calculus 5th ed_deborah hughes-hallet_chap18-sols

18.1 SOLUTIONS 1427

CHAPTER EIGHTEEN

Solutions for Section 18.1

Exercises

1. Positive, because the vectors are longer on the portion of the path that goes in the same direction as the vector field.

2. Negative because the vector field points in the opposite direction to the path.

3. Positive, because the vector field points in the same direction as the path.

4. Zero, because, by symmetry, the positive integral along the left half of the path cancels the negative integral along theright half.

5. Negative, because the vector field points in the opposite direction to the path.

6. Zero, because the positive contributions on the upper half of the path cancel the negative contributions on the lower halfof the path.

7. Since ~F is perpendicular to the curve at every point along it,∫

C

~F · d~r = 0.

8. At every point along the curve, ~F = 2~j and is parallel to the curve. Thus,∫

C

~F · d~r = 2 · Length of curve = 2 · 5 = 10.

9. Since ~F is a constant vector field and the curve is a line,∫C~F · d~r = ~F ·∆~r , where ∆~r = 7~j . Therefore,

∫

C

~F · d~r = (3~i + 4~j ) · 7~j = 28

10. Since ~F is perpendicular to the line, the line integral is 0.

11. Only the~i -component contributes to the integral, so∫

C

~F · d~r = 6 · Length of path = 6 · (7− 3) = 24.

12. The path is along the y-axis, so only the ~j -component contributes to the line integral. Since C is oriented in the −~jdirection, we have ∫

C

(x~i + 6~j − ~k ) · d~r = −6 · Length of path = −6 · 8 = −48.

13. At every point, the vector field is parallel to segments ∆~r = ∆x~i of the curve. Thus,

∫

C

~F · d~r =

∫ 6

2

x~i · dx~i =

∫ 6

2

xdx =x2

2

∣∣∣∣6

2

= 16.

14. The ~j -component of ~F does not contribute to the line integral. Since ∆~r = ∆x~i , we have

∫

C

~F · d~r =

∫ 6

2

(x~i + y~j ) · dx~i =

∫ 6

2

xdx =x2

2

∣∣∣∣6

2

= 16.

1428 Chapter Eighteen /SOLUTIONS

15. At every point on the path, ~F is parallel to ∆~r . Suppose r is the distance from the point (x, y) to the origin, so ‖~r ‖ = r.Then ~F ·∆~r = ‖ ~F ‖‖∆~r ‖ = r∆r. At the start of the path, r =

√22 + 22 = 2

√2 and at the end r = 6

√2. Thus,

∫

C

~F · d~r =

∫ 6√

2

2√

2

rdr =r2

2

∣∣∣∣6√

2

2√

2

= 32.

16. The path is parallel to the z-axis, so the vector field is perpendicular to the path at every point. Thus, the line integral is 0.

17. Since the curve is along the y-axis, only the ~j component of the vector field contributes to the integral:∫

C

(2~j + 3~k ) · d~r =

∫

C

2~j · d~r = 2 · Length of C = 2 · 10 = 20.

18. The vector field x~i + y~j + z~k points radially outward and is everywhere perpendicular to the unit circle. Thus, the lineintegral is 0.

Problems

19. (a) See Table 18.1.

Table 18.1

(x, y) ~F (x, y)

(0,−1) −~i(1,−1) −~i +~j

(2,−1) −~i + 4~j

(3,−1) −~i + 9~j

(4,−1) −~i + 16~j

(4, 0) 16~j

(4, 1) ~i + 16~j

(4, 2) 2~i + 16~j

(4, 3) 3~i + 16~j

1 2 3x

−1

1

2

3

y

I K O

O 6

�

�

�

Figure 18.1

(b) See Figure 18.1.(c) From the point (0,−1) to the point (4,−1), the x-component of the force field is always −1, i.e., it is pushing the

object backward with a constant force of 1. Thus, the work done on that part of the path is −1 · 4 = −4, becauseonly the horizontal component of the force field contributes to work.

From the point (4,−1) to the point (4, 3), the y-component of the force field is always 16, so it is pushing theobject forward with force of 16. Thus, the work done on that part of the path is 16 · 4 = 64, because only the verticalcomponent of the force field contributes to work.

So the total work done is −4 + 64 = 60.

20. The integral∫C~F · d~r is a sum of the line integrals of ~F over each of its three straight segments, which we can compute

separately:∫

PQ

~F · d~r = ~PQ · ~F = (4~i + 2~j ) ·~i = 4

∫

QR

~F · d~r = ~QR · ~F = (−~i + 2~j ) · (2~i −~j ) = −4

∫

RS

~F · d~r = ~RS · ~F = (−2~i − 2~j ) · (3~i +~j ) = −8

∫

C

~F · d~r = 4− 4− 8 = −8.

18.1 SOLUTIONS 1429

21. Since it appears that C1 is everywhere perpendicular to the vector field, all of the dot products in the line integral are zero,hence

∫C1

~F · d~r ≈ 0. Along the path C2 the dot products of ~F with ∆~ri are all positive, so their sum is positive and

we have∫C1

~F · d~r <∫C2

~F · d~r . For C3 the vectors ∆~ri are in the opposite direction to the vectors of ~F , so the dot

products ~F ·∆~ri are all negative; so,∫C3

~F · d~r < 0. Thus, we have∫

C3

~F · d~r <∫

C1

~F · d~r <∫

C2

~F · d~r

22. The force has no horizontal component. Therefore the (positive) work done in the first half of C1 will be exactly canceledby the (negative) work done in the second half, so the total work over the path C1 is zero. The same holds true for C2,again by virtue of the vertical symmetry of the path and the fact that ~F is constant and because the horizontal part of C2

contributes zero work. For C3, the total work will be greater than zero, since the diagonal part of C3 is in the same generaldirection as ~F and the horizontal part of C3 contributes zero work.

23. The line integral along C1 is positive; the line integrals along C2 and C3 appear to be zero.

24. The line integral along C1 appears to be zero, the line integral along C2 is positive, and the line integral along C3 isnegative.

25. The line integral along C1 is negative, the line integral along C2 is negative, and the line integral along C3 appears to bezero.

26. The line integral along C1 appears to be 0, the line integral along C2 is negative, and the line integral along C3 is positive.

27. The dot product of ~F and 10~i is positive if a > 0. There are no restrictions on b and c.

28. The ~k component of ~F does not contribute to the line integral. Since the line integral of y~i around C is negative, for theline integral of ~F to be positive, we need a < 0. No restriction on c is needed.

29. The vector field ~F is in the same direction as C if b > 0, so we want b < 0. No restriction is needed on c.

30. For any value of a, the vector field ay~i − ax~j is perpendicular to the vector~i +~j + ~k which is in the direction of C.Thus a can take any value. The ~k component of ~F is in the direction of C if the coefficient of ~k is positive, that is, ifc > 1.

31. The line C is parallel to the z-axis, so a~i + b~j does not contribute to the line integral. Thus, there are no restrictions ona and b. The dot product of ~F and −~k is negative if c > 3.

32. The vector field is F (~r ) = ~r . See Figure 18.2. The vector field is perpendicular to the circular arcs at every point, so∫

C1

~F · d~r =

∫

C3

~F · d~r = 0.

Also, since it is radially symmetric, ∫

C2

~F · d~r = −∫

C4

~F · d~r .

So, ∫

C

=

∫

C1

+

∫

C2

+

∫

C3

+

∫

C4

= 0.

1 2 3 4

−3

−2

−1

1

2

x

y

C4

C2

C3

C1

Figure 18.2

1 2 3

−2

−1

1

2

x

y

C4

C2

C3

C1

Figure 18.3


33. This vector field is illustrated in Figure 18.3. It is perpendicular to C2 and C4 at every point, since ~F (x, y) · ~r (x, y) = 0and C2 and C4 are radial line segments, then

∫

C2

~F · d~r =

∫

C4

~F · d~r = 0.

Since C3 is longer than C1, and the vector field is larger in magnitude along C3, the line integral along C3 has greaterabsolute value than that along C1. The line integral along C3 is positive and the line integral along C1 is negative, so

∫

C

~F · d~r =

∫

C3

~F · d~r +

∫

C1

~F · d~r > 0.

See Figure 18.3.

34. This vector field is illustrated in Figure 18.4. It is perpendicular to C2 and C4 at every point, since ~F (x, y) · ~r (x, y) = 0and C2 and C4 are radial line segments, then

∫

C2

~F · d~r =

∫

C4

~F · d~r = 0.

Since C3 is longer than C1, and the vector field is larger in magnitude along C3, the line integral along C3 has greaterabsolute value than that along C1. The line integral along C1 is positive and the line integral along C3 is negative, so

∫

C

~F · d~r =

∫

C3

~F · d~r +

∫

C1

~F · d~r < 0.

See Figure 18.4.

35. Since it does not depend on y, this vector field is constant along vertical lines, x = constant. Now let us consider twopoints P andQ on C1 which lie on the same vertical line. Because C1 is symmetric with respect to the x-axis, the tangentvectors at P and Q will be symmetric with respect to the vertical axis so their sum is a vertical vector. But ~F has onlyhorizontal component and thus ~F · (∆~r (P ) + ∆~r (Q)) = 0. As ~F is constant along vertical lines (so ~F (P ) = ~F (Q)),we obtain

~F (P ) ·∆~r (P ) + ~F (Q) ·∆~r (Q) = 0.

Summing these products and making ‖∆~r ‖ → 0 gives us∫

C1

~F · d~r = 0.

The same thing happens on C3, so∫C3

~F · d~r = 0.Now let P be on C2 and Q on C4 lying on the same vertical line. The respective tangent vectors are symmetric with

respect to the vertical axis hence they add up to a vertical vector and a similar argument as before gives

~F (P ) ·∆~r (P ) + ~F (Q) ·∆~r (Q) = 0

and ∫

C2

~F · d~r +

∫

C4

~F · d~r = 0

and so ∫

C

~F · d~r = 0.

See Figure 18.5.

18.1 SOLUTIONS 1431

1 2 3

−2

−1

1

2

x

y

C4

C2

C3

C1

Figure 18.4

1 2 3

−2

−1

1

2

x

y

C4

C2

C3

C1

Figure 18.5

1 2 3

−2

−1

1

2

x

y

C4

C2

C3

C1

Figure 18.6

36. First of all, ~F (x, y) is perpendicular to the position vector ~r (x, y) = x~i + y~j because

~F (x, y) · ~r (x, y) =−xy

x2 + y2+

xy

x2 + y2= 0.

Also the magnitude of ~F is inversely proportional to the distance from the origin because

‖~F (x, y)‖ =

√x2 + y2

x2 + y2=

1

‖~r (x, y)‖ .

So ~F is perpendicular to C2 and C4 and therefore∫

C2

~F · d~r =

∫

C4

~F · d~r = 0.

Suppose R is the radius of C3. On C3, the vector field ~F has the same direction as the tangent vector which is approxi-mated by ∆~r , so we have

~F ·∆~r = ‖ ~F ‖ · ‖∆~r ‖ =1

R‖∆~r ‖.

When all these products are summed and the limit is taken as ‖∆~r ‖ → 0, we get∫

C3

~F · d~r =1

R

∫

C3

‖d~r ‖

=1

R(length of C3) = measure of the arc C3 in radians.

Similarly, suppose r is the radius of C1. On C1, the vector field ~F is in the opposite direction to the tangent vectorwhich is approximated by ∆~r . Hence we have

∫

C1

~F · d~r = −1

r

∫

C1

‖d~r ‖

= −(1

r(length of C1)) = −(measure of C1 in radians).

Since C1 and C3 have the same measure in radians, we have∫

C

~F · d~r =

∫

C1

~F · d~r +

∫

C2

~F · d~r +

∫

C3

~F · d~r +

∫

C4

~F · d~r

= −π2

+ 0 + (+π

2) + 0 = 0.

See Figure 18.6.


37. (a) See Figure 18.7.

x

y(i)

x

y(ii)

x

y(iii)

x

y(iv)

Figure 18.7

(b) For (i) and (iii) a closed curve can be drawn; not for the others.

38. (a) The path C is a line segment, tangent to ~T =~i +~j at every point. Because the path C is on the line y = x we have~F (x, y) = 2~i − 2~j on C. Hence ~T · ~F = 0, which shows that C and ~F are perpendicular at every point of C. Theangle between them in π/2.

(b)∫C~F · d~r = 0 because ~F and C are perpendicular at every point of C.

39. (a) The path C is a line segment, tangent to ~T =~i +~j at every point. Because the path C is on the line y = x we have~F (x, y) = 2~i + 2~j = 2~T on C. Thus ~F is tangent to C at every point and points in the direction of the orientationof C. The angle between C and ~F is 0.

(b) On C we have ‖ ~F ‖ = ‖2~i + 2~j ‖ = 2√

2.(c) The path C has length 5

√2. Since the vector field ~F is everywhere tangent to C in the direction of the orientation

and of constant magnitude 2√

2 we have∫

C

~F · d~r = ‖ ~F ‖ · Length of C = 2√

2 · 5√

2 = 20.

40. The line integral is defined by chopping the curve C into little pieces, Ci, and forming the sum∑

Ci

~F ·∆~r .

When the pieces are small, ∆~r is approximately tangent to Ci, and its magnitude is approximately equal to the length ofthe little piece of curve Ci. This means that ~F and ∆~r are almost parallel, the dot product is approximately equal to theproduct of their magnitudes, i.e.,

~F ·∆~r ≈ m · (Length of Ci).When we sum all the dot products, we get

∑

Ci

~F ·∆~r ≈∑

Ci

m · (Length of Ci)

= m ·∑

Ci

(Length of Ci)

= m · (Length of C)

18.1 SOLUTIONS 1433

41. See Figure 18.8. The example chosen is the vector field ~F (x, y) = y~j and the path C is the line from (0,−1) to (0, 1).Since the vectors are symmetric about the x-axis, the dot products ~F · ∆~r cancel out along C to give 0 for the lineintegral. Many other answers are possible.

(0,−1)

(0, 1)

x

y

C

Figure 18.8

42. Suppose∫C~F · d~r = 0 for every closed curve C. Pick any two fixed points P1, P2 and curves C1, C2 each going from

P1 to P2. See Figure 18.9. Define −C2 to be the same curve as C2 except in the opposite direction. Therefore, the curveformed by traversing C1, followed by C2 in the opposite direction, written as C1 − C2, is a closed curve, so by ourassumption,

∫C1−C2

~F · d~r = 0. However, we can write

∫

C1−C2

~F · d~r =

∫

C1

~F · d~r −∫

C2

~F · d~r

since C2 and −C2 are the same except for direction. Therefore,∫

C1

~F · d~r −∫

C2

~F · d~r = 0,

so ∫

C1

~F · d~r =

∫

C2

~F · d~r .

SinceC1 andC2 are any two curves with the endpoints P1, P2, this gives the desired result – namely, that fixing endpointsand direction uniquely determines the value of

∫C~F · d~r . In other words, the value of the integral

∫C~F · d~r does not

depend on the path taken. We say the line integral is path-independent.

P1

P2

C2

C1

Figure 18.9

43. Pick any closed curve C. Choose two distinct points P1, P2 on C. Let C1, C2 be the two curves from P1 to P2 along C.See Figure 18.9. Let −C2 be the same as C2, except in the opposite direction. Thus, C1 − C2 = C. Therefore,

∫

C

~F · d~r =

∫

C1−C2

~F · d~r =

∫

C1

~F · d~r −∫

C2

~F · d~r


since C2 and −C2 differ only in direction. But C1 and C2 have the same endpoints (P1 and P2) and same direction (P1

to P2), so by assumption we have∫C1

~F · d~r =∫C2

~F · d~r . Therefore,∫

C

~F · d~r =

∫

C1

~F · d~r −∫

C2

~F · d~r = 0.

44. Figure 18.10 shows the wind velocity vectors on each side of the square, where the speed is v meter/sec on the southside and (v − 12) meter/sec on the north side. The circulation is the sum of the line integrals along the four sides of thesquare. The line integrals along the eastern and western edges are both zero, since the wind velocity is perpendicular tothese edges. The integral to the right along the south side equals (1000 km)(−v meter/sec) = −v× 106 meter2/sec, andthe integral to the left along the north side equals (1000 km)((v − 12) meter/sec) = (v − 12)× 106 meter2/sec.

Total circulation = −v × 106 + (v − 12)× 106 = −1.2× 107 meter2/sec.

6

N

�

�

�

�

1000 km

1000

km

(v − 12) meter/sec

v meter/sec

Figure 18.10: Wind velocity across asquare

45. Let r = ‖~r ‖. Since ∆~r points outward, in the opposite direction to ~F , we expect the answer to be negative.∫

C

~F · d~r =

∫

C

−GMm~r

r3· d~r =

∫ 10000

8000

−GMm

r2dr

=GMm

r

∣∣∣∣10000

8000

= GMm(

1

10000− 1

8000

)

= −2.5 · 10−5GMm.

46. Let r = ‖~r ‖. Since ∆~r points outward, in the opposite direction to ~F , we expect a negative answer. We take the upperlimit to be r =∞, so the integral is improper.

∫

C

~F · d~r =

∫

C

−GMm~r

r3· d~r =

∫ ∞

8000

−GMm

r2dr

= limb→∞

∫ b

8000

−GMm

r2dr = lim

b→∞

GMm

r

∣∣∣∣b

8000

= limb→∞

GMm(

1

b− 1

8000

)

= −GMm

8000

47. The force of the field on the particle at each point is ~E , so the force applied in moving the particle against the field is− ~E , so

φ(P ) = −∫

C

~E · d~r

where C is a path from P0 to P .

18.1 SOLUTIONS 1435

48. Any point P which is a units from the origin can be reached from P0 by a path C lying on the sphere of radius a. Since ~Eis perpendicular to the sphere,

∫C~E ·d~r = 0, so φ(P ) = 0. On the other hand, if P does not lie on the sphere of radius a,

it can be reached by a path consisting of two pieces, C1 and C2, one lying on the sphere of radius a and one going straightalong a line radiating from the origin (see Figure 18.11).

∫C1

~E ·d~r = 0 as before, but∫C2

~E ·d~r 6= 0, since ~E is parallel

to C2 and always points out. Thus, if C is the path consisting of C1 followed by C2, we have∫C~E · d~r =

∫C2

~E · d~r .

Thus∫C~E · d~r is always positive or always negative along the path C which joins P0 to P . Hence the set of points with

potential zero is the sphere of radius a.

P

P0

C1

C2

x

y

z

Figure 18.11

49. In Problem 48 we saw that the surface where the potential is zero is a sphere of radius a. Let S be any sphere centered atthe origin, and let P1 be a point on S, and C1 a path from P0 to P1. If P is any point on S, then P can be reached fromP0 by a path, C, consisting of C1 followed by C2, where C2 is a path from P1 to P lying entirely on the sphere, S. Then∫C2

~E · d~r = 0, since ~E is perpendicular to the sphere. So

φ(P ) = −∫

C

~E · d~r = −∫

C1

~E · d~r −∫

C2

~E · d~r = −∫

C1

~E · d~r = φ(P1).

Thus, φ is constant on S. The equipotential surfaces are spheres centered at the origin.

50. (a) Suppose P is b units from the origin. Then P can be reached by a path, C, consisting of two pieces, C1 and C2, onelying on the sphere of radius a and one going straight along a line radiating from the origin (see Figure 18.11). Wehave ~E ·∆~r = 0 on C1, and, writing r = ‖~r ‖, we have ~E ·∆~r = || ~E ||∆r on C2, so

φ(P ) = −∫

C

~E · d~r = −∫

C1

~E · d~r −∫

C2

~E · d~r

= 0−∫ b

a

|| ~E || dr = 0−∫ b

a

Q

4πε

1

r2dr

=Q

4πε

1

r

∣∣∣b

a=

Q

4πε

1

b− Q

4πε

1

a.

Let P be the point with position vector ~r . Then

φ(~r ) = − Q

4πε

1

a+

Q

4πε

1

||~r || .

(b) If we let a→∞ in the formula for φ, the first term goes to zero and we get the simpler expression

φ(~r ) =Q

4πε

1

||~r || .



Exercises

1. Only the~i -component contributes to the line integral, so d~r =~i dx and∫

C

(2x~i + 3y~j ) · d~r =

∫ (5,0,0)

(1,0,0)

(2x~i + 3y~j ) ·~i dx =

∫ 5

1

2x dx = x2

∣∣∣∣5

1

= 24.

2. Only the y-component of the vector field contributes to the line integral. On the curve, d~r = ~j dy, so

∫

C

(3~i + (y + 5)~j ) · d~r =

∫ 3

0

(y + 5) dy =

(y2

2+ 5y

)∣∣∣∣3

0

=39

2.

3. We will find the line integral from (0, 0) to (3, 1) and then take the negative. The line segment is parameterized by

x = 3t y = t, for 0 ≤ t ≤ 1.

Then ~r ′(t) = 3~i +~j , so∫

C

(2y2~i + x~j ) · d~r = −∫ 1

0

(2t2~i + 3t~j

)· (3~i +~j ) dt = −

∫ 1

0

(6t2 + 3t) dt = −(

2t3 +3

2t2) ∣∣∣∣

1

0

= −7

2.

4. The semicircle has radius 1 and is centered at (2, 0). It can be parameterized by

x = 2 + cos t y = sin t, for 0 ≤ t ≤ π.Then ~r ′(t) = − sin t~i + cos t~j , so

∫

C

(x~i + y~j ) · d~r =

∫ π

0

((2 + cos t)~i + sin t~j

)· (− sin t~i + cos t~j ) dt

=

∫ π

0

(−2 sin t− cos t sin t+ sin t cos t) dt = 2 cos t

∣∣∣∣π

0

= −4.

5. Since ~F = (x2 + y)~i + y3~j , the line integral along the third segment, which is parallel to the z-axis, is zero. On the firstsegment, which is parallel to the y-axis, only the ~j -component contributes. On the second segment, which is parallel tothe x-axis, only the~i -component contributes. On the first segment x = 4 and y varies from 0 to 3; on the second segmenty = 3 and x varies from 4 to 0. Thus, we have

∫

C

~F · d~r =

∫ 3

0

((42 + y)~i + y3~j ) ·~j dy +

∫ 0

4

((x2 + 3)~i + 33~j ) ·~i dx

=

∫ 3

0

y3 dy +

∫ 0

4

(x2 + 3) dx =y4

4

∣∣∣3

0−(x3

3+ 3x

)∣∣∣4

0=

81

4− 64

3− 12 = −157

12.

6. Only the~i component of ~F contributes to the line integral. Since C goes a distance of 3 in the −~i direction, we have∫

C

~F · d~r = (2~i ) · (−3~i ) = −6.

7. Parameterizing C by x(t) = t, y(t) = t for 1 ≤ t ≤ 5, we have ~r ′(t) =~i +~j ,∫

~F · d~r =

∫ 5

1

(3~j −~i ) · (~i +~j ) dt =

∫ 5

1

2 dt = 8.

8. The curve C is parameterized by (x, y) = (t, t) for 0 ≤ t ≤ 3. Thus,∫

C

~F · d~r =

∫ 3

0

(t~i + t~j ) · (~i +~j )dt =

∫ 3

0

2tdt = t2∣∣∣∣3

0

= 9.

18.2 SOLUTIONS 1437

9. The line can be parameterized by (1 + 2t, 2 + 2t), for 0 ≤ t ≤ 1, so the integral looks like∫

C

~F · d~r =

∫ 1

0

~F (1 + 2t, 2 + 2t) · (2~i + 2~j ) dt

=

∫ 1

0

[(1 + 2t)2~i + (2 + 2t)2~j ] · (2~i + 2~j ) dt

=

∫ 1

0

2(1 + 4t+ 4t2) + 2(4 + 8t+ 4t2) dt

=

∫ 1

0

(10 + 24t+ 16t2) dt

= (10t+ 12t2 + 16t3/3)∣∣10

= 10 + 12 + 16/3− (0 + 0 + 0) = 82/3

10. Parameterize the curve: ~r (t) = sin t~i + cos t~j , 0 ≤ t ≤ π. Then∫

C

~F · d~r =

∫ π

0

(cos t~i − sin t~j

)·(cos t~i − sin t~j

)dt

=

∫ π

0

((cos t)2 + (− sin t)2

)dt =

∫ π

0

1 dt = π.

11. Use x(t) = t, y(t) = t2, so x′(t) = 1, y′(t) = 2t, with 0 ≤ t ≤ 2. Then∫

~F · d~r =

∫ 2

0

(−t2 sin t~i + cos t~j ) · (~i + 2t~j ) dt

=

∫ 2

0

(−t2 sin t+ 2t cos t) dt = t2 cos t∣∣20

= 4 cos 2.

12. The parameterization is given, so∫

C

~F · d~r =

∫ 4

2

~F (2t, t3) · (2~i + 3t2~j ) dt

=

∫ 4

2

[(ln(t3)~i + ln(2t)~j ] · (2~i + 3t2~j ) dt

=

∫ 4

2

(2 ln(t3) + 3t2 ln(2t)) dt

=

∫ 4

2

(6 ln(t) + 3t2 ln(2t)) dt since ln(t3) = 3 ln(t).

This integral can be computed numerically, or using integration by parts or the integral table, giving∫

C

~F · d~r =

∫ 4

2

(6 ln(t) + 3t2 ln(2t)) dt

= (6(t ln(t)− t) + t3 ln(2t)− t3/3)∣∣42

= 240 ln 2− 136

3− (28 ln 2− 44

3)

= 212 ln 2− 92/3 ≈ 116.28.

The expression containing ln 2 was obtained using the properties of the natural log.


13. The curve C is parameterized by~r = cos t~i + sin t~j , for 0 ≤ t ≤ 2π,

so,~r ′(t) = − sin t~i + cos t~j .

Thus,∫

C

~F · d~r =

∫ 2π

0

(2 sin t~i − sin (sin t)~j ) · (− sin t~i + cos t~j )dt

=

∫ 2π

0

(−2 sin2 t− sin (sin t) cos t)dt

= sin t cos t− t+ cos (sin t)

∣∣∣∣2π

0

= −2π.

14. Parameterizing C by x(t) = 3t, y(t) = 2t for 0 ≤ t ≤ 1, we have ~r ′(t) = 3~i + 2~j , so∫

C

~F · d~r =

∫ 1

0

((2t)3~i + (3t)2~j ) · (3~i + 2~j ) dt

=

∫ 1

0

(24t3 + 18t2) dt = 6t4 + 6t3∣∣∣∣1

0

= 12.

15. Parameterizing C by x(t) = t, y(t) = t, z(t) = t for 0 ≤ t ≤ 2, we have ~r ′(t) =~i +~j + ~k , so∫

C

(x~i + 6~j − ~k ) · d~r =

∫ 2

0

(t~i + 6~j − ~k ) · (~i +~j + ~k ) dt

=

∫ 2

0

(t+ 6− 1) dt =t2

2+ 5t

∣∣∣∣2

0

= 12.

16. The path can be broken into three line segments: C1, from (1, 0) to (−1, 0), and C2, from (−1, 0) to (0, 1), and C3, from(0, 1) to (1, 0). (See Figure 18.12.)

(−1, 0) (1, 0)

(0, 1)

C1

C2 C3

x

y

Figure 18.12

Along C1 we have y = 0 so the vector field xy~i + (x − y)~j is perpendicular to C1; Thus, the line integral alongC1 is 0.

C2 can be parameterized by (−1 + t, t), for 0 ≤ t ≤ 1 so the integral is∫

C2

~F · d~r =

∫ 1

0

~F (−1 + t, t) · (~i +~j ) dt

=

∫ 1

0

[t(−1 + t)~i + (−1)~j ] · (~i +~j ) dt

=

∫ 1

0

(−t+ t2 − 1) dt

= (−t2/2 + t3/3− t)∣∣10

= −1/2 + 1/3− 1− (0 + 0 + 0) = −7/6

18.2 SOLUTIONS 1439

C3 can be parameterized by (t, 1− t), for 0 ≤ t ≤ 1 so the integral is∫

C3

~F · d~r =

∫ 1

0

~F (t, 1− t) · (~i −~j ) dt

=

∫ 1

0

(t(1− t)~i + (2t− 1)~j ) · (~i −~j ) dt

=

∫ 1

0

(−t2 − t+ 1) dt

= (−t3/3− t2/2 + t)∣∣10

= −1/3− 1/2 + 1− (0 + 0 + 0) = 1/6

So the total line integral is∫

C

~F · d~r =

∫

C1

~F · d~r +

∫

C2

~F · d~r +

∫

C3

~F · d~r = 0 + (−7

6) +

1

6= −1

17. Since ~r = x(t)~i + y(t)~j + z(t)~k = t~i + t2~j + t3~k , for 1 ≤ t ≤ 2,we have ~r ′(t) = x′(t)~i + y′(t)~j + z′(t)~k =~i + 2t~j + 3t2~k . Then

∫

C

~F · d~r =

∫ 2

1

(t~i + 2t3t2~j + t~k ) · (~i + 2t~j + 3t2~k ) dt

=

∫ 2

1

(t+ 4t6 + 3t3) dt

=t2

2+

4t7

7+

3t4

4

∣∣∣2

1=

2389

28≈ 85.32

18. We parameterize C by~r = 2t~i + 3t~j + 4t~k , for 0 ≤ t ≤ 1.

Then ~r ′(t) = 2~i + 3~j + 4~k and so∫

C

~F · d~r =

∫ 1

0

((2t)3~i + (3t)2~j + (4t)~k

)· (2~i + 3~j + 4~k )dt

=

∫ 1

0

(16t3 + 27t2 + 16t)dt

= 4t4 + 9t3 + 8t2∣∣∣∣1

0

= 21.

19. Since C is given by ~r = cos t~i + sin t~j + t~k , we have ~r ′(t) = − sin t~i + cos t~j + ~k . Thus,∫

C

~F · d~r =

∫ 4π

0

(− sin t~i + cos t~j + 5~k ) · (− sin t~i + cos t~j + ~k )dt

=

∫ 4π

0

(sin2 t+ cos2 t+ 5)dt =

∫ 4π

0

6dt = 24π.

20. The first step is to parameterize C by

(x(t), y(t), z(t)) = (0, 2 cos t,−2 sin t), 0 ≤ t ≤ 2π.

Thus, we have~r ′(t) = x′(t)~i + y′(t)~j + z′(t)~k = −2 sin t~j − 2 cos t~k .


So we have∫

C

~F · d~r =

∫ 2π

0

(e2 cos t~i + ~k ) · ((−2 sin t)~j + (−2 cos t)~k )dt

=

∫ 2π

0

−2 cos tdt

= −2 sin t∣∣∣2π

0

= 0

21.∫C

3xdx− y sinxdy

22.∫Cy2dx+ z2dy + (x2 − 5)dz

23. ~F = (x+ 2y)~i + x2y~j

24. ~F = e−3y~i − yz(sinx)~j + (y + z)~k

25. From x = t2 and y = t3 we get dx = 2tdt and dy = 3t2dt. Hence∫

C

ydx+ xdy =

∫ 5

1

t3(2t)dt+ t2(3t2)dt =

∫ 5

1

5t4dt = 55 − 1 = 3124.

26. Fromx = cos t, y = sin t, z = 3t

we getdx = − sin t dt dy = cos t dt, dz = 3dt.

Hence∫

C

dx+ ydy + zdz =

∫ 2π

0

− sin t dt+ sin t cos t dt+ 3t(3dt)

= cos t+1

2sin2 t+

9

2t2|2π0 = 18π2.

27. Parameterize C:x = 1 + 4t, y = 3 + 6t, 0 ≤ t ≤ 1

so that dx = 4dt and dy = 6dt. Hence∫

C

3ydx+ 4xdy =

∫ 1

0

3(3 + 6t)4dt+ 4(1 + 4t)6dt =

∫ 1

0

60 + 168tdt = 144.

28. Parameterize C:x = 0, y = 3 cos t, z = 3 sin t, 0 ≤ t ≤ 2π

so thatdx = 0dt, dy = −3 sin t dt, dz = 3 cos t dt.

Hence∫

C

xdx+ zdy − ydz =

∫ 2π

0

0dt+ 3 sin t(−3 sin t)dt− 3 cos t(3 cos t)dt =

∫ 2π

0

−9dt = −18π.

Problems

29. (a) Figure 18.13 shows the curves.

18.2 SOLUTIONS 1441

−1

1

C2

C1

x

y

Figure 18.13

(b) On C1, only the ~j -component of ~F contributes to the integral. There d~r = ~j dy, so

∫

C1

~F · d~r =

∫ 1

−1

y~j ·~j dy =

∫ 1

−1

y dy =y2

2

∣∣∣∣∣

1

−1

= 0.

On C2, we have ~r ′(t) = − sin t~i + cos t~j , so

∫

C2

~F · d~r =

∫ 3π/2

π/2

((cos t+ 3 sin t)~i + sin t~j ) · (− sin t~i + cos t~j ) dt

=

∫ 3π/2

π/2

− cos t sin t− 3 sin2 t+ cos t sin t dt =

∫ 3π/2

π/2

−3 sin2 t dt

= −3(t

2− sin t cos t

2

) ∣∣∣∣∣

3π/2

π/2

= −3π

2.

30. (a) Since ~r (t) = t~i + t2~j , we have ~r ′(t) =~i + 2t~j . Thus,∫

C

~F · d~r =

∫ 1

0

~F (t, t2) · (~i + 2t~j ) dt

=

∫ 1

0

[(3t− t2)~i + t~j ] · (~i + 2t~j ) dt

=

∫ 1

0

(3t+ t2) dt

= (3t2

2+t3

3)

∣∣∣∣1

0

=3

2+

1

3− (0 + 0) =

11

6

(b) Since ~r (t) = t2~i + t~j , we have ~r ′(t) = 2t~i +~j . Thus,∫

C

~F · d~r =

∫ 1

0

~F (t2, t) · (2t~i +~j ) dt

=

∫ 1

0

[(3t2 − t)~i + t2~j ] · (2t~i +~j ) dt

=

∫ 1

0

(6t3 − t2) dt

= (3t4

2− t3

3)

∣∣∣∣1

0

=3

2− 1

3− (0− 0) =

7

6


31. (a) The unit circle centered at the origin has equation x2 + y2 = 1. At any point in the plane, the magnitude of ~F isgiven by ‖ ~F ‖ =

√(−y)2 + x2. Along the unit circle, ‖ ~F ‖ = 1.

(b) Suppose ~r = x~i + y~j is a radius vector to a point (x, y) on the unit circle centered at the origin. See Figure 18.14.Then

~r · ~F = (x~i + y~j ) · (−y~i + x~j ) = −xy + xy = 0.

So the vector field is perpendicular to any corresponding radius vector, that is, the vector field is tangent to the circleat every point.

x

y

x2 + y2 = 1

~r

~F = −y~i + x~j

Figure 18.14

(c) We can parameterize C by (cos t, sin t), for 0 ≤ t ≤ 2π. Then∫

C

~F · d~r =

∫ 2π

0

~F (cos t, sin t) · (− sin t~i + cos t~j ) dt

=

∫ 2π

0

(− sin t~i + cos t~j ) · (− sin t~i + cos t~j ) dt

=

∫ 2π

0

(sin2 t+ cos2 t) dt

=

∫ 2π

0

1 dt

= 2π

Thus, ∫

C

~F · d~r = 2π = Circumference of the unit circle.

32. We parameterize the helical staircase by observing that

x = 5 cos t, y = 5 sin t, z = t

has the correct radius, but climbs 2π in one revolution. To make it climb 4 meters in one revolution, we write:

x = 5 cos t, y = 5 sin t, z =4t

2π=

2t

π.

Thus,

~r ′(t) = −5 sin t~i + 5 cos t~j +2

π~k .

The gravitational force is given by ~F = −70g~k , and we want to go around 2 turns of the staircase, so we take 0 ≤ t ≤4π. Thus,

Work done by gravity =

∫~F · d~r =

∫ 4π

0

−70g~k · (−5 sin t~i + 5 cos t~j +2

π~k )dt

=

∫ 4π

0

−140g

πdt = −140g

πt

∣∣∣∣4π

0

= −560g.

18.2 SOLUTIONS 1443

Notice that the result can also be obtained by multiplying the force by the vertical distance:

Gravitational force · Vertical distance moved = (−70g)8 = −560g.

NowWork done by person = −Work done by gravity = 560g.

33. (a) The line integral∫C

(xy~i + x~j ) · d~r is positive. This follows from the fact that all of the vectors of xy~i + x~jat points along C point approximately in the same direction as C (meaning the angles between the vectors and thedirection of C are less than π/2).

(b) Using the parameterization x(t) = t, y(t) = 3t, with x′(t) = 1, y′(t) = 3, we have∫

C

~F · d~r =

∫ 4

0

~F (t, 3t) · (~i + 3~j ) dt

=

∫ 4

0

(3t2~i + t~j ) · (~i + 3~j ) dt

=

∫ 4

0

(3t2 + 3t) dt

=(t3 +

3

2t2) ∣∣∣∣

4

0

= 88.

(c) Figure 18.15 shows the oriented path C ′, with the “turn around” points P and Q. The particle first travels from theorigin to the point P (call this path C1), then backs up from P to Q (call this path C2), then goes from Q to the point(4, 12) in the original direction (call this path C3). See Figure 18.16. Thus, C ′ = C1 + C2 + C3. Along the parts ofC1 and C2 that overlap, the line integrals cancel, so we are left with the line integral over the part of C1 that does notoverlap with C2, followed by the line integral over C3. Thus, the line integral over C ′ is the same as the line integralover the direct route from the point (0, 0) to the point (4, 12).

Q

P

(4, 12)

1 2 3 40123456789

101112

x

y

Figure 18.15

C1

C2

C3

Figure 18.16

(d) The parameterization

(x(t), y(t)) =(

1

3(t3 − 6t2 + 11t), (t3 − 6t2 + 11t)

)

has (x(0), y(0)) = (0, 0) and (x(4), y(4)) = (4, 12). The form of the parameterization we were given shows thatthe second coordinate is always three times the first. Thus all points on the parameterized curve lie on the line y = 3x.

We have to do a bit more work to guarantee that all points on the curve lie on the line between the point (0, 0)and the point (4, 12); it is possible that they might shoot off to, say, (100, 300) before returning to (4, 12). Let’sinvestigate the maximum and minimum values of f(t) = t3−6t2 +11t on the interval 0 ≤ t ≤ 4. We can do this ona graphing calculator or computer, or use single-variable calculus. We already know the values of f at the endpoints,namely 0 and 12. We’ll look for local extrema:

0 = f ′(t) = 3t2 − 12t+ 11


which has roots at t = 2 ± 1√3

. These are the values of t where the particle changes direction: t = 2 − 1√3

corresponds to point P and t = 2+ 1√3

corresponds to pointQ of C ′. At these values of t we have f(2− 1√3) ≈ 6.4,

and f(2 + 1√3) ≈ 5.6. The fact that these values are between 0 and 12 shows that f takes on its maximum and

minimum values at the endpoints of the interval and not in between.(e) Using the parameterization given in part (d), we have

~r ′(t) = x′(t)~i + y′(t)~j =1

3(3t2 − 12t+ 11)~i + (3t2 − 12t+ 11)~j .

Thus,∫

C′

~F · d~r

=

∫ 4

0

~F (1

3(t3 − 6t2 + 11t), t3 − 6t2 + 11t) · ( 1

3(3t2 − 12t+ 11)~i + (3t2 − 12t+ 11)~j ) dt

=

∫ 4

0

(1

3(t3 − 6t2 + 11t)2~i +

1

3(t3 − 6t2 + 11t)~j ) · (1

3(3t2 − 12t+ 11)~i + (3t2 − 12t+ 11)~j ) dt

=

∫ 4

0

1

3(t3 − 6t2 + 11t)(3t2 − 12t+ 11)

{((t3 − 6t2 + 11t)~i +~j ) · (1

3~i +~j )

}dt

=

∫ 4

0

1

3(t3 − 6t2 + 11t)(3t2 − 12t+ 11)

{1

3(t3 − 6t2 + 11t) + 1

}dt

=1

9

∫ 4

0

(t3 − 6t2 + 11t)(3t2 − 12t+ 11)(t3 − 6t2 + 11t+ 3) dt

Numerical integration yields an answer of 88, which agrees with the answer found in part b).

34. First, check that each of these gives a parameterization of L: each has both coordinates equal (as do all points on L) andeach begins at (0, 0) and ends at (1, 1). Now we calculate the line integral of the vector field ~F = (3x− y)~i + x~j usingeach parameterization.

(a) Using B(t) gives

∫

L

~F · d~r =

∫ 1/2

0

((6t− 2t)~i + 2t~j ) · (2~i + 2~j ) dt =

∫ 1/2

0

12t dt = 6t2∣∣∣∣1/2

0

=3

2.

(b) Now we use C(t):∫

L

~F · d~r =

∫ 2

1

((3(t2 − 1)

3− (t2 − 1)

3

)~i +

t2 − 1

3~j

)·(

2t

3~i +

2t

3~j)dt

=

∫ 2

1

2t

3(t2 − 1) dt =

2

3

∫ 2

1

(t3 − t) dt

=2

3

(t4

4− t2

2

)∣∣∣∣2

1

=3

2.

35. The integral corresponding to A(t) = (t, t) is ∫ 1

0

3t dt.

The integral corresponding to B(t) = (2t, 2t) is ∫ 1/2

0

12t dt.

The substitution s = 2t has ds = 2 dt and s = 0 when t = 0 and s = 1 when t = 1/2. Thus, substituting t =s

2into the

integral corresponding to B(t) gives∫ 1/2

0

12t dt =

∫ 1

0

12(s

2)(

1

2ds) =

∫ 1

0

3s ds.

18.3 SOLUTIONS 1445

The integral on the right-hand side is now the same as the integral corresponding to A(t). Therefore we have∫ 1/2

0

12t dt =

∫ 1

0

3s ds =

∫ 1

0

3t dt.

Alternatively, a similar calculation shows that the substitution t = 2w converts the integral corresponding to A(t)into the integral corresponding to B(t).


0

3t dt.

The integral corresponding to C(t) = ( t2−13, t

2−13

) is

2

3

∫ 2

1

(t3 − t) dt.

The substitution s =t2 − 1

3has ds =

2

3t dt. Also s = 0 when t = 1 and s = 1 when t = 2. Thus, substituting into the

integral corresponding to C(t) gives

2

3

∫ 2

1

(t3 − t) dt =

∫ 2

1

(t2 − 1)2

3t dt =

∫ 1

0

3s ds.

The integral on the right-hand side is the same as the integral corresponding to A(t). Therefore we have

2

3

∫ 2

1

(t3 − t) dt =

∫ 1

0

3s ds =

∫ 1

0

3t dt.

Alternatively, the substitution t =w2 − 1

3converts the integral corresponding toA(t) into the integral corresponding

to C(t).


0

3t dt.

The integral corresponding to D(t) = (et − 1, et − 1) is

3

∫ ln 2

0

(e2t − et) dt.

The substitution s = et − 1 has ds = et dt. Also s = 0 when t = 0 and s = 1 when t = ln 2. Thus, substituting into theintegral corresponding to D(t) and using the fact that e2t = et · et gives

3

∫ ln 2

0

(e2t − et) dt = 3

∫ ln 2

0

(et − 1)et dt =

∫ 1

0

3s ds.

The integral on the right-hand side is the same as the integral corresponding to A(t). Therefore we have

3

∫ ln 2

0

(e2t − et) dt =

∫ 1

0

3s ds =

∫ 1

0

3t dt.

Alternatively, the substitution t = ew − 1 converts the integral corresponding to A(t) into the integral corresponding toB(t).


Exercises

1. Since ~F is a gradient field, with ~F = grad f where f(x, y) = x2 + y4, we use the Fundamental Theorem of LineIntegrals. The starting point of the path C is (2, 0) and the end is (0, 2). Thus,

∫

C

~F · d~r = f(0, 2)− f(2, 0) = 16− 4 = 12.


2. Since, if f(x, y, z) = sin(xy) + ez , we have gradf = ~F , we use the Fundamental Theorem of Line Integrals. Thestarting point of the path is (0, 0, 0) and the end is (

√2,√

5, 2) so∫

C

~F · d~r = f(√

2,√

5, 2)− f(0, 0, 0) = sin√

10 + e2 − 1.

Notice that since ~F is a gradient field, the intermediate points on the path do not affect the answer.

3. Path independent

4. Path independent

5. Path-independent, because the vector field appears constant.

6. Not path-independent, because the line integral around a closed curve around the origin is not zero.

7. Path independent

8. Path dependent

9. Since ~F = 3x2~i + 4y3~j = grad(x3 + y4), we take f(x, y) = x3 + y4. Then by the Fundamental Theorem of LineIntegrals, ∫

C

~F · d~r = f(−1, 0)− f(1, 0) = (−1)3 − 13 = −2.

10. Since ~F = (x+ 2)~i + (2y + 3)~j = grad(x2

2+ 2x+ y2 + 3y

), the Fundamental Theorem of Line Integrals gives

∫

C

~F · d~r =

(x2

2+ 2x+ y2 + 3y

)∣∣∣∣(3,1)

(1,0)

=(

9

2+ 6 + 1 + 3

)−(

1

2+ 2)

= 12.

11. Since ~F = 2x~i − 4y~j + (2z − 3)~k = grad(x2 − 2y2 + z2 − 3z), the Fundamental Theorem of Line Integrals gives∫

C

~F · d~r = (x2 − 2y2 + z2 − 3z)

∣∣∣∣(2,3,−1)

(1,1,1)

= (4− 2 · 32 + (−1)2 + 3)− (12 − 2 · 12 + 12 − 3) = −7.

12. Since ~F = 2 sin(2x+ y)~i + sin(2x+ y)~j = grad(− cos(2x+ y)), we take

f(x, y) = − cos(2x+ y).

Then, using the Fundamental Theorem of Line Integrals,∫

C

~F · d~r = f(0, 5π)− f(π, 0) = − cos(5π)− (− cos(2π)) = −(−1)− (−1) = 2.

Notice that only the endpoints of the curve affect the answer.

13. Since ~F = y sin(xy)~i + x sin(xy)~j = grad(− cos(xy)), the Fundamental Theorem of Line Integrals gives∫

C

~F · d~r = − cos(xy)

∣∣∣∣(3,18)

(1,2)

= − cos(54) + cos(2) = cos(2)− cos(54).

14. Since ~F = x2/3~i + e7y~j = grad(

35x5/3 + 1

7e7y)

, we see ~F is a gradient vector field. Therefore,∫

C

(x2/3~i + e7y~j ) · d~r = 0.

15. Since ~F = x2/3~i + e7y~j = grad(

35x5/3 + 1

7e7y)

, we have

∫

C

(x2/3~i + e7y~j ) · d~r =3

5x5/3 +

1

7e7y

∣∣∣∣(0,1)

(1,0)

=3

5· 05/3 +

1

7e7·1 − 3

5· 15/3 − 1

7e7·0

=1

7(e7 − 1)− 3

5.

18.3 SOLUTIONS 1447

16. Since ~F = grad(exy + sin z), we take f(x, y, z) = exy + sin z and use the Fundamental Theorem of Line Integrals∫

C

~F · d~r = f(3, 1, 9π)− f(0, 0, π) = e3 + sin (9π)− e0 − sinπ

= e3 − 1.

17. Since ~F = 2xy2zex2y2z~i +2x2yzex

2y2z~j +x2y2ex2y2z~k = grad(ex

2y2z) and the curveC is closed, the FundamentalTheorem of Line Integrals tells us that

∫C~F · d~r = 0, since

∫

C

~F · d~r = ex2y2z

∣∣∣∣(1,0,1)

(1,0,1)

= e0 − e0 = 0.

18. Since ~F = grad f is a gradient vector field, the Fundamental Theorem of Line Integrals give us∫

C

~F · d~r = f (end)− f (start) =(x2 + 2y3 + 3z4

) ∣∣∣(0,0,5)

(4,0,0)= 3 · 54 − 42 = 1859.

Problems

19. The vector field ~F points radially outward, and so is everywhere perpendicular to A; thus,∫A~F · d~r = 0.

Along the first half of B, the terms ~F ·∆~r are negative; along the second half the terms ~F ·∆~r are positive. Bysymmetry the positive and negative contributions cancel out, giving a Riemann sum and a line integral of 0.

The line integral is also 0 along C, by cancellation. Here the values of ~F along the x-axis have the same magnitudeas those along the y-axis. On the first half of C the path is traversed in the opposite direction to ~F ; on the second half ofC the path is traversed in the same direction as ~F . So the two halves cancel.

20. We parameterize A by x = t, y = t, where 0 ≤ t ≤ 1. Then∫

A

~F · d~r =

∫ 1

0

(t~i + t~j ) · (~i +~j ) dt

=

∫ 1

0

2t dt = t2∣∣∣∣1

0

= 1.

The path B has the parameterization x = t, y = t2, where 0 ≤ t ≤ 1. Then we have∫

B

~F · d~r =

∫ 1

0

(t~i + t2~j ) · (~i + 2t~j ) dt

=

∫ 1

0

(t+ 2t3) dt =t2

2+

2t4

4

∣∣∣∣1

0

= 1.

We have to break the path C into two separate parameterizations: x = t, y = 0, where 0 ≤ t ≤ 1 and x = 1, y = t,where 0 ≤ t ≤ 1. Then

∫

C

~F · d~r =

∫ 1

0

(t~i ·~i ) dt+

∫ 1

0

(~i + t~j ) ·~j dt

=

∫ 1

0

t dt+

∫ 1

0

t dt =1

2+

1

2= 1.

21. Yes. If f(x, y) = 12x2, then grad f = x~i .

22. Yes. If f(x, y) = 13x3 − xy2, then grad f = (x2 − y2)~i − 2xy~j .

23. No. Suppose there were a function f such that grad f = ~F . Then we would have

∂f

∂x=

−z√x2 + z2

.


Hence we would have∂2f

∂y∂x=

∂

∂y(−z√x2 + z2

) = 0.

In addition, since grad f = ~F , we have that∂f

∂y=

y√x2 + z2

.

Thus we also know that∂2f

∂x∂y=

∂

∂x

(y√

x2 + y2

)= −xy(x2 + z2)−3/2.

Notice that∂2f

∂y∂x6= ∂2f

∂x∂y.

Since we expect ∂2f∂y∂x

= ∂2f∂x∂y

, we have got a contradiction. The only way out of this contradiction is to conclude there

is no function f with grad f = ~F . Thus ~F is not a gradient vector field.

24. Yes. Letf(~r ) = −1

r= −(x2 + y2 + z2)−1/2

Then

∂f

∂x= x(x2 + y2 + z2)−3/2

∂f

∂y= y(x2 + y2 + z2)−3/2

∂f

∂z= z(x2 + y2 + z2)−3/2

So grad f = (x2 + y2 + z2)−3/2(x~i + y~j + z~k ) = ~r /r3

25. (a) To find the change in f by computing a line integral, we first choose a path C between the points; the simplest is aline. We parameterize the line by (x(t), y(t)) = (t, πt/2), with 0 ≤ t ≤ 1. Then (x′(t), y′(t)) = (1, π/2), so theFundamental Theorem of Line Integrals tells us that

f(1,π

2)− f(0, 0) =

∫

C

grad f · d~r

=

∫ 1

0

grad f(t,πt

2

)·(~i +

π

2~j)dt

=

∫ 1

0

(2tet

2

sin(πt

2

)~i + et

2

cos(πt

2

)~j)·(~i +

π

2~j)dt

=

∫ 1

0

(2tet

2

sin(πt

2

)+πet

2

2cos(πt

2

))dt

=

∫ 1

0

d

dt

(et

2

sin(πt

2

))dt

= et2

sin(πt

2

) ∣∣∣∣1

0

= e = 2.718.

This integral can also be approximated numerically.(b) The other way to find the change in f between these two points is to first find f . To do this, observe that

2xex2

sin y~i + ex2

cos y~j =∂

∂x

(ex

2

sin y)~i +

∂

∂y

(ex

2

sin y)~j = grad

(ex

2

sin y).

So one possibility for f is f(x, y) = ex2

sin y. Thus,

Change in f

∣∣∣∣(1,π/2)

(0,0)

= ex2

sin y

∣∣∣∣(1,π/2)

(0,0)

= e1 sin(π

2

)− e0 sin 0 = e.

The exact answer confirms our calculations in part (a) which show that the answer is e.

18.3 SOLUTIONS 1449

26. (a) For path (i), we have x(t) = t, y(t) = t2, so x′(t) = 1, y′(t) = 2t. Thus,∫

C

~F · d~r =

∫ 1

0

~F (t, t2) · (~i + 2t~j ) dt

=

∫ 1

0

[(t+ t2)~i + t~j ] · (~i + 2t~j ) dt

=

∫ 1

0

(t+ 3t2) dt

= (t2

2+ t3)

∣∣∣∣1

0

=1

2+ 1− (0 + 0) =

3

2.

For path (ii), we have x(t) = t2, y(t) = t, so x′(t) = 2t, y′(t) = 1. Thus,∫

C

~F · d~r =

∫ 1

0

~F (t2, t) · (2t~i +~j ) dt

=

∫ 1

0

[(t2 + t)~i + t2~j ] · (2t~i +~j ) dt

=

∫ 1

0

(2t3 + 3t2) dt

= (t4

2+ t3)

∣∣∣∣1

0

=1

2+ 1− (0 + 0) =

3

2.

For path (iii), we have x(t) = t, y(t) = tn, so x′(t) = 1, y′(t) = ntn−1. Thus,∫

C

~F · d~r =

∫ 1

0

~F (t, tn) · (~i + ntn−1~j ) dt

=

∫ 1

0

[(t+ tn)~i + t~j ] · (~i + ntn−1~j ) dt

=

∫ 1

0

(t+ tn + ntn) dt

=

∫ 1

0

(t+ (n+ 1)tn) dt

=

(t2

2+ tn+1

)∣∣∣∣1

0

=1

2+ 1− (0 + 0) =

3

2.

(b) If f(x, y) = xy + x2/2, we have ~F = grad f . Each path goes from (0, 0) to (1, 1). Thus in each case∫

C

~F · d~r = f(1, 1)− f(0, 0) =3

2.

27. The unit circle cuts the negative x-axis at (−1, 0, 0), and it cuts the negative y-axis at (0,−1, 0). There is a quarter circlebetween these points if the circle is traversed counterclockwise.

(a) Since 2πx~i + y2~j = grad(πx2 + y3/3), we use the Fundamental Theorem of Line Integrals:

∫

C

(2πx~i + y2~j ) · d~r =

(πx2 +

y3

3

)∣∣∣∣(0,−1,0)

(−1,0,0)

=

(π(02) +

(−1)3

3

)−(π(−1)2 +

03

3

)= −1

3− π.


(b) Since ~F is not a gradient field, we parameterize c. If x = cos t and y = sin t, then π ≤ t ≤ 3π/2 parametrizes C.Thus

∫

C

(−2y~i + x~j ) · d~r =

∫ 3π/2

π

(−2 sin t~i + cos t~j ) · (− sin t~i + cos t~j ) dt

=

∫ 3π/2

π

(2 sin2 t+ cos2 t) dt =

∫ 3π/2

π

(1 + sin2 t) dt

=(t+

t

2− sin t cos t

2

) ∣∣∣∣3π/2

π

=3π

4.

28. Since ~F = grad

(x2 + y2

2

), the line integral can be calculated using the Fundamental Theorem of Line Integrals:

∫

c

~F · d~r =x2 + y2

2

∣∣∣∣(3/√

2,3/√

2)

(0,0)

=9

2.

29. This vector field is not a gradient field, so we evaluate the line integral directly. Let C1 be the path along the x-axis from(0, 0) to (3, 0) and let C2 be the path from (3, 0) to (3/

√2, 3/√

2) along x2 + y2 = 9. Then∫

C

~H · d~r =

∫

C1

~H · d~r +

∫

C2

~H · d~r .

On C1, the vector field has only a ~j component (since y = 0), and ~H is therefore perpendicular to the path. Thus,∫

C1

~H · d~r = 0.

On C2, the vector field is tangent to the path. The path is one eighth of a circle of radius 3 and so has length 2π(3/8) =3π/4. ∫

C2

~H · d~r = ‖ ~H ‖ · Length of path = 3 ·(

3π

4

)=

9π

4.

Thus, ∫

C

~H · d~r =9π

4.

30. Since ~F = grad(y ln(x+ 1)), we evaluate the line integral using the Fundamental Theorem of Line Integrals:

∫

C

~F · d~r = y ln(x+ 1)

∣∣∣∣(3/√

2,3/√

2)

(0,0)

=3√2

ln

(3√2

+ 1

)− 0 ln 1 =

3√2

ln

(3√2

+ 1

).

31. Since ~G = grad(exy +sin(x+y)), the line integral can be calculated using the Fundamental Theorem of Line Integrals:

∫

c

~F · d~r = exy + sin(x+ y)

∣∣∣∣(3/√

2,3/√

2)

(0.0)

= e9/2 + sin

(6√2

)− e0 = e9/2 + sin(3

√2)− 1.

32. Since ~F = yz2exyz2~i + xz2exyz

2~j + 2xyzexyz2~k = grad(exyz

2

), we can use the Fundamental Theorem of LineIntegrals. The start of the path, where t = 0, is (1, 0, 0). The end of the path is (cos(1.25π), sin(1.25π), 1.25π) =(−1/

√2,−1/

√2, 1.25π). Thus

∫

C

~F · d~r = exyz2

∣∣∣∣∣

(−1/√

2,−1/√

2,1.25π)

(1,0,0)

= e(1.25π)2/2 − 1.

18.3 SOLUTIONS 1451

33. Although this curve is complicated, the vector field is a gradient field since

~F = sin(x

2

)sin(y

2

)~i − cos

(x

2

)cos(y

2

)~j = grad

(−2 cos

(x

2

)sin(y

2

)).

Thus, only the endpoints of the curve, P and Q, are needed. Since P = (−3π/2, 3π/2) and Q = (−3π/2,−3π/2) and~F = grad(−2 cos(x/2) sin(y/2)), we have

∫

C

~F · d~r = −2 cos(x

2

)sin(y

2

) ∣∣∣∣∣

Q=(−3π/2,−3π/2)

P=(−3π/2,3π/2)

= −2 cos(−3π

4

)sin(−3π

4

)+ 2 cos

(−3π

4

)sin(

3π

4

)

= 2 cos(

3π

4

)sin(

3π

4

)+ 2 cos

(3π

4

)sin(

3π

4

)

= −2 · 1√2· 1√

2− 2 · 1√

2· 1√

2= −2.

34. (a) Three possible paths are shown in Figure 18.17. Since ~F is perpendicular to the horizontal axis everywhere, ~F · ~dr =0 along C1.

Since C2 starts out in the direction of ~F , the first leg of C2 will have a positive line integral. The secondhorizontal part of C2 will have a 0 line integral, and the third leg that ends at Q will have a positive line integral. Thusthe line integral along C2 is positive.

A similar argument shows that the line integral along C3 < 0.(b) No, ~F is not a gradient field, since the line integrals along these three paths joining P and Q do not have the same

value.

C1

C3

C2

Q Px

y

Figure 18.17

P

Q

135 7

9

x

y

Figure 18.18

35. (a) See Figure 18.18.(b) Vector at P is shorter than vector at Q.(c) By the Fundamental Theorem of Line Integrals

∫

C

grad f · d~r = f(Q)− f(P ) = 9− 3 = 6.

36. Since ~F is a gradient vector field, we use the Fundamental Theorem of line integrals, giving∫

C

~F · d~r =

∫

C

grad f · d~r = f(end)− f(start).

(a) The line integral∫C2

~F · d~r is 0, since the curve begins and ends on the same contour, so f(end) = f(start).(b) Since C1 crosses more contours than C4, and since both curves are oriented in the direction of increasing f ,

0 <

∫

C4

~F · d~r <∫

C1

~F · d~r .


Since C3 goes from higher to lower values of f ,∫

C3

~F · d~r < 0 =

∫

C2

~F · d~r .

Thus, we have ∫

C3

~F · d~r <∫

C2

~F · d~r <∫

C4

~F · d~r <∫

C1

~F · d~r .

(c) Since C3 and C4 have endpoints on the same contours, but with start and finish reversed,∫

C3

~F · d~r = −∫

C4

~F · d~r .

The line integral∫C3

~F · d~r is negative because Q3 is on a contour of lower value than P3.

37. (a) By the Fundamental Theorem of Line Integrals∫ (3,4)

(0,2)

grad f · d~r = f(3, 4)− f(0, 2) = 66− 57 = 9.

(b) By the Fundamental Theorem of Line Integrals, since C is a closed path,∫C

grad f · d~r = 0.

38. (a) The integral is positive, because the portion of the path that goes with the vector field is longer than the portion of thepath that goes against it, and in addition the vectors are larger in magnitude along the former and smaller in magnitudealong the latter.

(b) If it were true that ~F = grad f for some function f , then the integral around every closed path would be zero. Butin part (a) we saw that the integral around one closed path was not zero, so ~F cannot be a gradient vector field.

(c) The region shown is in the first quadrant. In that quadrant, the vectors of ~F1 point away from the origin, so ~F1 doesnot fit. The vectors of both ~F2 and ~F3 point up and to the left, so they are both possibilities; of these, ~F2 fits bestbecause its vectors get larger in magnitude as you move away from the origin, which fits the diagram. The vectors in~F 3 shrink as you move away from the origin.

39. Since the vector field is path independent, the line integral around the closed curve (0, 0) to (1, 0) to (1, 1) to (0, 1) to(0, 0) is zero. Thus

∫ (0,0)

(0,1)

~F · d~r = −(∫ (1,0)

(0,0)

~F · d~r +

∫ (1,1)

(1,0)

~F · d~r +

∫ (0,1)

(1,1)

~F · d~r)

= −(5.1 + 3.2− 4.7) = −3.6.

40. (a) To maximize the line integral, we choose C to be parallel to grad f = 3~i + 4~j . Thus C has parametric equation~r = (2~i +~j ) + t~v where ~v = 3~i + 4~j , so

x = 2 + 3t y = 1 + 4t.

If the other end of C is at (x1, y1), since the length of C is 10, we have√

(x1 − 2)2 + (y1 − 1)2 = 10√(3t)2 + (4t)2 = 10

t√

32 + 42 = 10

5t = 10

t = 2.

Thus t = 2 at (x1, y1), sox1 = 2 + 2 · 3 = 8 and y1 = 1 + 2 · 4 = 9.

Thus C ends at the point (8, 9).(b) By the Fundamental Theorem of Line Integrals,

∫

C

grad f · d~r = f(8, 9)− f(2, 1) = (3 · 8 + 4 · 9)− (3 · 2 + 4 · 1) = 50.

Alternately, since grad f and C are parallel,∫

C

grad f · d~r = ‖ grad f‖ · Length of C = 5 · 10 = 50.

18.3 SOLUTIONS 1453

41. (a) Work done by the force is the line integral, so

Work done against force = −∫

C

~F · d~r = −∫

C

(−mg~k ) · d~r .

Since ~r = (cos t)~i + (sin t)~j + t~k , we have ~r ′ = −(sin t)~i + (cos t)~j + ~k ,

Work done against force =

∫ 2π

0

mg~k · (− sin t~i + cos t~j + ~k )dt

=

∫ 2π

0

mg dt = 2πmg.

(b) We know from physical principles that the force is conservative. (Because the work done depends only on the verticaldistance moved, not on the path taken.) Alternatively, we see that

~F = −mg~k = grad(−mgz),

so ~F is a gradient field and therefore path independent, or conservative.

42. (a) We parameterize the path by (cos t, sin t) for π/2 ≤ t ≤ π. Since t = π/2 gives the end point, (0, 1) and t = πgives the starting point (−1, 0), we have

∫

C

~F · ~dr =

∫ π/2

π

~F (cos t, sin t) · (− sin t~i + cos t~j ) dt

= −∫ π

π/2

(sin t~i − cos t~j ) · (− sin t~i + cos t~j ) dt

= −∫ π

π/2

(− sin2 t− cos2 t) dt

=

∫ π

π/2

1 dt = t

∣∣∣∣π

π/2

= π/2.

The work done by the force is +π/2. The work is positive since the force is always in the direction of the path (infact it is always tangent to C since ~F · ~r = 0).

(b) If we redo our computations using the entire unit circle, the only change will be the limits of integration: they’llchange to 0 to 2π. This yields an answer of 2π (or −2π, depending on orientation). Since the work around a closedpath is not zero, the force is not path-independent.

43. (a) Since ~r · ~a = a1x+ a2y + a3z, we have

grad(~r · ~a ) = a1~i + a2

~j + a3~k = ~a .

(b) By the Fundamental Theorem of Line Integrals, if ~r0 = x0~i + y0

~j + z0~k , we have

∫

C

grad(~r · ~a ) · d~r = ~r · ~a∣∣∣∣(x0,y0,z0)

(0,0,0)

= ~r 0 · ~a .

(c) Since ~r 0 · ~a = ||~r 0||||~a || cos θ = 10||~a || cos θ, where the angle between ~r 0 and ~a , the maximum value of ~r 0 · ~aoccurs if ~r 0 is parallel to ~a . Then θ = 0 and

∫

C

grad(~r · ~a ) · d~r = 10||~a || cos 0 = 10||~a ||.

44. (a) Since ~F (x, y) − ~G (x, y) is parallel to gradh(x, y), it is perpendicular to the level curves of h. Since the orientedpath C is on a level curve of h, ~F (x, y)− ~G (x, y) is perpendicular to C at every point of C. Hence

∫

C

(~F (x, y)− ~G (x, y)) · d~r = 0.

Therefore ∫

C

~F · d~r =

∫

C

~G · d~r .


(b) By the Fundamental Theorem of Calculus for Line Integrals we have∫

C

~G · d~r =

∫

C

gradφ · d~r = φ(Q)− φ(P ).

Using part (a) we have ∫

C

~F · d~r = φ(Q)− φ(P ).

45. (a) We have

gradh = ~i

gradφ = 2y~i + 2x~j

~F − gradφ = −y~i = −y gradh.

Thus, ~F − gradφ is a multiple of gradh.(b) By part (a) the vector fields ~F and gradφ have the same components perpendicular to gradh, which is to say the

same components in the direction of the level curve C of h. Thus, the line integrals of ~F and gradφ along C areequal. Using the Fundamental Theorem of Calculus for Line Integrals, we have

∫

C

~F · d~r =

∫

C

gradφ · d~r = φ(Q)− φ(P ) = 60− 30 = 30.

46. (a) We have

gradh = ~j

gradφ = y~i + x~j

~F − gradφ = x~j = x gradh.



∫

C

~F · d~r =

∫

C

gradφ · d~r = φ(Q)− φ(P ) = 80− 30 = 50.

47. (a) We have

gradh = −2x~i +~j

gradφ = (2x2 + y)~i + x~j

~F − gradφ = −2x2~i + x~j = x gradh.



∫

C

~F · d~r =

∫

C

gradφ · d~r = φ(Q)− φ(P ) = 384− 0 = 384.

48. (a) We have

gradh = ~i +~j

gradφ = (x+ 3y)~i + (3x+ 2y)~j

~F − gradφ = (−x− 2y)~i + (−x− 2y)~j = −(x+ 2y) gradh.



∫

C

~F · d~r =

∫

C

gradφ · d~r = φ(Q)− φ(P ) = 1800− 1850 = −50.

18.4 SOLUTIONS 1455

49. (a) By the chain ruledh

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt= fxx

′(t) + fyy′(t),

which is the result we want.(b) Using the parameterization of C that we were given,

∫

C

grad f · d~r =

∫ b

a

(fx(x(t), y(t))~i + fy(x(t), y(t))~j ) · (x′(t)~i + y′(t)~j )dt

=

∫ b

a

(fx(x(t), y(t))x′(t) + fy(x(t), y(t))y′(t))dt.

Using the result of part (a), this gives us∫

C

grad f · d~r =

∫ b

a

h′(t)dt

= h(b)− h(a) = f(Q)− f(P ).

50. (a) The level surfaces are horizontal planes given by gz = c, so z = c/g. The potential energy increases with the heightabove the earth. This means that more energy is stored as “potential to fall” as height increases.

(b) The gradient of φ points upward (in the direction of increasing potential energy), so ∇φ = g~k . The gravitationalforce acts toward the earth in the direction of −~k . So, ~F = −g~k . The negative sign represents the fact that thegravitational force acts in the direction of the decreasing potential energy.

51. (a) We have

ϕ(~r ) =p1x+ p2y + p3z

(x2 + y2 + z2)3/2.

Taking partial derivatives gives

ϕx(~r ) =p1(x2 + y2 + z2)3/2 − (3/2)(p1x+ p2y + p3z)(2x)(x2 + y3 + z2)1/2

(x2 + y2 + z3)3

=p1

||~r ||3 − 3(~p · ~r )x

||~r ||5= −D1(~r ).

Similar computations give ϕy = −D2 and ϕz = −D3, so gradϕ = − ~D .(b) The field ~D is necessarily path-independent since it is a gradient vector field.


Exercises

1. We know that∂f

∂x= 2xy and

∂f

∂y= x2,

so, integrating with respect to x, thinking of y as a constant gives

f(x, y) = x2y + C(y).

Differentiating with respect to y gives∂f

∂y= x2 + C′(y),

so we take C(y) = k for some constant K2. Thus

f(x, y) = x2y +K.


2. We know that∂f

∂x= 2xy and

∂f

∂y= x2 + 8y3

Now think of y as a constant in the equation for ∂f/∂x and integrate, giving

f(x, y) = x2y + C(y).

Since the constant of integration may depend on y, it is written C(y). Differentiating this expression for f(x, y) withrespect to y and using the fact that ∂f/∂y = x2 + 8y3, we get

∂f

∂y= x2 + C′(y) = x2 + 8y3.

ThereforeC′(y) = 8y3 so C(y) = 2y4 +K.

for some constant K. Thus,f(x, y) = x2y + 2y4 +K.

3. Integrating∂f

∂x= yzexyz + z2 cos(xz2)

with respect to x and thinking of y and z as constant gives

f(x, y, z) = exyz + sin(xz2) + C(y, z).

Differentiating with respect to y and using the fact that ∂f/∂y = xzexyz gives

∂f

∂y= xzexyz +

∂C

∂y= xzexyz.

Thus, ∂C/∂y = 0. This means C does not depend on y and can be written C(z), giving:

f(x, y, z) = exyz + sin(xz2) + C(z).

Differentiating with respect to z, we get

∂f

∂z= xyexyz + 2zx cos(xz2) + C ′(z).

The expression for grad f tells us that∂f

∂z= xyexyz + 2xz cos(xz2).

Thus, we have C ′(z) = 0 so C = constant, giving

f(x, y, z) = exyz + sin(xz2) + C.

4. We have∂Fi∂y

= 1 and∂F2

∂x= −1, so

∂F1

∂y6= ∂F2

∂xand this cannot be a gradient vector field.

5. Yes, since ~F = 2xy~i + x2~j = grad(x2y).

6. The domain of the vector field ~F (x, y) = y~i + y~j is the whole xy-plane. In order to see if ~F is a gradient let us applythe curl test:

∂F1

∂y= 1

and∂F2

∂x= 0

So ~F is not the gradient of any function.

7. No, since if ~F = 2xy~i + 2xy~j ,∂F2

∂x= 2y and

∂F1

∂y= 2x,

so∂F2

∂x− ∂F1

∂y6= 0.

18.4 SOLUTIONS 1457

8. The domain of the vector field ~F (x, y) = (x2 + y2)~i + 2xy~j is the whole xy-plane. Let us apply the curl test:

∂F1

∂y= 2y =

∂F2

∂x

so ~F is the gradient of some function f . In order to compute f we first integrate

∂f

∂x= x2 + y2

with respect to x, thinking of y as a constant.We get

f(x, y) =x3

3+ xy2 + C(y)

Differentiating with respect to y and using the fact that ∂f/∂y = 2xy gives

∂f

∂y= 2xy + C ′(y) = 2xy

Thus C ′(y) = 0 so C is a constant and

f(x, y) =x3

3+ xy2 + C.

9. The domain of the vector field ~F = (2xy3 + y)~i + (3x2y2 + x)~j is the whole xy-plane. We apply the curl test:

∂F1

∂y= 6xy2 + 1 =

∂F2

∂x

so ~F is the gradient of a function f . In order to compute f we first integrate

∂f

∂x= 2xy3 + y

with respect to x thinking of y as a constant. We get

f(x, y) = x2y3 + xy + C(y)

Differentiating with respect to y and using the fact that ∂f/∂y = 3x2y2 + x gives

∂f

∂y= 3x2y2 + x+ C ′(y) = 3x2y2 + x

Thus C ′(y) = 0 so C is constant andf(x, y) = x2y3 + xy + C.

10. The domain of the vector field ~F =~i

x+~j

y+~k

zis the set of points (x, y, z) in the three space such that x 6= 0, y 6= 0

and z 6= 0. This is what is left in the three space after removing the coordinate planes. This domain has the property thatevery closed curve is the boundary of a surface entirely contained in it, hence we can apply the curl test.

curl ~F =

∣∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

1x

1y

1z

∣∣∣∣∣∣∣

So curl ~F = ~0 and thus ~F is the gradient of a function f . In order to compute f we first integrate

∂f

∂x=

1

x

with respect to x, thinking of y and z as constants. We get

f(x, y, z) = ln |x|+ C(y, z)


Differentiating with respect to y and using the fact that ∂f/∂y = 1/y gives

∂f

∂y=∂C

∂y=

1

y

We integrate this relation with respect to y thinking of z as a constant. We get

f(x, y, z) = ln |xy|+K(z)

Differentiating with respect to z and using the fact that ∂f/∂z = 1/z gives

∂f

∂z= K′(z) =

1

z

Now we integrate with respect to z and getf(x, y, z) = lnA|xyz|

where A is a positive constant.

11. The domain of the vector field ~F =~i

x+~j

y+~k

xyis the set of points in the three space, (x, y, z) such that x 6= 0 and

y 6= 0. This is the set of points in the three space left after removing the planes x = 0 and y = 0. This domain has theproperty that every closed curve is the boundary of a surface entirely contained in it, hence we can apply the curl test.

curl ~F =

∣∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

1x

1y

1xy

∣∣∣∣∣∣∣

=~i

(∂

∂y(

1

xy)− ∂

∂z(1

y)

)−~j

(∂

∂x(

1

xy)− ∂

∂z(1

x)

)+ ~k

(∂

∂x(1

y)− ∂

∂y(1

x)

)

= − 1

xy2~i +

1

x2y~j 6= 0

Therefore ~F is not the gradient of any function.

12. The domain of the vector field ~F = 2x cos(x2 + z2)~i + sin(x2 + z2)~j + 2z cos(x2 + z2)~k is the whole three spaceso we can apply the curl test.

curl ~F =

∣∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

2x cos(x2 + y2) sin(x2 + y2) 2z cos(x2 + y2)

∣∣∣∣∣∣∣= −4yz sin(x2 + y2)~i + 4xz sin(x2 + y2)~j + (2x cos(x2 + y2) + 4xy sin(x2 + y2))~k 6= 0

As curl ~F 6= ~0 , ~F is not the gradient of any function.

13. We have

∂F1

∂y=

(x2 + y2)1− y(2y)

(x2 + y2)2=

x2 − y2

(x2 + y2)2

∂F2

∂x= − (x2 + y2)1− x(2x)

(x2 + y2)2= − y2 − x2

(x2 + y2)2=

x2 − y2

(x2 + y2)2.

Thus∂F1

∂y=∂F2

∂x. However, the domain of the vector field contains a “hole” at the origin, so the curl test does not apply.

This is not a gradient field. See Example 7 on page 957 of the text.

14. By Green’s Theorem, with R representing the interior of the circle,∫

C

~F · d~r =

∫

R

(∂

∂x(−x)− ∂

∂y(y)

)dA = −2

∫

R

dA

= −2 · Area of circle = −2π(12) = −2π.

18.4 SOLUTIONS 1459

15. By Green’s Theorem, with R representing the interior of the square,∫

C

~F · d~r =

∫

R

(∂

∂x(xy)− ∂

∂y(0)

)dA =

∫

R

y dA

=

∫ 1

0

∫ 1

0

y dy dx =

∫ 1

0

y2

2

∣∣∣∣1

0

dx =1

2.

16. By Green’s Theorem, with R representing the interior of the triangle,∫

C

~F · ~r =

∫

R

(∂

∂x(2x+ 3y2)− ∂

∂y(2x2 + 3y)

)dA =

∫

R

(2− 3) dA = −∫

R

dA

= − Area of triangle = −1

2· 4 · 3 = −6.

17. By Green’s Theorem, with R representing the interior of the circle,∫

C

~F · d~r =

∫

R

(∂

∂x(xy)− ∂

∂y(3y)

)dA =

∫

R

(y − 3) dA.

The integral of y over the interior of the circle is 0, by symmetry, because positive contributions of y from the top half ofthe circle cancel those from the bottom half. Thus ∫

R

y dA = 0.

So ∫

C

~F · d~r =

∫

R

(y − 3) dA =

∫

R

−3 dA = −3 · Area of circle = −3 · π(1)2 = −3π.

18. Green’s theorem gives∫

C

((3x+ 5y)~i + (2x+ 7y)~j ) · d~r =

∫ ∫

R

(− ∂

∂y(3x+ 5y) +

∂

∂x(2x+ 5y)

)dA

=

∫ ∫

R

−3dA = −3 · Area of R = −3πm2.

19. (a) The vector field points in the opposite direction to the orientation of the curve, hence the circulation is negative. SeeFigure 18.19.

1−1

1

−1

x

y

Figure 18.19

(b) Since ~F = y~i , we have ∂F1/∂y = 1 and ∂F2/∂x = 0 and ∂F1/∂y = 1. Thus, using Green’s Theorem if R isthe region enclosed by the closed curve C (the unit circle centered at the origin and traversed counterclockwise), wehave ∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dx dy =

∫

R

(−1) dx dy = −Area of R = −π.


Problems

20. The perimeter of the rectangle is a closed curve, C, so we can use Green’s Theorem. See Figure 18.20. The curve istraversed in the correct direction to apply Green’s Theorem directly. Let R be the interior of the rectangle,

∫

C

~F · d~r =

∫

R

(∂(x+ y)

∂x− ∂(sinx)

∂y

)dxdy =

∫

R

1 dxdy = Area of rectangle = 4 · 5 = 20.

(−1, 0) (3, 0)

(3, 5)(−1, 5)

-� 4

6

?

5 R

x

y

Figure 18.20

21. The curve is closed, so we can use Green’s Theorem. If R represents the interior of the region∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dA =

∫

R

(∂(x)

∂x− ∂(x− y)

∂y

)dA

=

∫

R

(1− (−1)) dA =

∫

R

2 dA = 2 · Area of sector.

Since R is 1/8 of a circle, R has area π(32)/8. Thus,∫

C

~F · d~r = 2 · 9π

8=

9π

4.

22. The curve is closed, so we can use Green’s Theorem. If R represents the interior of the region∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dA =

∫

R

(∂(sin y)

∂x− ∂(x+ y)

∂y

)dA

=

∫

R

(−1) dA = (−1) · Area of sector.

Since R is 1/8 of a circle, R has area π(32)/8. Thus,∫

C

~F · d~r = (−1) · 9π

8= −9π

8.

23. (a) Since ~F = grad(x2ey) is a gradient vector field and C is a closed curve,∫C~F · d~r = 0.

(b) Since∂G2

∂x− ∂G1

∂y=

∂

∂x(x+ y)− ∂

∂y(x− y) = 2,

by Green’s Theorem,∫

C

~G · d~r =

∫

R

(∂G2

∂x− ∂G1

∂y

)dA = 2 · Area of triangle =

2 · 3 · 82

= 24.

18.4 SOLUTIONS 1461

24. (a) C1 is a line along the vertical axis; C2 is a half circle from the positive y to the negative y-axis. See Figure 18.21.

−1

1

C1

C2

x

y

Figure 18.21

(b) Either use Green’s Theorem or calculate directly. Using Green’s Theorem, with R as the region inside C, we get∫

C1+C2

~F · d~r =

∫

R

(∂

∂x(y)− ∂

∂y(x+ 3y)

)dA

=

∫

R

−3 dA = −3(Area of region) = −3π · 12

2= −3π

2.

25. Since ~F = x~j , we have ∂F2/∂x = 1 and ∂F1/∂y = 0. Thus, using Green’s Theorem if R is the region enclosed by theclosed curve C, we have

∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dx dy =

∫

R

1 dx dy = Area of R

26. Using ~F = x~j = a cos t~j and ~r ′(t) = −a sin t~i + b cos t~j , we have

A =

∫

C

~F · d~r =

∫ 2π

0

(a cos t)(b cos t) dt

= ab

∫ 2π

0

cos2 t dt

= ab

∫ 2π

0

1 + cos 2t

2dt

= πab+ab

4sin 2t

∣∣2π0

= πab

The ellipse is shown in Figure 18.22.

a

b

x

y

Figure 18.22:x2

a2+y2

b2= 1


27. Using ~F = x~j = a cos3 t and ~r ′(t) = −3a cos2 t sin t~i + 3a sin2 t cos t~j , we have

A =

∫

C

~F · d~r =

∫ 2π

0

(a cos3 t)(3a sin2 t cos t) dt

= 3a2

∫ 2π

0

cos4 t sin2 t dt = 3a2

∫ 2π

0

cos2 t(sin t cos t)2 dt = 3a2

∫ 2π

0

cos2 tsin2 2t

4dt

=3a2

16

∫ 2π

0

(1 + cos 2t)(1− cos 4t) dt

=3a2

16

∫ 2π

0

(1 + cos 2t− cos 4t− cos 2t cos 4t) dt

=3a2

16

∫ 2π

0

(1 + cos 2t− cos 4t− 1

2cos 6t− 1

2cos 2t) dt

=3a2

16(t− 1

2sin 2t− 1

4sin 4t+

1

12sin 6t+

1

4sin 2t)

∣∣∣∣2π

0

=3πa2

8

For the last integral we use the trigonometric formula cos 2t cos 4t = 12(cos 6t+ cos 2t).

The hypocycloid is shown in Figure 18.23.

a

a

x

y

Figure 18.23: x2/3 + y2/3 = a2/3

2

2

x

y

Figure 18.24: x3 + y3 = 3xy

28. Using ~F = x~j =3t2

1 + t3~j and ~r ′(t) =

1− 2t3

(1 + t3)2~i +

3t(2− t3)

(1 + t3)2~j , we have

A =

∫

C

~F · d~r =

∫ ∞

0

3t

1 + t3· 3t(2− t3)

(1 + t3)2dt

= 9

∫ ∞

0

t2(2− t3)

(1 + t3)3dt

We make the change of variables u = 1 + t3 so du = 3t2dt and 2− t3 = 3− u. So

A = 3

∫ ∞

1

3− uu3

du.

This is an improper integral, so it can be computed as follows

A = 3

∫ ∞

1

3− uu3

du = limb→∞

3

∫ b

1

(3

u3− 1

u2

)du

= limb→∞

[9(−1

2

)u−2

∣∣∣∣b

1

+ 31

u

∣∣∣∣b

1

]

18.4 SOLUTIONS 1463

= limb→∞

[−9

2

(1

b2− 1)

+ 3(

1

b− 1)]

= −9

2(0− 1) + 3(0− 1) =

3

2.

The Folium of Descartes is shown in Figure 18.24.

29. (a) The curve, C, is closed and oriented in the correct direction for Green’s Theorem. See Figure 18.25. Writing R forthe interior of the circle, we have

∫

C

((x2 − y)~i + (y2 + x)~j

)· d~r =

∫

R

(∂(y2 + x)

∂x− ∂(x2 − y)

∂y

)dxdy

=

∫

R

(1− (−1)) dxdy = 2

∫

R

dxdy

= 2 · Area of circle = 2(π · 32) = 18π.

(b) The circle given has radius R and center (a, b). The argument in part (a) works for any circle of radius R, orientedcounterclockwise. So the line integral has the value 2πR2.

5

4 R C

(x− 5)2 + (y − 4)2 = 9

x

y

Figure 18.25

−2

2

5

−5

y

x

x2 + y2 = 25

Figure 18.26

30. Suppose C encloses a region R. Then, using Green’s Theorem, we have∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dA

=

∫

R

∂

∂x(4x(1− y2) + x sin(xy))− ∂

∂y(−y3 + y sin(xy)) dA

=

∫

R

4(1− y2) + sin(xy) + xy cos(xy) + 3y2 − sin(xy)− xy cos(xy) dA

=

∫

R

(4− y2) dA

This integral over R is largest if C encloses the maximum possible region where 4− y2 > 0, that is, where −2 ≤ y ≤ 2.Therefore C should be the curve with two sides along the lines y = −2 and y = 2, as well as two arcs of the circlex2 + y2 = 25. See Figure 18.26.

31. Since the level curves must be perpendicular to the gradient vectors, if there were a contour diagram fitting this gradientfield, it would have to look like Figure 18.27. However, this diagram could not be the contour diagram because the originis on all contours. This means that f(0, 0) would have to take on more than one value, which is impossible. At a point Pother than the origin, we have the same problem. The values on the contours increase as you go counterclockwise around,since the gradient vector points in the direction of greatest increase of a function. But, starting at P , and going all the wayaround the origin, you would eventually get back to P again, and with a larger value of f , which is impossible.

An additional problem arises from the fact that the vectors in the original vector field are longer as you go away fromthe origin. This means that if there were a potential function f then || grad f || would increase as you went away from theorigin. This would mean that the level curves of f would get closer together as you go outward which does not happen inthe contour diagram in Figure 18.27.


x

y

Figure 18.27

32. The drawing of the contour diagrams in Figure 18.28 fitting this gradient field would look like Figure 18.28. The valueson the contours would increase both as y increases (for positive x) and as y decreases (for negative x), following the rulethat the gradient vector points in the direction of greatest increase of a function. Therefore, it is impossible for this to be acontour diagram.

x

y

o

?

6Increasingf -values

Increasingf -values

Figure 18.28

33. (a) We see that ~F , ~G , ~H are all gradient vector fields, since

grad(xy) = ~F for all x, y

grad(arctan(x/y)) = ~G except where y = 0

grad((x2 + y2)1/2

)= ~H except at (0, 0).

Other answer are possible. For example grad(− arctan(y/x)) = ~G for x 6= 0.(b) Parameterizing the unit circle, C, by x = cos t, y = sin t, 0 ≤ t ≤ π, we have ~r ′(t) = − sin t~i + cos t~j , so

∫

C

~F · d~r =

∫ 2π

0

((sin t)~i + (cos t)~j ) · ((− sin t)~i + (cos t)~j ) dt =

∫ 2π

0

cos(2t) dt = 0.

The vector field ~G is tangent to the circle, pointing in the opposite direction to the parameterization, and of length 1everywhere. Thus ∫

C

~G · d~r = −1 · Length of circle = −2π.

The vector field ~H points radially outward, so it is perpendicular to the circle everywhere. Thus∫

C

~H · d~r = 0.

(c) Green’s Theorem does not apply to the computation of the line integrals for ~G and ~H because their domains do notinclude the origin, which is in the interior, R, of the circle. Green’s Theorem does apply to ~F = y~i + x~j .

∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dx dy =

∫0 dx dy = 0.

18.4 SOLUTIONS 1465

34. (a) I Green’s Theorem can be used. The curve is closed and the vector field is smooth throughout the interior of theregion enclosed

II Green’s Theorem cannot be used. The vector field is not defined at the origin which is inside the curve.III Green’s Theorem cannot be used. The curve is not closed.

(b) For the integral in [I], let R be the region enclosed by C. See Figure 18.29. Green’s Theorem gives∫

C

(x2 + y2) dx+ (x2 + y2) dy =

∫

R

(∂F2

∂x− ∂F1

∂y

)dA =

∫

R

(∂

∂x(x2 + y2)− ∂

∂y(x2 + y2)

)dA

=

∫

R

(2x− 2y) dA =

∫ 1

0

∫ x

x2

(2x− 2y) dy dx

=

∫ 1

0

(2xy − y2)

∣∣∣∣x

x2

dx =

∫ 1

0

(2x2 − x2 − (2x3 − x4)) dx

=

∫ 1

0

(x2 − 2x3 + x4) dx =

(x3

3− 2

4x4 +

x5

5

)∣∣∣∣1

0

=1

30.

1

1

y = x

y = x2

x

y

Figure 18.29

35. Green’s theorem says that for a closed curve C oriented counterclockwise, bounding region R,∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dA =

∫

R

(x2 + y2 − 1) dA.

If R is a region contained strictly inside the unit circle, then x2 + y2 < 1 for any point (x, y) in R, so x2 + y2 − 1 < 0,which gives ∫

R

(x2 + y2 − 1) dA < 0, which implies that∫

C

~F · d~r < 0.

Now, let C be the curve C1 − C2. Since∫

C

~F · d~r =

∫

C1−C2

~F · d~r =

∫

C1

~F · d~r −∫

C2

~F · d~r = L1 − L2 < 0,

we have L1 < L2. Similarly, if we let C = C2 − C3, then∫

C

~F · d~r =

∫

C2

~F · d~r −∫

C3

~F · d~r = L2 − L3 < 0,

which gives L2 < L3. ThusL1 < L2 < L3.


36. (a) Writing R1 for the interior of the circle, Green’s Theorem gives∫

C1

~F · d~r =

∫

R1

(∂F2

∂x− ∂F1

∂y

)dA =

∫

S

3 dA = 3 · Area of disk = 3 · π12 = 3π.

(b) Writing R2 as the interior of the rectangle, Green’s Theorem gives∫

C2

~F · d~r =

∫

R2

3 dA = 3 · Area of rectangle = 3 · 3 · 2 = 18.

(c) In parts (a) and (b), we see that the line integral is three times the area enclosed in the curve. Since C3 encloses a diskof radius 7 and area π · 72 = 153.9, and C4 encloses a disk of radius 8 and area π · 82 = 201.1, and C5 encloses asquare of side 14 and area 142 = 196, we have

∫

C3

~F · d~r <∫

C5

~F · d~r <∫

C4

~F · d~r .

37. (a) We use Green’s Theorem. Let R be the region enclosed by the circle C. Then∫

C

~F · d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dA =

∫

R

(∂

∂x(ey

2

+ 12x)− ∂

∂y(3x2y + y3 + ex)

)dA

=

∫

R

(12− (3x2 + 3y2) dA =

∫

R

(12− 3(x2 + y2)) dA.

Converting to polar coordinates, we have∫

C

~F · d~r =

∫ 2π

0

∫ 1

0

(12− 3r2)r dr dθ = 2π(

6r2 − 3

4r4) ∣∣∣∣

1

0

= 2π(

6− 3

4

)=

21π

2.

(b) The integrand of the integral over the disk R is 12− 3(x2 + y2). Since the integrand is positive for x2 + y2 < 4 andnegative for x2 + y2 > 4, the integrand is positive inside the circle of radius 2 and negative outside that circle. Thus,the integral over R increases with a until a = 2 and then decreases. The maximum value of the line integral occurswhen a = 2.

38. (a) We can show

curl ~E =

∣∣∣∣∣∣∣

~ı ~ ~k∂∂x

∂∂y

∂∂z

qx||~r ||3

qy||~r ||3

qz||~r ||3

∣∣∣∣∣∣∣= ~0 .

Let’s check, for instance, the~ı component of curl ~E :

∂

∂y

qz

(x2 + y2 + z2)3/2− ∂

∂z

qy

(x2 + y2 + z2)3/2=

(−3/2)2qyz − (−3/2)2qzy

(x2 + y2 + z2)5/2= 0.

The vector field ~E is a gradient vector field, as curl ~E = 0 and ~E is defined everywhere in 3-space except at theorigin. This domain satisfies the criteria for the curl test in 3-space. Every closed curve in 3-space which does notpass through 0 bounds a surface not containing the origin.

(b) The function ϕ(~r ) = q/||~r || is a potential for ~E , since

∂ϕ

∂x= q

∂

∂x(x2 + y2 + z2)−1/2 = −qx(x2 + y2 + z2)−3/2 = −E1

and similarly for ∂ϕ/∂y and ∂ϕ/∂z; hence ~E = − gradϕ.

Solutions for Chapter 18 Review

Exercises

1. The angle between the vector field and the curve is more than 90◦ at all points on C, so the line integral is negative.

SOLUTIONS to Review Problems for Chapter Eighteen 1467

2. On the top half of the circle, the angle between the vector field and the curve is less than 90◦, so the line integral is positive.On the bottom half of the circle, the angle is more than 90◦, so the line integral is negative. However the magnitude of thevector field is larger on the top half of the curve, so the positive contribution to the line integral is larger than the negative.Thus the line integral

∫C~F · d~r is positive.

3. (a) The line integral around A is zero, because the curve is perpendicular to the field everywhere.(b) The line integral along C1 or C3 is zero because the curves are everywhere perpendicular to the vector field. Along

C2, the line integral is negative, since ~F points along the opposite direction to the curve. Along C4, the line integralis positive, since ~F points in the same direction as the curve.

(c) The line integral around C is zero because C1 and C3 are perpendicular to the field and the contributions from C2

and C4 cancel out.

4. (a) The line integral around A is negative, because the vectors of the field are all pointing in the opposite direction to thedirection of the path.

(b) Along C1, the line integral is positive, since ~F points in the same direction as the curve. Along C2 or C4, the lineintegral is zero, since ~F is perpendicular to the curve everywhere. Along C3, the line integral is negative, since ~Fpoints in the opposite direction to the curve.

(c) The line integral around C is negative because C3 is longer than C1 and the magnitude of the field is bigger alongC3 than C1.

5. Since ~F = 6~i − 7~j , consider the function ff(x, y) = 6x− 7y.

Then we see that grad f = 6~i − 7~j , so we use the Fundamental Theorem of Calculus for Line Integrals:∫

C

~F · d~r =

∫

C

grad f · d~r

= f(4, 4)− f(2,−6) = (−4)− (54) = −58.

6. Only the~i -component of the vector field contributes to the integral. This component, 5~i , points in the opposite directionto the orientation of the path, which has length 8. Thus,

∫

C

(5~i + 7~j ) · d~r = −5 · Length of path = −5 · 8 = −40.

7. We know that if f(x, y) =x2

2+y2

2, then grad f = x~i + y~j = ~F . Thus,

∫

C

~F · d~r = 0.

8. Since ~F is a gradient field, ~F = grad

(x2

2+y2

2

), we have

∫

C

~F · d~r =

(x2

2+y2

2

)∣∣∣∣(0,10)

(0,0)

=100

2− 0 = 50.

9. Only the ~j component contributes to the integral. On the y-axis, x = 0, so

∫

C

~F · d~r =

∫ 5

3

y2~j ·~j dy =y3

3

∣∣∣∣∣

5

3

=98

3.

10. Only the~i -component contributes to the integral, so

∫

C

~F · d~r =

∫ 3

2

x2~i ·~i dx =

∫ 3

2

x2 dx =x3

3

∣∣∣∣∣

3

2

=19

3.


11. We can parameterize the curve C by (t, t2 + 1), for 0 ≤ t ≤ 1. Then∫

C

~F · d~r =

∫ 1

0

~F (t, t2 + 1) · (~i + 2t~j )dt =

∫ 1

0

((−1)~i + (t4 + 2t2 + t+ 1)~j ) · (~i + 2t~j )dt

=

∫ 1

0

(−1 + 2t(t4 + 2t2 + t+ 1))dt =

∫ 1

0

(−1 + 2t5 + 4t3 + 2t2 + 2t)dt

=

(−t+

2t6

6+

4t4

4+

2t3

3+ t2

)∣∣∣∣1

0

= 2

12. Since ~F = grad

(x2

2+y2

2+z2

2


∫

C

~F · d~r =

(x2

2+y2

2+z2

2

)∣∣∣∣(0,0,7)

(2,3,0)

=72

2−(

22

2+

32

2

)= 18.

13. The triangle C consists of the three paths shown in Figure 18.30.

(0, 0) (3, 0)

(3, 2)

C1

C2C3

Figure 18.30

Write C = C1 + C2 + C3 where C1, C2, and C3 are parameterized by

C1 : (t, 0) for 0 ≤ t ≤ 3; C2 : (3, t) for 0 ≤ t ≤ 2; C3 : (3− 3t, 2− 2t) for 0 ≤ t ≤ 1.

Then ∫

C

~F · d~r =

∫

C1

~F · d~r +

∫

C2

~F · d~r +

∫

C3

~F · d~r

where∫

C1

~F · d~r =

∫ 3

0

~F (t, 0) ·~i dt =

∫ 3

0

(2t+ 4)dt = (t2 + 4t)∣∣30

= 21

∫

C2

~F · d~r =

∫ 2

0

~F (3, t) ·~j dt =

∫ 2

0

(5t+ 3)dt = (5t2/2 + 3t)∣∣20

= 16

∫

C3

~F · d~r =

∫ 1

0

~F (3− 3t, 2− 2t) · (−3~i − 2~j )dt

=

∫ 1

0

((−4t+ 8)~i + (−19t+ 13)~j ) · (−3~i − 2~j )dt

= 50

∫ 1

0

(t− 1)dt = −25.

So ∫

C

~F d~r = 21 + 16− 25 = 12.

14. Using x as the parameter we have dy = 2xdx. Thus∫

C

3x2dx+ 4ydy =

∫ 5

1

3x2dx+ 4x2(2xdx) =

∫ 5

1

x2 + 8x3 dx = x3 + 2x4|51 = 1372.


15. Using x as the parameter we have dy = cosx dx. Thus∫

C

ydx+ xdy =

∫ π/2

0

sinx dx+ x(cosx dx) =

∫ π/2

0

sinx+ x cosx dx

= − cosx+ cosx+ x sinx|π/20 =π

2.

16. We have F1 = y2 and F2 = x. By Green’s Theorem∫

C

(y2~i + x~j

)· d~r =

∫

R

(∂F2

∂x− ∂F1

∂y

)dx dy =

∫ 3

0

∫ 2

0

(1− 2y) dx dy = −12.

17. The domain is all 3-space. Since F1 = y,

curl y~i =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k = −~k 6= ~0 ,

so ~F is not path-independent.

18. The domain is all 3-space. Since F2 = y,

curl y~j =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k = ~0 ,

so ~F is path-independent.

19. The domain is all 3-space. Since F3 = z,

curl z~k =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k = ~0 ,


20. Since F2 = F3 = z,

curl (z~j + z~k ) =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k = −~i 6= ~0 ,


21. The domain is all 3-space. Since F1 = y, F2 = x,

curl y~i + x~j =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k = ~0 ,

so ~F is path-independent

22. The domain is all 3-space. Since F1 = x+ y,

curl (x+ y)~i =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k = −~k 6= ~0 ,


23. The domain is all 3-space. Since F1 = yz, F2 = zx, F3 = xy,

curl (yz~i + zx~j + xy~k ) =

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k

= (x− x)~i + (y − y)~j + (z − z)~k = ~0 ,



24. Since the line is parallel to the y-axis, only the ~j -component contributes to the line integral. On C, we have x = 2, so~F = 10~i + 6~j and d~r = ~j dy. Thus,

∫

C

~F · d~r =

∫ 8

3

6~j ·~j dy = 6 · 5 = 30.

25. Since the line is parallel to the x-axis, only the~i -component contributes to the line integral. On C, we have d~r = ~i dx,so ∫

C

~F · d~r =

∫ 12

2

5x~i ·~i dx =5

2x2

∣∣∣∣∣

12

2

= 350.

26. Parameterizing C by x(t) = t, y(t) = t2, with 1 ≤ t ≤ 2, we have ~r ′(t) =~i + 2t~j . Thus,

∫

C

~F · d~r =

∫ 2

1

(5t~i + 3t~j ) · (~i + 2t~j ) dt =

∫ 2

1

(5t+ 6t2) dt =

(5t2

2+ 2t3

)∣∣∣∣∣

2

1

=43

2.

27. Parameterizing C by x(t) = 3 cos t, y = 3 sin t, with 0 ≤ t ≤ π, we have ~r ′(t) = −(3 sin t)~i + (3 cos t)~j . Thus,∫

C

~F · d~r =

∫ π

0

(15 cos t~i + 9 cos t~j ) · (−3 sin t~i + 3 cos t~j ) dt

= 9

∫ π

0

(−5 cos t sin t+ 3 cos2 t) dt

= 9(

5

2cos2 t+

3

2(cos t sin t+ t)

) ∣∣∣∣∣

π

0

=27π

2.

The integral∫

cos2 t was calculated using Formula IV-18.

28. We use Green’s Theorem:∫

C

~F · d~r =

∫

C

(5x~i + 3x~j ) · d~r =

∫

R

(∂(3x)

∂x− ∂(5x)

∂y

)dx dy

=

∫

R

3 dx dy = 3 · Area of region = 3(3 · 2 +1

23 · 3) =

63

2.

29. We can calculate this line integral either by calculating a separate line integral for each side, or by adding a line segment,C1, from (1, 4) to (1, 1) to form the closed curveC+C1. Since we now have a closed curve, we can use Green’s Theorem:

∫

C+C1

~F · d~r =

∫

C+C1

(5x~i + 3x~j ) · d~r =

∫

R

(∂

∂x(3x)− ∂

∂y(5y)

)dx dy

=

∫

R

3 dx dy = 3 · Area of region = 3(

2 · 3 +1

23 · 4

)= 36.

Since d~r = −~j dy on C1, we have∫

C1

~F · d~r =

∫ 1

4

3 · 1~j · (−~j dy) = −3 · 3 = −9.

Since ∫

C+C1

~F · d~r =

∫

C

~F · d~r +

∫

C1

~F · d~r =

∫

C

~F · d~r − 9 = 36

we have ∫

C

~F · d~r = 45.


Problems

30. If a vector field is a gradient vector field, it has zero circulation around every closed curve. Vector fields (i) and (iii) donot have this property. Therefore, (ii) and (iv) could represent gradient vector fields.

31. (a) The curves C1 and C3 give line integrals which we expect to be zero because at every point, the curve looks perpen-dicular to the vector field.

(b) The curve C4 gives a negative line integral because the path is traversed in the direction opposite to the vector field.(c) The line integrals along C2, C5, C6 and C7 are all positive. The vector field is path-independent; it is the gradient

of a function f whose contours appear to be equally spaced circles centered at the origin; the value of f increasesgoing outward. By the Fundamental Theorem of Line Integrals, the value of a line integral is the difference betweenthe values of f at the two endpoints. The difference between the radii of the circles containing the endpoints of C2

and the difference between the radii of the circles containing the endpoints of C6 look about the same, so the lineintegrals along C2 and C6 are approximately equal. Since C6 and C7 have the same endpoints, their line integrals arealso equal. The difference between the radii of the circles containing the endpoints of C2 is less than the differencebetween the radii of the circles containing the endpoints of C5, so the line integral along C2 is smaller than the lineintegral along C5. Thus

C2 = C6 = C7 < C5.

32. (a) Since ~F = grad(yex), we can use the Fundamental Theorem which says that

∫

C

~F · d~r = yex

∣∣∣∣∣

(3,7)

(1,2)

= 7e3 − 2e1.

It does not matter how the curve goes because the Fundamental Theorem gives the line integral in terms of the valuesof the function f(x, y) = yex at the end points only.

(b) The line is given by ~r =~i + 2~j + t(2~i + 5~j ) = (1 + 2t)~i + (2 + 5t)~j . Thus, r′(t) = 2~i + 5~j , so∫

C

~F · d~r =

∫ 1

0

((2 + 5t)e1+2t~i + e1+2t~j

)· (2~i + 5~j ) dt =

∫ 1

0

(4 + 10t+ 5)e1+2tdt

=9e1+2t

2

∣∣∣∣1

0

+ 10

∫ 1

0

te1+2tdt.

Using integration by parts with u = t, u′ = 1, v′ = e1+2t, v = e1+2t/2 for the second integral, we get∫

C

~F · d~r =9

2(e3 − e1) + 10

(te1+2t

2

∣∣∣∣1

0

−∫ 1

0

e1+2t

2dt

)

=9

2(e3 − e1) + 5e3 − 5e1+2t

2

∣∣∣∣1

0

=9

2(e3 − e1) + 5e3 − 5

2(e3 − e1)

= 7e3 − 2e.

33. (a) The vector field is everywhere perpendicular to the radial line from the origin to (2, 3), so the line integral is 0.(b) Since the path is parallel to the x-axis, only the~i component of the vector field contributes to the line integral. The~i

component is −3~i on this line, and the displacement along this line is −2~i , so

Line integral = (−3~i ) · (−2~i ) = 6.

(c) The circle of radius 5 has equation x2 + y2 = 25. On this curve, || ~F || =√

(−y2) + x2 =√

25 = 5. In addition,~F is everywhere tangent to the circle, and the path is 3/4 of the circle. Thus

Line integral = || ~F || · Length of curve = 5 · 3

4· 2π(5) =

75

2π.

(d) Use Green’s Theorem. Writing C for the curve around the boundary of the triangle, we have

∂F2

∂x− ∂F1

∂y= 1− (−1) = 2,

so ∫

C

~F · d~r =

∫

Triangle2 dA = 2 · Area of triangle = 2 · 7 = 14.


34. Since ~F (x, y, z) = grad(ex2+yz), we use the Fundamental Theorem of line integrals

∫

C

~F · d~r =

∫

C

grad(ex

2+yz)· d~r = ex

2+yz∣∣∣(3,0,0)

(0,0,0)= e9 − e0 = e9 − 1.

35. (a) Since ~F = grad(x2ey), the Fundamental Theorem of Line Integrals gives

∫

C

~F · d~r = x2ey∣∣∣∣(2,4)

(0,0)

= 4e4.

(b) Since∂G2

∂x− ∂G1

∂y=

∂

∂x(x+ y)− ∂

∂y(x− y) = 2,

we know that ~G is not a gradient field. Parameterizing C by x(t) = t, y(t) = 2t for 0 ≤ t ≤ 2, we have~r ′(t) =~i + 2~j , so

∫

C

~G · d~r =

∫ 2

0

((t− 2t)~i + (t+ 2t)~j ) · (~i + 2~j ) dt

=

∫ 2

0

((t− 2t) + (2t+ 4t)) dt =

∫ 2

0

5t dt =5

2t2∣∣∣∣2

0

= 10.

36. (a) Since ~F = grad(x3/3 + x3y4


∫

C1

~F · d~r =x3

3+ x3y4

∣∣∣∣(−2,0)

(2,0)

= −8

3+ 0−

(8

3+ 0)

= −16

3.

(b) Since a gradient field is path-independent, and the endpoints of C1 and C2 are the same:∫

C2

~F · d~r =

∫

C1

~F · d~r = −16

3.

(c) The vector field ~G is not a gradient vector field, so we parameterize C1. Using x(t) = t, y(t) = 0, from t = 2 tot = −2 gives ∫

C1

~G · d~r =

∫ −2

2

(t4 + 0)~i + (0~j ) ·~i dt =

∫ −2

2

t4 dt =t5

5

∣∣∣∣−2

2

= −64

5.

(d) Parameterizing C2 by x(t) = 2 cos t, y(t) = 2 sin t for 0 ≤ t ≤ π gives∫

C2

~G · d~r =

∫ π

0

(((2 cos t)4 + (2 cos t)3(2 sin t)2

)~i + (2 cos t)2(2 sin t)3~j

)· (−2 sin t~i + 2 cos t~j ) dt

= 32

∫ π

0

(− cos4 t sin t− 2 cos3 t sin3 t+ 2 cos3 t sin3 t) dt

= −32

∫ π

0

cos4 t sin t dt = 32cos5 t

5

∣∣∣∣π

0

= −64

5.

37. (a) Since ~F = (6x+ y2)~i + 2xy~j = grad(3x2 + xy2), the vector field ~F is path independent, so∫

C1

~F · d~r = 0.

(b) Since C1 is closed, we use Green’s Theorem, so∫

C1

~G · d~r =

∫

Interior of C1

(∂

∂x(x+ y)− ∂

∂y(x− y)

)dA

= 2

∫

C1

dA = 2 · Area inside C1 = 2 · 1

2· 2 · 2 = 4.


(c) Since ~F = grad(3x2 + xy2), using the Fundamental Theorem of Line Integrals gives∫

C2

~F · d~r = (3x2 + xy2)

∣∣∣(0,−2)

(2,0)= 0− 3 · 22 = −12.

(d) Parameterizing the circle by

x = 2 cos t y = 2 sin t 0 ≤ t ≤ 3π

2,

givesx′ = −2 sin t y′ = 2 cos t,

so the integral is∫

C2

~G · d~r =

∫ 3π/2

0

((2 cos t− 2 sin t)~i + (2 cos t+ 2 sin t)~j

)· (−2 sin t~i + 2 cos t~j ) dt

=

∫ 3π/2

0

4(− cos t sin t+ sin2 t+ cos2 t+ sin t cos t) dt

= 4

∫ 3π/2

0

dt = 4 · 3π

2= 6π.

38. (a) Since ~F = x~i + y~j = grad

(x2 + y2

2

), we know that ~F is a gradient vector field. Thus, by the Fundamental

Theorem of Line Integrals, ∫

OA

~F · d~r =x2 + y2

2

∣∣∣∣(3,0)

(0,0)

=9

2.

(b) We know that ~F is path independent. If C is the closed curve consisting of the line in part (a) followed by thetwo-part curve in part (b), then ∫

C

~F · d~r = 0.

Thus, if ABO is the two-part curve of part (b) and OA is the line in part (a),∫

ABO

~F · d~r = −∫

OA

~F · d~r = −9

2.

39. Yes, the line integral over C1 is the negative of the line integral over C2. One way to see this is to observe that the vectorfield x~i + y~j is symmetric in the y-axis and that C1 and C2 are reflections in the y axis (except for orientation). SeeFigure 18.31. Since the orientation of C2 is the reverse of the orientation of a mirror image of C1, the two line integralsare opposite in sign.

(1, 0)(−1, 0)

(0, 2)

C2 C1

x

y

Figure 18.31


40. (a) See Figure 18.32. Notice that C = C1 + C2 + C3 is a closed curve.(b) See Figure 18.33.(c) (i) Since the component of ~F in the direction of C1 is −~i ,

∫

C1

~F · d~r = −Length of C1 = −1.

(ii) Since ~F is parallel to C2 and in the same direction, and || ~F || =√

2,

∫

C2

~F · d~r =√

2 · Length of C2 =√

2 ·√

2 = 2.

(iii) Since the component of ~F in the direction of C3 is ~j , and a vector in the direction of C3 is −~j ,∫

C3

~F · d~r = −Length of C3 = −1.

(iv) Since ~F is constant, it is a gradient field and C is closed,∫

C

~F · d~r = 0.

C1

C2C3

(1, 0)

(0, 1)

x

y

Figure 18.32

x

y

Figure 18.33

41. (a) Since∂

∂x(y5 +x)− ∂

∂y(x3−y) = 1+1 = 2, any closed curve oriented counterclockwise will do. See Figure 18.34.

x

y

Figure 18.34

x

y

Figure 18.35

(b) Since∂

∂x(y5 − xy) − ∂

∂y(x3) = −y, any closed curve in the lower half-plane oriented counterclockwise or any

closed curve in the upper half-plane oriented clockwise will do. See Figure 18.35. Other answers are possible.


42. (a) The vector field∇f is perpendicular to the level curves, in direction of increasing f . The length of∇f is the rate ofchange of f in that direction. See Figure 18.36

1 2

1

22.7

23

23.3

Q

P

x

y

Figure 18.36

(b) Longer.(c) Using the Fundamental Theorem of Calculus for Line Integrals, we have

∫

C

∇f · d~r = f(Q)− f(P ) = 22.7− 23 = −0.3.

43. A contour of f is a set on which f does not change, so the total change of f from P to Q, f(P )− f(Q), is zero. If C isa part of a contour of f , we know that grad f is perpendicular to C. This means that the line integral of grad f along C,which also computes the total change in f between its endpoints, must be zero, since the dot products in its definition areall zero.

44. (a) (i) The curveC is the line given by ~r = x~i +y~j , which we can parameterize by x = t, y = −t+1 for 0 ≤ t ≤ 1.Then ~r ′(t) =~i −~j so

∫

C

~v · d~r =

∫ 1

0

((1− t)~i + 2t~j ) · (~i −~j )dt =

∫ 1

0

(1− 3t)dt = −1

2.

(ii) The curve C is the circle given by ~r = x~i + y~j where x = sin t, y = cos t for 0 ≤ t ≤ π2

. Thus ~r ′(t) =

cos t~i − sin t~j and∫

C

~v · d~r =

∫ π/2

0

(cos t~i + 2 sin t~j ) · (cos t~i − sin t~j )dt =

∫ π/2

0

(cos2 t− 2 sin2 t)dt = −π4.

(b) Since the value of the integral along two paths gives different results, ~v is not path independent.

45. Since || ~F || ≤ 7, the line integral cannot be larger than 7 times the length of the curve. Thus∫

C

~F · d~r ≤ 7 · Circumference of circle = 7 · 2π = 14π.

The line integral is equal to 14π if ~F is everywhere of magnitude 7, tangent to the curve, and pointing in the direction inwhich the curve is traversed.

The smallest possible value occurs if the vector field is everywhere of magnitude 7, tangent to the curve and pointingopposite to the direction in which the curve is transversed. Thus

∫

C

~F · d~r ≥ −14π.

46. We’ll assume that the rod is positioned along the z-axis, and look at the magnetic field ~B in the xy-plane. If C is a circleof radius r in the plane, centered at the origin, then we are told that the magnetic field is tangent to the circle and hasconstant magnitude ‖ ~B ‖. We divide the curve C into little pieces Ci and then we sum ~B ·∆~r computed on each pieceCi. But ∆~r points nearly in the same direction as ~B , that is, tangent to C, and has magnitude nearly equal to the length


of Ci. So the dot product is nearly equal to ‖ ~B ‖× length of Ci. When all of these dot products are summed and the limitis taken as ‖∆~r ‖ → 0, we get

∫

C

~B · d~r = ‖ ~B ‖ × length of C = ‖ ~B ‖ × 2πr

Now Ampere’s Law also tells us that ∫

C

~B · d~r = kI

Setting these expressions for the line integral equal to each other and solving for ‖ ~B ‖ gives kI = 2πr‖ ~B ‖, so

‖ ~B ‖ =kI

2πr.

47. (a) An example of a central field is in Figure 18.37.

x

y

Figure 18.37

x

y

Figure 18.38

(b) The vectors of ~F are radial and the contours of f must be perpendicular to the vectors. Therefore, every contourmust be a circle centered at the origin. Sketching some contours results in a diagram like that in Figure 18.38.

(c) No, not every gradient field is a central field, because there are gradient fields which are not perpendicular to circles.An example is the gradient of f(x, y) = y, where grad f = ~j , so the gradient is parallel to the y axis. Thus, ~F = ~jis an example of a gradient field which is not a central field.

(d) When a particle moves around a circle centered at O, no work is done, because ~F is tangent to the circle. Thus theonly work done in moving from P to Q is in moving between the circles. Since ~F is central, the work done on anyradial line between C1 and C3, for example, depends on only the radii of C1 and C3 (~F is parallel to this path andits magnitude is a function of the distance to the center of the circle only). For that reason, on a path which goes fromC1 to C2 and then from C2 to C3, the same amount of work will be done as on a path direct from C1 to C3.

(e) Pick any two points P and Q. Any path between them can be well-approximated by a path which is partly radial andpartly around a circle centered at O. By the answer to part d), the work along any such path depends only on the radiiof the circles on which P and Q sit, not on the path. Thus, the work done is independent of the path. Hence, ~F mustbe path-independent and therefore a gradient field.

48. (a) Since −y~i + x~j is a counterclockwise rotation, both ω and K must be positive. In order to find the values of ω andK, we must look at the velocity field where we know the magnitude. At a radius of 100 m from the center, we knowthat√x2 + y2 = 100, and that ‖~v ‖ = 3 · 105. Thus, using ~v = ω(−y~i + x~j ) we get

‖~v ‖ = ω√

(−y)2 + x2 = 100ω = 3 · 105 meters/hr,

soω = 3000 rad/hr.

Using ~v = K(x2 + y2)−1(−y~i + x~j ) gives

‖~v ‖ = |K|(x2 + y2)−1√

(−y)2 + x2 =K100

1002= 3 · 105 meters/hr,

so K = 3 · 107 meters2·rad/hr.


(b)

Figure 18.39 Figure 18.40

The vector field in Figure 18.39 shows the velocity vectors inside the tornado, (i.e. r < 100 meters). The vectorfield in Figure 18.40 shows the velocity vectors as seen from a great distance (i.e. r >> 100 meters) with the tornadoat the origin.

(c) Let C be the circle of radius r around the origin. If r < 100 meters, the velocity vectors at distance r from the originhave magnitude ωr. Since they are tangent, and point counterclockwise, the circulation is

∫

C

~v · d~r = ‖~v ‖ · Length of C = 2ωπr2.

If r ≥ 100 meters, the vectors at distance r from the origin have magnitude K/r and are again tangent to the circle.The circulation here is ∫

C

~v · d~r = ‖v‖ · Length of C = (K

r)2πr = 2Kπ.

49. The free vortex appears to starts at about r = 200 meters (that’s where the graph changes its behavior) and the tangentialvelocity at this point is about 200 km/hr = 2 · 105 meters/hr.

Since ~v = ω(−y~i + x~j ) for√x2 + y2 ≤ 200, at r = 200 we have

‖~v ‖ = ω√

(−y)2 + x2 = ω(200) = 2 · 105 meters/hr,

soω = 103 rad/hr.

Since ~v = K(x2 + y2)−1(−y~i + x~j ) for√x2 + y2 ≥ 200, at r = 200 we have

‖~v ‖ = K(2002)−1(200) =K

200= 2 · 105 meters/hr

soK = 4 · 107 m2·rad/hr.

50. By Green’s Theorem, if Ra is the interior of Ca∫

Ca

~F · d~r =

∫

Ra

(∂F2

∂x− ∂F1

∂y

)dA.

The quantity∂F2

∂x− ∂F1

∂yis positive for points (x, y) near the origin and negative farther away. This quantity changes

sign where

∂F2

∂x− ∂F1

∂y= 3(x2 + y2)− (x2 + y2)3/2 = 0

3(x2 + y2) = (x2 + y2)3/2

(x2 + y2)1/2 = 3.


Thus∂F2

∂x− ∂F1

∂yis positive within C3, the circle of radius 3, and negative outside. The maximum value of the line

integral occurs when a = 3. Converting to polars,∫

C3

~F · d~r =

∫

R3

(3(x2 + y2)− (x2 + y2)3/2

)dA

=

∫ 2π

0

∫ 3

0

(3r2 − r3)r dr dθ

= 2π

(3r4

4− r5

5

)∣∣∣∣3

0

=35π

10.

CAS Challenge Problems

51. (a) We parameterize Ca by ~r (t) = a cos t~i + a sin t~j . Then, using a CAS, we find∫

Ca

~F (~r (t)) · ~r ′(t) dt =

∫ 2π

0

a cos t

(2a cos t− a3cos t3

3+ a3 cos tsin t2

)

−a sin t

(−(a sin t) +

2a3sin t3

3

)dt

= −π2

(−6a2 + a4)

The derivative of the expression on the right with respect to a is −(2π)(−3a + a3), which is zero at a = 0,±√

3.Checking at a = 0 and as a→∞, we find the maximum is at a =

√3.

(b) We have∂F2

∂x− ∂F1

∂y= (2− x2 + y2)− (−1 + 2y2) = 3− x2 − y2.

So, by Green’s theorem, ∫

Ca

~F · d~r =

∫ ∫

Da

(3− x2 − y2) dA,

where Da is the disk of radius a centered at the origin. The integrand is positive for x2 + y2 < 3, so it is positiveinside the disk of radius

√3 and negative outside it. Thus the integral has its maximum value when a =

√3.

52. We parameterize the line from (0, 0) to (x, y) by ~r (t) = t(x~i + y~j ). Using a CAS to compute the integral, we get

(a)

f(x, y) =

∫ 1

0

~F (~r (t)) · ~r ′(t) dt =

∫ 1

0

2axyt dt = axy + Constant

(b)

f(x, y) =

∫ 1

0

~F (~r (t)) · ~r ′(t)dt

=

∫ 1

0

(abebt2xytxy + (c+ abebt

2xytx)y) dt = aebxy + cy + Constant

53. We have∫

C1

~F · d~r =

∫ 3

0

~F (~r (t)) · ~r ′(t)dt

=

∫ 3

0

(2(2at+ bt2

)+ 2t

(2ct+ dt2

)) dt = 18a+ 18b+ 36c+ (81d/2)

and∫

C2

~F · d~r =

∫ 3

0

~F (~r (t)) · ~r ′(t)dt

CHECK YOUR UNDERSTANDING 1479

=

∫ 3

0

(−2(2a (3− t) + b(3− t)2

)− 2(2c (3− t) + d(3− t)2

)(3− t)

)dt

= −18a− 18b− 36c− (81d/2)

The second integral is the negative of the first. This is because C2 is the same curve as C1 but traveling in the oppositedirection.

CHECK YOUR UNDERSTANDING

1. A path-independent vector field must have zero circulation around all closed paths. Consider a vector field like ~F (x, y) =|x|~j , shown in Figure 18.41.

y

x

Figure 18.41

A rectangular path that is symmetric about the y-axis will have zero circulation: on the horizontal sides, the field isperpendicular, so the line integral is zero. The line integrals on the vertical sides are equal in magnitude and opposite insign, so they cancel out, giving a line integral of zero. However, this field is not path-independent, because it is possible tofind two paths with the same endpoints but different values of the line integral of ~F . For example, consider the two points(0, 0) and (0, 1). The path C1 in Figure 18.42 along the y axis gives zero for the line integral, because the field is 0 alongthe y axis, whereas a path like C2 will have a nonzero line integral. Thus the line integral depends on the path betweenthe points, so ~F is not path-independent.

C1C2

x

y

(0, 0)

(0, 1)

Figure 18.42


2. This is false, because the line integral yields a scalar whereas the total change in ~F would be a vector. In the special casewhen ~F happens to be the gradient of a scalar function f , the line integral does give the total change of the scalar functionf along the path—but not of the vector function ~F .

3. You can easily come up with counterexamples: suppose that ~F 6= ~G but that both are gradient fields. For example,~F =~i and ~G = ~j . Then, if C is a closed curve, the line integral around C of both ~F and ~G will equal to zero. But thisdoes not mean that ~F = ~G .

4. The total change of f along C depends only on the endpoints of C. If f has the same values at each endpoint (or if C isclosed, so that the endpoints coincide) then the total change will be zero. This in no way restricts the shape of the curveC. For example, take f(x, y) = x2 + y2 and C to be the straight line from the point (1, 0) to the point (0, 1). Thenf(1, 0) = f(0, 1) = 1 so the change in f along C is zero, but C is not a contour of f .

5. False. Because ~F ·∆~r is a scalar quantity,∫C~F · d~r is also a scalar quantity.

6. False. The left side is a scalar, the right side is a vector. A true statement is: If ~F = grad f and P andQ are the endpointsof C, then

∫C~F · d~r = f(Q)− f(P ).

7. False. The line integral of a gradient vector field around this circle would be 0, but the converse is not necessarily true.That is, the fact that the line integral around this one circle is zero does not mean ~F is necessarily a gradient field.

8. True. You can trace out C2 using the same subdivisions, but each ∆~r will have the opposite sign as before and will betraced out twice, so

∫C2

~F · d~r = −2∫C1

~F · d~r = −6.

9. False. We can calculate a line integral of any vector field.

10. True. The vector field ~F = x~i + y~j = ~r has radial direction, pointing everywhere perpendicular to the path ofintegration, so the line integral is zero.

11. True. Since x~i + y~j = grad(

12(x2 + y2)


∫

C

(x~i + y~j ) · d~r =1

2(x2 + y2)

∣∣∣∣(a,b)

(0,0)

1

2(a2 + b2).

12. True. The line integral is the limit of a sum of dot products, hence is a scalar.

13. False. The relative sizes of the line integrals along C1 and C2 depend on the behavior of the vector field ~F along thecurves. As a counterexample, take the vector field ~F = ~i , and C1 to be the line from the origin to (0, 2), while C2 isthe line from the origin to (1, 0). Then the length of C1 is 2, which is greater than the length of C2, which is 1. However∫C1

~F · d~r = 0 (since ~F is perpendicular to C1) while∫C2

~F · d~r > 0 (since ~F points along C2).

14. False. For example, the vector field ~F could be perpendicular to C everywhere. For instance, let ~F = ~j and C be thecurve t~i , for 0 ≤ t ≤ 1. Alternatively, ~F might point along part of C and in the opposite direction on another part ofC and so that the sum cancels out, yielding a zero line integral. For instance, let ~F = x~i and C be the curve t~i , for−1 ≤ t ≤ 1.

15. True. All of the dot products ~F (~r i) ·∆~r i in this line integral are positive, since the vector field (the constant~i ) pointsin the same direction as ∆~r i.

16. False. All of the dot products ~F (~r i) · ∆~r i in this line integral are zero, since the vector field (the constant ~i ) pointsperpendicular to ∆~r i.

17. False. The relation between these two line integrals depends on the behavior of the vector field along each of the curves,so there is no reason to expect one to be the negative of the other. As an example, if ~F (x, y) = y~i , then, by symmetry,both line integrals are equal to the same negative number.

18. False. The relation between these two line integrals depends on the behavior of the vector field along each of the curves, sothere is no reason to expect one to be larger than the other. If, for example, the line integral along C2 is negative, then theline integral along both taken together (C1 + C2) will be less than the line integral over C1 by itself. A specific exampleis given by ~F = ~i , with C1 the line from (0, 0) to (1, 0), and C2 the line from (1, 0) to (0, 1). Then

∫C1

~F · d~r =∫ 1

0~i ·~i dt = 1/2, and

∫C1+C2

~F · d~r = 1/2 +∫ 1

0~i · (−~i +~j ) dt = 1/2− 1 = −1/2.

19. False. The vector field swirls counterclockwise about the origin, and the path is oriented clockwise, so the line integral isnegative.

20. False. The line integrals of many vector fields (so called path independent or conservative fields) are zero around closedcurves, but this is not true of all fields. For example, a vector field that is flowing in the same direction as the curve C allalong the curve has a positive line integral. A specific example is given by ~F = −y~i + x~j , where C is the unit circle

CHECK YOUR UNDERSTANDING 1481

centered at the origin, oriented counterclockwise.

21. True. The dot product of the integrand 4~i with ~r ′(t) =~i + 2t~j is 4, so the integral has value∫ 2

04 dt = 8.

22. False. The curves C1 and C2 are different. The curve C1 starts at the point (1, 0) and travels around the unit circlecounterclockwise to (−1, 0). The curveC2 starts at the point (1, 0) and travels around the unit circle clockwise to (−1, 0).

23. True. The curvesC1 andC2 both parameterize the upper unit semicircle with the same orientation (but at different speeds).Since the line integral is independent of parameterization, the integrals over C1 and C2 are the same.

24. False. As a counterexample, consider the unit circle C, centered at the origin, oriented counterclockwise and the vectorfield ~F = −y~i + x~j . The vector field is always tangent to the circle, and in the same direction as C, so the line integralis positive.

25. False. As a counterexample, consider the vector field ~F = x~i . Then if we parameterizeC1 by ~r (t) = t~i , with 0 ≤ t ≤ 1,we get ∫

C1

x~i · d~r =

∫ 1

0

t~i ·~i dt =

∫ 1

0

t dt =t2

2

∣∣∣∣1

0

=1

2.

A similar computation for C2 gives a line integral with value 2.

26. True. If we parameterize C by ~r (t) = a cos t~i + a sin t~j , with 0 ≤ t ≤ 2π, then∫

C

(2x~i + y~j ) · d~r =

∫ 2π

0

(2a cos t~i + a sin t~j ) · (−a sin t~i + a cos t~j ) dt =

∫ 2π

0

−a2 cos t sin t dt

=a2 cos2 t

2

∣∣∣∣2π

0

= 0.

27. False. If we parameterize C by ~r (t) = a cos t~i + a sin t~j , with 0 ≤ t ≤ 2π, then∫

C

(2y~i+x~j )·d~r =

∫ 2π

0

(2a sin t~i+a cos t~j )·(−a sin t~i+a cos t~j ) dt =

∫ 2π

0

(−2a2 sin2 t+ a2 cos2 t

)dt = −πa2.

28. True. The curves C1 and C2 are the same (they follow the graph of y = x2 between (0, 0) and (2, 4)), except that theirorientations are opposite.

29. True. By the Fundamental Theorem for Line Integrals, if C is a path from P to Q, then∫

C

grad f · d~r = f(Q)− f(P ),

so the value of the line integral∫C

grad f · d~r depends only on the endpoints and not the path.

30. False. The statement is true if C1 and C2 have the same initial and final points. For example, ~F (x, y) = ~i is path-independent (since it is the gradient of f(x, y) = x), but the line integral of ~F over a path from (0, 0) to (1, 0) isf(1, 0)− f(0, 0) = 1, but the line integral over a path from (0, 0) to (0, 1) is f(0, 1)− f(0, 0) = 0.

31. False. However, if ~F is a gradient field, the line integral gives the total change in the potential function f , where ~F =grad f .

32. True. The construction at the end of Section 18.3 shows how to make a potential function from a path-independent vectorfield.

33. True. Since a gradient field is path-independent, andC1 andC2 have the same initial and final points, the two line integralsare equal.

34. True. Since the curve can be thought of as having the same initial and final points, if f is a potential function for ~F wehave

∫C~F · d~r =

∫C

grad f · d~r = f(P )− f(P ) = 0.

35. False. If there were a potential function f , then fx = y2, so f(x, y) = xy2 + g(y), where g(y) is a function of y only.Then fy = 2xy + g′(y), which depends on x no matter what g′(y) is. Thus fy cannot be equal to a constant k, and sothere is no potential function f such that ~F = grad f .

36. False. For example, take ~F = y~i . By symmetry, the line integral of ~F over any circle centered at the origin is zero.But the curve consisting of the upper semicircle connecting (−a, 0) to (a, 0) has a positive line integral, while the lineconnecting these points along the x-axis has a zero line integral, so the field cannot be path-independent.

37. True. The value of∂F2

∂x− ∂F1

∂yis 0− 0 = 0, so the field is path-independent.

38. False. The fact that ~F = grad f means that ~F is a potential field, hence ~F is path-independent. Thus∫C~F · d~r = 0

since C is closed.


39. True. Since ~F and ~G are both path-independent, we know ~F = grad f and ~G = grad g for some scalar functions fand g. Then grad(f + g) = grad f + grad g = ~F + ~G , so ~F + ~G is a gradient field, hence path-independent.

40. False. As a counterexample, consider ~F = x~j and ~G = y~i . Then both of these are path-dependent (they each havenonzero curl), but the curl of ~F + ~G = y~i + x~j is zero everywhere, so ~F + ~G is path-independent.

41. True. This vector field has components F1 = x, F2 = y, and F3 = z. Using the 3-space curl test gives zero for all of thecomponents of curl ~F , so the field is path-independent.

42. True. The value of∂F2

∂x− ∂F1

∂yis 0− 0 = 0, so the field is path-independent.

43. True. Since ~F is path-independent, we know ~F = grad f for some scalar function f . Then grad(kf) = k grad f = k ~F ,so k ~F is a gradient field, hence path-independent.

44. False. As a counterexample, consider the vector field ~F = 2x~i , which is path-independent, since it is the gradient off(x, y) = x2. Multiplying ~F by the function h(x, y) = y gives the field y ~F = 2xy~i . The curl of this vector field is−2x 6= 0, so y ~F is path-dependent.

PROJECTS FOR CHAPTER EIGHTEEN

1. (a) Since ‖~v (t)‖2 = ~v (t) · ~v (t) and since ~v (t) = ~r ′(t) = x′(t)~i + y′(t)~j + z′(t)~k , we have

1

2

d

dt‖~v (t)‖2 =

1

2

d

dt(~v (t) · ~v (t))

=1

2

d

dt(x′(t)2 + y′(t)2 + z′(t)2)

=1

2(2x′(t)x′′(t) + 2y′(t)y′′(t) + 2z′(t)z′′(t))

= x′(t)x′′(t) + y′(t)y′′(t) + z′(t)z′′(t)

= (x′(t)~i + y′(t)~j + z′(t)~k ) · (x′′(t)~i + y′′(t)~j + z′′(t)~k )

= ~v (t) · ~a (t) (Since ~a (t) = x′′(t)~i + y′′(t)~j + z′′(t)~k .)

= ~a (t) · ~v (t).

(b) (i) We use ~F = m~a and the parameterization of C given by r(t) for t0 ≤ t ≤ t1. In addition, we needthe fact that 1

2ddt‖~v (t)‖2 = ~a · ~v :

∫

C

~F · d~r =

∫

C

m~a · d~r =

∫ t1

t0

m~a · ~r ′dt

=

∫ t1

t0

m (~a · ~v ) dt

=

∫ t1

t0

m

2

(d

dt‖~v (t)‖2

)dt

=m

2‖~v (t)‖2

∣∣∣∣t1

t0

=m

2‖~v (t1)‖2 − m

2‖~v (t0)‖2

= Kinetic energy at Q− Kinetic energy at P.

(ii) Since ~F = −∇f we use the Fundamental Theorem of Line Integrals:∫

C

~F · d~r =

∫

C

−∇f · d~r = −(f(~r (t1))− f(~r (t0)))

= −(Potential energy at Q− Potential energy at P )

= Potential energy at P − Potential energy at Q.

PROJECTS FOR CHAPTER EIGHTEEN 1483

(iii) In parts (a) and (b) we derived two expressions for the work done by ~F as the particle moves from Pto Q. These two expressions must be equal, so

Kinetic energy at Q− Kinetic energy at P = Potential energy at P − Potential energy at Q.

Rewriting this equation we have,

(Potential energy + Kinetic energy) at P = (Potential energy + Kinetic energy) at Q.

This shows that the total energy is the same at P as at Q. Since P and Q are arbitrary points in space,the total energy of a particle moving in a force vector field ~F = −∇f is a constant.

2. (a) We have~m = (x− a)~i + y~j .

Since ~m has magnitude L we have(x− a)2 + y2 = L2

and sox− a =

√L2 − y2

where we have used the fact that a < x. Thus

~k × ~m = −y~i + (x− a)~j .

Using once more the fact that ~m has magnitude L, we see that the unit vector ~F in the direction of ~k × ~mis

~F =1

L(~k × ~m ) =

−yL~i +

1

L

√L2 − y2~j .

(b) We have

curl ~F =∂F2

∂x− ∂F1

∂y

=1

L

∂√L2 − y2

∂x− 1

L

∂(−y)

∂y=

1

L.

(c) By Green’s theorem and part (b), we have∫

C

~F · d~r =

∫

R

curl ~F dA

=

∫

R

1

LdA =

1

L· (Area of R).

Since ~F is by definition in part (a) a vector field of unit vectors in the direction of the wheel vector alongC, we have

(Area of R) = L

∫

C

~F · d~r = L · (Total roll of planimeter wheel).

3. (a) We take the surface to be a disk of radius r, parallel to the xy-plane and centered on the z-axis. Theboundary of the disk is the circle C. We know that the magnetic field, ~B , is tangent to the circle and hasconstant magnitude, ‖ ~B ‖, along each circle. Thus, for any such circle,

∫

C

~B · d~r = ‖ ~B ‖ · Length of C = 2πr‖ ~B ‖.

If r ≥ r0 (where r0 is the radius of the wire) then the current through the surface is I . Therefore

2πr‖ ~B ‖ =

∫

C

~B · d~r = kI,


so‖ ~B ‖ =

kI

2πr.

If r < r0, then the current flowing through the surface is not I , but the amount of current passing througha disk of radius r. Such a disk has an area which is (πr2)/(πr2

0) of the cross-sectional area of the wire. Sothe current to use in Ampere’s Law is (πr2)/(πr2

0) I . Thus,

2πr‖ ~B ‖ =

∫

C

~B · d~r = kπr2

πr20

I.

Solving for ‖ ~B ‖ gives

‖ ~B ‖ =kIr

2πr20

.

(b) We again use Ampere’s Law on a disk of radius r, lying perpendicular to, and centered on, the z-axis (SeeFigure 18.43). If the boundary, C, of this disk lies inside the torus, then the wire goes through the disk Ntimes, and so the net current through the disk is NI . Thus,

2πr‖ ~B ‖ =

∫

C

~B · d~r = kNI,

which gives

‖ ~B ‖ =kNI

2πr.

On the other hand, if the boundary, C, lies outside the torus, then the net current through the disk is 0. (Thewire goes into the disk N times and out of the disk N times, and so the currents cancel.) Hence we have

2πr‖ ~B ‖ =

∫

C

~B · d~r = 0.

So ‖ ~B ‖ = 0.

x

y

z

(0, β, 0)

�

(β, 0, 0)

RαI

CircleC of radius r;boundary of horizontaldisk

R

Wires formingsolenoid

Figure 18.43

calculus 5th ed_deborah hughes-hallet_chap18-sols

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