axis from (1, 0, 0) to the origin, followed by the parabola

from there to

z

y

x

(1, 0, 1)

(1, 0, 0)

(0, 0, 0)

1

p

21, , 0

y 5 x 2p

2

s1, p >2, 0 d y = p x2>2, z = 0

z

y

x

(1, 0, 1)

(0, 0, 0)

1

1

p

2

p

21, , 0

s1, p >2, 0 d

z

y

x

(1, 0, 1)

1

p

2

p

21, , 01

x = 1, y = p t >2, z = 1 - t , 0 t 1s1, p >2, 0 d.

e yz i + s xze yz + z cos yd j + s xye yz + sin ydk F =

z

y

x

(1, 0, 1)

(0, 0, 0)

1

(1, 0, 0)

z 5 x 2

z = x2, y = 0

r st d = scos t di + ssin t d j + st >2p dk , 0 t 2p x = 1, y = 0, 0 z 1

s x2 + ydi + s y 2 + xd j + ze z k F =


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31. Evaluating a work integral two ways Let and letC be the path in the xy-plane from to (1, 1) that consistsof the line segment from to (0, 0) followed by the line

segment from (0, 0) to (1, 1). Evaluate in two ways.a. Find parametrizations for the segments that make up C and

evaluate the integral.

b. Use as a potential function for F .

32. Integral along different paths Evaluate the line integralalong the following paths C in the

xy-plane.

a. The parabola from (1, 0) to (0, 1)

b. The line segment from to (1, 0)

c. The x-axis from to (1, 0)

d. The astroidcounterclockwise from (1, 0) back to (1, 0)

33. a. Exact differential form How are the constants a, b , and crelated if the following differential form is exact?

b. Gradient field For what values of b and c will

be a gradient field?

F = s y 2 + 2czxdi + ysbx + cz d j + s y 2 + cx 2dk

say 2 + 2czxd dx + ysbx + cz d dy + say 2 + cx 2d dz

x

y

(0, 1)

(0, 1)

(1, 0)(1, 0)

r st d = scos 3 t di + ssin3 t d j , 0 t 2p ,s - 1, 0 d

s - 1, p d y = s x - 1d2

sin y dy1C 2 x cos y dx - x2s x, yd = x3 y 2

1C F #d rs - 1, 1 d

s - 1, 1 dF = s x3 y 2d

16.4 Greens Theorem in

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Thomas Calculus !6th Chapter