16.1 Line Integrals 923 - University of California, Davis

[Pages:25]16.1 Line Integrals 923

The formulas in Table 16.1 then give

p

M = d ds = s2 - zd ds = s2 - sin td dt = 2p - 2

LC

LC

L0

p

Mxy = zd ds = zs2 - zd ds = ssin tds2 - sin td dt

LC

LC

L0

=

p

s2 sin t L0

-

sin2 td dt

=

8

2

p

# z

=

Mxy M

=

8

2

p

1 2p -

2

=

8-p 4p - 4

L

0.57.

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

z height f (x, y)

y

t

a (x,

y)

sk Plane curve C

x

t b

FIGURE 16.5 The line integral 1C ds gives the area of the portion of the

cylindrical surface or "wall" beneath z = sx, yd ? 0.

Line Integrals in the Plane

There is an interesting geometric interpretation for line integrals in the plane. If C is a smooth curve in the xy-plane parametrized by rstd = xstdi + ystdj, a ... t ... b, we generate a cylindrical surface by moving a straight line along C orthogonal to the plane, holding the line parallel to the z-axis, as in Section 12.6. If z = sx, yd is a nonnegative continuous function over a region in the plane containing the curve C, then the graph of is a surface

that lies above the plane. The cylinder cuts through this surface, forming a curve on it that

lies above the curve C and follows its winding nature. The part of the cylindrical surface

that 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 the curve, the height of the wall is sx, yd. We show the wall in Figure 16.5, where the "top" of the wall is the curve lying on the surface z = sx, yd. (We do not display the surface formed by the graph of in the figure, only the curve on it that is cut out by the cylinder.)

From the definition

n

ds LC

=

lim

n:q

a

k=1

sxk,

ykd

?sk,

where ?sk : 0 as n : q , we see that the line integral 1C ds is the area of the wall shown in the figure.

Exercises 16.1

Graphs of Vector Equations Match the vector equations in Exercises 1?8 with the graphs (a)?(h) given here.

a.

b.

z

z

?1

2

1 y

x

1

y

x

c. z

1

1

x

d. z

(2, 2, 2)

y

2

x

2 y

924

Chapter 16: Integration in Vector Fields

e. z

f. z 2

(1, 1, 1)

?1

1

1

y

y

x

x (1, 1, ?1)

?2

g. z

h. z

2

2

?2

2

y

2

y

x

x

1. rstd = ti + s1 - tdj, 0 ... t ... 1 2. rstd = i + j + t k, - 1 ... t ... 1 3. rstd = s2 cos tdi + s2 sin tdj, 0 ... t ... 2p 4. rstd = ti, - 1 ... t ... 1 5. rstd = ti + tj + tk, 0 ... t ... 2 6. rstd = t j + s2 - 2tdk, 0 ... t ... 1 7. rstd = st2 - 1dj + 2tk, - 1 ... t ... 1 8. rstd = s2 cos tdi + s2 sin tdk, 0 ... t ... p

Evaluating Line Integrals over Space Curves 9. Evaluate 1C sx + yd ds where C is the straight-line segment

x = t, y = s1 - td, z = 0, from (0, 1, 0) to (1, 0, 0). 10. Evaluate 1C sx - y + z - 2d ds where C is the straight-line seg-

ment x = t, y = s1 - td, z = 1, from (0, 1, 1) to (1, 0, 1). 11. Evaluate 1C sxy + y + zd ds along the curve rstd = 2ti +

t j + s2 - 2tdk, 0 ... t ... 1. 12. Evaluate 1C 2x2 + y 2 ds along the curve rstd = s4 cos tdi +

s4 sin tdj + 3tk, -2p ... t ... 2p. 13. Find the line integral of sx, y, zd = x + y + z over the straight-

line segment from (1, 2, 3) to s0, -1, 1d. 14. Find the line integral of sx, y, zd = 23>sx2 + y 2 + z2d over

the curve rstd = ti + tj + tk, 1 ... t ... q. 15. Integrate sx, y, zd = x + 1y - z2 over the path from (0, 0, 0)

to (1, 1, 1) (see accompanying figure) given by

C1: rstd = ti + t 2j, 0 ... t ... 1 C2: rstd = i + j + tk, 0 ... t ... 1

z

z

(0, 0, 1)

C2

(0, 1, 1)

(0, 0, 0)

(1, 1, 1)

y

C2

x

C1

(1, 1, 0)

(a)

C1 (0, 0, 0)

x

C3 (1, 1, 1)

y

(b)

The paths of integration for Exercises 15 and 16.

16. Integrate sx, y, zd = x + 1y - z2 over the path from (0, 0, 0) to (1, 1, 1) (see accompanying figure) given by

C1: rstd = tk, 0 ... t ... 1 C2: rstd = tj + k, 0 ... t ... 1 C3: rstd = ti + j + k, 0 ... t ... 1

17. Integrate sx, y, zd = sx + y + zd>sx2 + y 2 + z2d over the path rstd = ti + tj + tk, 0 6 a ... t ... b.

18. Integrate sx, y, zd = - 2x2 + z2 over the circle

rstd = sa cos tdj + sa sin tdk, 0 ... t ... 2p.

Line Integrals over Plane Curves

19. Evaluate 1C x ds, where C is a. the straight-line segment x = t, y = t>2, from (0, 0) to (4, 2).

b. the parabolic curve x = t, y = t2, from (0, 0) to (2, 4).

20. Evaluate 1C 2x + 2y ds, where C is a. the straight-line segment x = t, y = 4t, from (0, 0) to (1, 4).

b. C1 ? C2; C1 is the line segment from (0, 0) to (1, 0) and C2 is the line segment from (1, 0) to (1, 2).

21. Find the line integral of sx, yd = yex2 along the curve rstd = 4ti - 3tj, -1 ... t ... 2.

22. Find the line integral of sx, yd = x - y + 3 along the curve rstd = (cos t)i + (sin t)j, 0 ... t ... 2p.

23.

Evaluate

LC

x2 y 4>3

ds,

where

C

is

the

curve

x

=

t 2, y

=

t 3,

for

1 ... t ... 2.

24. Find the line integral of sx, yd = 2y>x along the curve rstd = t 3i + t 4j, 1>2 ... t ... 1.

25. Evaluate 1C Ax + 2yB ds where C is given in the accompanying

figure.

y

(0, 0)

(1, 1) C y 5 x y 5 x2

x

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

925

26.

Evaluate figure.

LC

x2

+

1 y2

+

ds where C is given in the accompanying 1

y

(0, 1)

(1, 1)

(0, 0)

x (1, 0)

In Exercises 27?30, integrate over the given curve.

27. sx, yd = x3>y, C: y = x2>2, 0 ... x ... 2 28. sx, yd = sx + y 2d> 21 + x2, C: y = x2>2 from (1, 1>2) to

(0, 0)

29. sx, yd = x + y, C: x2 + y 2 = 4 in the first quadrant from (2, 0) to (0, 2)

30. sx, yd = x2 - y, C: x2 + y 2 = 4 in the first quadrant from (0, 2) to s 12, 12d

31. Find the area of one side of the "winding wall" standing orthogonally on the curve y = x2, 0 ... x ... 2, and beneath the curve on the surface sx, yd = x + 2y.

32. Find the area of one side of the "wall" standing orthogonally on the curve 2x + 3y = 6, 0 ... x ... 6, and beneath the curve on the surface sx, yd = 4 + 3x + 2y.

Masses and Moments 33. Mass of a wire Find the mass of a wire that lies along the curve

rstd = st2 - 1dj + 2tk, 0 ... t ... 1, if the density is d = s3>2dt.

34. Center of mass of a curved wire A wire of density dsx, y, zd = 15 2y + 2 lies along the curve rstd = st2 - 1dj + 2tk, -1 ... t ... 1. Find its center of mass. Then sketch the curve and center of mass together.

35. Mass of wire with variable density Find the mass of a thin wire lying along the curve rstd = 22ti + 22tj + s4 - t 2dk, 0 ... t ... 1, if the density is (a) d = 3t and (b) d = 1.

36. Center of mass of wire with variable density Find the center of mass of a thin wire lying along the curve rstd = ti + 2tj + s2>3dt 3>2k, 0 ... t ... 2, if the density is d = 3 15 + t.

37. Moment of inertia of wire hoop A circular wire hoop of constant density d lies along the circle x 2 + y 2 = a 2 in the xy-plane. Find the hoop's moment of inertia about the z-axis.

38. Inertia of a slender rod A slender rod of constant density lies along the line segment rstd = tj + s2 - 2tdk, 0 ... t ... 1, in the

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

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

rstd = scos tdi + ssin tdj + tk, 0 ... t ... 2p.

a. Find Iz.

b. Suppose that you have another spring of constant density d that is twice as long as the spring in part (a) and lies along the helix for 0 ... t ... 4p. Do you expect Iz for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calculating Iz for the longer spring.

40. Wire of constant density A wire of constant density d = 1 lies along the curve

rstd = st cos tdi + st sin tdj + A 2 22>3 B t 3>2k, 0 ... t ... 1.

Find z and Iz.

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

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

rstd

=

ti

+

2

22 3

t

3>2j

+

t2 2

k,

if the density is d = 1>st + 1d.

0 ... t ... 2,

COMPUTER EXPLORATIONS In Exercises 43?46, use a CAS to perform the following steps to evaluate the line integrals.

a. Find ds = vstd dt for the path rstd = gstdi + hstdj + kstdk.

b. Express the integrand sgstd, hstd, kstdd vstd as a function of the parameter t.

c. Evaluate 1C ds using Equation (2) in the text.

43. sx, y, zd = 21 + 30x2 + 10y ; rstd = ti + t 2j + 3t 2k, 0...t...2

44. sx, y, zd = 21 + x3 + 5y 3 ;

rstd

=

ti

+

1 3

t

2j

+

1tk,

0...t...2

45. sx, y, zd = x 1y - 3z2 ; rstd = scos 2tdi + ssin 2tdj + 5tk, 0 ... t ... 2p

46.

sx, y, zd

=

a1

+

9 4

1>4

z1>3 b ;

rstd = scos 2tdi + ssin 2tdj +

t 5>2k, 0 ... t ... 2p

16.2

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

Gravitational and electric forces have both a direction and a magnitude. They are represented by a vector at each point in their domain, producing a vector field. In this section we show how to compute the work done in moving an object through such a field by using a line integral involving the vector field. We also discuss velocity fields, such as the vector

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

935

Calculating Flux Across a Smooth Closed Plane Curve

sFlux of F = Mi + Nj across Cd = M dy - N dx

(7)

F

C

The integral can be evaluated from any smooth parametrization x = gstd, y = hstd,

a ... t ... b, that traces C counterclockwise exactly once.

EXAMPLE 8 Find the flux of F = sx - ydi + xj across the circle x2 + y2 = 1 in the xy-plane. (The vector field and curve were shown previously in Figure 16.19.)

Solution The parametrization rstd = scos tdi + ssin tdj, 0 ... t ... 2p, traces the circle counterclockwise exactly once. We can therefore use this parametrization in Equation (7). With

M = x - y = cos t - sin t, dy = dssin td = cos t dt

N = x = cos t,

dx = dscos td = - sin t dt,

we find

2p

Flux = M dy - N dx =

scos2 t - sin t cos t + cos t sin td dt

LC

L0

=

2p

cos2 t dt L0

=

2p 1 L0

+

cos 2

2t

dt

=

c2t

+

sin 4

2t

d

2p 0

=

p.

Eq. (7)

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

Exercises 16.2

Vector Fields Find the gradient fields of the functions in Exercises 1?4. 1. sx, y, zd = sx2 + y 2 + z2d-1>2 2. sx, y, zd = ln 2x2 + y 2 + z2 3. gsx, y, zd = ez - ln sx2 + y 2d 4. gsx, y, zd = xy + yz + xz 5. Give a formula F = Msx, ydi + Nsx, ydj for the vector field in

the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from (x, y) to the origin. (The field is not defined at (0, 0).)

6. Give a formula F = Msx, ydi + Nsx, ydj for the vector field in the plane that has the properties that F = 0 at (0, 0) and that at any other point (a, b), F is tangent to the circle x 2 + y 2 = a2 + b 2 and points in the clockwise direction with magnitude F = 2a2 + b 2.

Line Integrals of Vector Fields In Exercises 7?12, find the line integrals of F from (0, 0, 0) to (1, 1, 1) over each of the following paths in the accompanying figure.

a. The straight-line path C1: rstd = ti + tj + tk, 0 ... t ... 1 b. The curved path C2: rstd = ti + t 2j + t 4k, 0 ... t ... 1

c. The path C3 ? C4 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. F = 3yi + 2xj + 4zk

8. F = [1>sx2 + 1d]j

9. F = 1zi - 2xj + 1yk 10. F = xyi + yzj + xzk

11. F = s3x2 - 3xdi + 3zj + k

12. F = s y + zdi + sz + xdj + sx + ydk

z

(0, 0, 0)

C1

(1, 1, 1)

C2

C4

y

C3

x (1, 1, 0)

936

Chapter 16: Integration in Vector Fields

Line Integrals with Respect to x, y, and z In Exercises 13?16, find the line integrals along the given path C.

13. (x - y) dx, where C: x = t, y = 2t + 1, for 0 ... t ... 3 LC

14.

LC

x y

dy,

where

C:

x

=

t, y

=

t 2, for 1

...

t

...

2

15. (x2 + y 2) dy, where C is given in the accompanying figure. LC y

(0, 0)

(3, 3) C

x (3, 0)

16. 2x + y dx, where C is given in the accompanying figure. LC y

C

(0, 3)

(1, 3)

y 5 3x

x (0, 0)

17. Along the curve r(t) = ti - j + t2k, 0 ... t ... 1, evaluate each of the following integrals.

a. (x + y - z) dx LC

b. (x + y - z) dy LC

c. (x + y - z) dz LC

18. Along the curve r(t) = (cos t)i + (sin t)j - (cos t)k, 0 ... t ... p, evaluate each of the following integrals.

a. xz dx b. xz dy c. xyz dz

LC

LC

LC

Work In Exercises 19?22, find the work done by F over the curve in the direction of increasing t.

19. F = xyi + yj - yzk rstd = ti + t 2j + tk, 0 ... t ... 1

20. F = 2yi + 3xj + sx + ydk rstd = scos tdi + ssin tdj + st>6dk, 0 ... t ... 2p

21. F = zi + xj + yk rstd = ssin tdi + scos tdj + tk, 0 ... t ... 2p

22. F = 6zi + y 2j + 12xk rstd = ssin tdi + scos tdj + st>6dk, 0 ... t ... 2p

Line Integrals in the Plane 23. Evaluate 1C xy dx + sx + yd dy along the curve y = x2 from

s -1, 1d to (2, 4). 24. Evaluate 1C sx - yd dx + sx + yd dy counterclockwise around

the triangle with vertices (0, 0), (1, 0), and (0, 1).

# 25. Evaluate 1C F T ds for the vector field F = x2i - yj along the

curve x = y 2 from (4, 2) to s1, - 1d .

26. Evaluate 1C F # dr for the vector field F = yi - xj counter-

clockwise along the unit circle x2 + y 2 = 1 from (1, 0) to (0, 1).

Work, Circulation, and Flux in the Plane 27. Work Find the work done by the force F = xyi + sy - xdj

over the straight line from (1, 1) to (2, 3). 28. Work Find the work done by the gradient of sx, yd = sx + yd2

counterclockwise around the circle x2 + y 2 = 4 from (2, 0) to itself.

29. Circulation and flux Find the circulation and flux of the fields

F1 = xi + yj and F2 = - yi + xj

around and across each of the following curves. a. The circle rstd = scos tdi + ssin tdj, 0 ... t ... 2p b. The ellipse rstd = scos tdi + s4 sin tdj, 0 ... t ... 2p 30. Flux across a circle Find the flux of the fields

F1 = 2xi - 3yj and F2 = 2xi + sx - ydj

across the circle

rstd = sa cos tdi + sa sin tdj, 0 ... t ... 2p.

In Exercises 31?34, find the circulation and flux of the field F around

and across the closed semicircular path that consists of the semicircu-

lar arch r1std = sa cos tdi + sa sin tdj, 0 ... t ... p, followed by the line segment r2std = ti, - a ... t ... a.

31. F = xi + yj

32. F = x2i + y 2j

33. F = - yi + xj

34. F = - y 2i + x2j

35. Flow integrals Find the flow of the velocity field F = sx + ydi - sx2 + y 2dj along each of the following paths from (1, 0) to s - 1, 0d in the xy-plane. a. The upper half of the circle x2 + y 2 = 1

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

c. The line segment from (1, 0) to s0, -1d followed by the line segment from s0, -1d to s -1, 0d

36. Flux across a triangle Find the flux of the field F in Exercise 35 outward across the triangle with vertices (1, 0), (0, 1), s -1, 0d.

37. Find the flow of the velocity field F = y 2i + 2xyj along each of the following paths from (0, 0) to (2, 4).

a.

y

b.

y

(2, 4)

(2, 4)

y 5 2x

y 5 x2

(0, 0)

x 2

(0, 0)

x 2

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

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

937

38. Find the circulation of the field F = yi + (x + 2y)j around each of the following closed paths.

a.

y

(?1, 1)

(1, 1)

x

(?1, ?1) b.

(1, ?1)

y x2 1 y2 5 4

x

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

Vector Fields in the Plane 39. Spin field Draw the spin field

F

=

y - 2x2 +

i y2

+

x 2x2 +

j y2

(see Figure 16.12) along with its horizontal and vertical components at a representative assortment of points on the circle x2 + y 2 = 4.

40. Radial field Draw the radial field

F = xi + yj

(see Figure 16.11) along with its horizontal and vertical components at a representative assortment of points on the circle x2 + y 2 = 1. 41. A field of tangent vectors

a. Find a field G = Psx, ydi + Qsx, ydj in the xy-plane with the property that at any point sa, bd Z s0, 0d, G is a vector of magnitude 2a2 + b 2 tangent to the circle x2 + y 2 = a2 + b 2 and 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 G = Psx, ydi + Qsx, ydj in the xy-plane with the property that at any point sa, bd Z s0, 0d, G is a unit vector tangent to the circle x2 + y 2 = a2 + b 2 and pointing in the clockwise direction.

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

43. Unit vectors pointing toward the origin Find a field F = Msx, ydi + Nsx, ydj in the xy-plane with the property that at each point sx, yd Z s0, 0d, F is a unit vector pointing toward the origin. (The field is undefined at (0, 0).)

44. Two "central" fields Find a field F = Msx, ydi + Nsx, ydj in the xy-plane with the property that at each point sx, yd Z s0, 0d, F points toward the origin and F 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 positive for a ... t ... b. Let C be the path rstd = ti + stdj, a ... t ... b, and F = yi. Is there any relation between the value of the work integral

F # dr

LC

and the area of the region bounded by the t-axis, the graph of , and the lines t = a and t = b? Give reasons for your answer. 46. Work done by a radial force with constant magnitude A particle moves along the smooth curve y = sxd 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

#F T ds = k C sb 2 + ssbdd2d1>2 - sa2 + ssadd2d1>2 D .

LC

Flow Integrals in Space In Exercises 47?50, F is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing t.

47. F = - 4xyi + 8yj + 2k rstd = ti + t2j + k, 0 ... t ... 2

48. F = x2i + yzj + y 2k rstd = 3tj + 4tk, 0 ... t ... 1

49. F = sx - zdi + xk rstd = scos tdi + ssin tdk, 0 ... t ... p

50. F = - yi + xj + 2k rstd = s -2 cos tdi + s2 sin tdj + 2tk, 0 ... t ... 2p

51. Circulation Find the circulation of F = 2xi + 2zj + 2yk around the closed path consisting of the following three curves traversed in the direction of increasing t.

C1: rstd = scos tdi + ssin tdj + tk, 0 ... t ... p>2

C2: rstd = j + sp>2ds1 - tdk, 0 ... t ... 1

C3: rstd = ti + s1 - tdj, 0 ... t ... 1

z

0,

1,

2

(1, 0, 0) x

C1 C3

C2 (0, 1, 0)

y

938

Chapter 16: Integration in Vector Fields

52. Zero circulation Let C be the ellipse in which the plane 2x + 3y - z = 0 meets the cylinder x2 + y 2 = 12. Show, without evaluating either line integral directly, that the circulation of the field F = xi + yj + zk around C in either direction is zero.

53. Flow along a curve The field F = xyi + yj - yzk 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 y = x2 and the plane z = x . (Hint: Use t = x as the parameter.)

z

z x

(1, 1, 1) y

y x2 x

54. Flow of a gradient field Find the flow of the field F = ?sxy 2z3d: 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.

COMPUTER EXPLORATIONS In Exercises 55?60, use a CAS to perform the following steps for finding the work done by force F over the given path:

a. Find dr for the path rstd = gstdi + hstdj + kstdk. b. Evaluate the force F along the path.

c. Evaluate F # dr.

LC 55. F = xy 6 i + 3xsxy 5 + 2dj;

0 ... t ... 2p

rstd = s2 cos tdi + ssin tdj,

56.

F

=

1

3 +

x2

i

+

1

2 +

y

2

j;

rstd = scos tdi + ssin tdj,

0...t...p

57. F = s y + yz cos xyzdi + sx2 + xz cos xyzdj + sz + xy cos xyzdk; rstd = (2 cos t)i + (3 sin t)j + k, 0 ... t ... 2p

58. F = 2xyi - y2j + zex k; rstd = - ti + 1tj + 3tk, 1...t...4

59. F = s2y + sin xdi + sz2 + s1>3dcos ydj + x4 k; rstd = ssin tdi + scos tdj + ssin 2tdk, - p>2 ... t ... p>2

60.

F

=

s x 2y di

+

1 3

x

3j

+

xyk;

rstd = scos tdi + ssin tdj +

s2 sin2 t - 1dk, 0 ... t ... 2p

16.3

Path Independence, Conservative Fields, and Potential Functions

A gravitational field G is a vector field that represents the effect of gravity at a point in space 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 in space that represents the effect of electric forces on a charged particle placed within it. The force on a body of charge q placed in the field is given by F = qE. In gravitational and electric 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 object--not on which path is taken between these positions. In this section we study vector fields with this property and the calculation of 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 Section 16.1. For some special fields, however, the integral's value is the same for all paths from A to B.

DEFINITIONS and suppose that

Let F for any

be a two

vector points

field A and

defined on B in D the

an open region line integral 1C

FD#

in space, dr along

a1CpaFth#

C from A to B in D is the same over all paths dr is path independent in D and the field F

from A to B. Then the integral is conservative on D.

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

Exercises 16.3

16.3 Path Independence, Conservative Fields, and Potential Functions

947

EXAMPLE 6 Show that y dx + x dy + 4 dz is exact and evaluate the integral

s2,3, -1d

y dx + x dy + 4 dz Ls1,1,1d over any path from (1, 1, 1) to s2, 3, - 1d.

Solution We let M = y, N = x, P = 4 and apply the Test for Exactness:

0P 0y

=

0

=

00Nz ,

0M 0z

=

0

=

00Px ,

0N 0x

=

1

=

0M 0y

.

These equalities tell us that y dx + x dy + 4 dz is exact, so

y dx + x dy + 4 dz = d

for some function , and the integral's value is s2, 3, -1d - s1, 1, 1d. We find up to a constant by integrating the equations

0 0x

=

y,

0 0y

=

x,

0 0z

=

4.

(4)

From the first equation we get

sx, y, zd = xy + gsy, zd.

The second equation tells us that

0 0y

=

x

+

0g 0y

=

x,

or

0g 0y

=

0.

Hence, g is a function of z alone, and

sx, y, zd = xy + hszd.

The third of Equations (4) tells us that

0 0z

=

0

+

dh dz

=

4,

or hszd = 4z + C.

Therefore,

sx, y, zd = xy + 4z + C.

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

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

Testing for Conservative Fields Which fields in Exercises 1?6 are conservative, and which are not?

1. F = yzi + xzj + xyk 2. F = s y sin zdi + sx sin zdj + sxy cos zdk 3. F = yi + sx + zdj - yk 4. F = - yi + xj 5. F = sz + ydi + zj + s y + xdk 6. F = sex cos ydi - sex sin ydj + zk

Finding Potential Functions In Exercises 7?12, find a potential function for the field F.

7. F = 2xi + 3yj + 4zk

8. F = s y + zdi + sx + zdj + sx + ydk 9. F = ey+2zsi + xj + 2xkd

10. F = s y sin zdi + sx sin zdj + sxy cos zdk 11. F = sln x + sec2sx + yddi +

asec2sx

+

yd

+

y2

y +

z2 bj

+

y2

z +

z2 k

12.

F=

y 1 + x2 y2 i +

a 1

x + x2 y2

+

z

bj +

21 - y 2 z2

a 21

y -

y2 z2

+

1 z

b

k

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