But this last integral is precisely the line integral in (10). Therefore we have

(1)

Section 16.2 Line Integrals

8. Evaluate the line integral, where C is the given plane curve. R

Cx²dx + y²dy.

SECTION 16.2 Line Integrals 1141

But this last integral is precisely the line integral in (10). Therefore we have

y

C

F dr − y

C

P dx 1 Q dy 1 R dz where F − P i 1 Q j 1 R k

For example, the integral y

C

y dx 1 z dy 1 x dz in Example 6 could be expressed as y

C

F dr, where

F

sx, y, zd − y i 1 z j 1 x k A similar result holds for vector fields F on R

²

:

14

y

C

F dr − y

C

P dx 1 Q dy

where F − P i 1 Q j.

16.2 Exercises

1–8 Evaluate the line integral, where C is the given plane curve.

1. yC y ds, C: x − t², y − 2t, 0 < t < 3 2. yC sxyyd ds, C: x − t³, y − t⁴, 1 < t < 2

3. yC xy⁴ds, C is the right half of the circle x²1y²− 16 4. yC xe^y ds, C is the line segment from s2, 0d to s5, 4d 5. yC sx²y 1sin xd dy,

C is the arc of the parabola y − x² from s0, 0d to s, ²d 6. yC e^x dx,

C is the arc of the curve x − y³ from s21, 21d to s1, 1d 7. yC sx 1 2yd dx 1 x² dy

0 y

x C

(2, 1)

(3, 0)

8. yC x² dx 1 y² dy

≈+¥=4

0 y

x C

(2, 0) (0, 2)

(_1, 1)

9–18 Evaluate the line integral, where C is the given space curve.

9. yC x²y ds,

C: x − cos t, y − sin t, z − t, 0 < t < y2 10. yC y²z ds,

C is the line segment from s3, 1, 2d to s1, 2, 5d 11. yC xe^yz ds,

C is the line segment from (0, 0, 0) to (1, 2, 3) 12. yC sx²1y²1z²d ds,

C: x − t, y − cos 2t, z − sin 2t, 0 < t < 2

13. yC xye^yz dy, C: x − t, y − t², z − t³, 0 < t < 1 14. yC ye^z dz 1 x ln x dy 2 y dx,

C: x − e^t, y − 2t, z − ln t, 1 < t < 2 15. yC z dx 1 xy dy 1 y² dz,

C: x − sin t, y − cos t, z − tan t, 2y4 < t < y4 16. yC y dx 1 z dy 1 x dz,

C: x −st, y − t, z − t², 1 < t < 4 17. yC z² dx 1 x² dy 1 y² dz,

C is the line segment from s1, 0, 0d to s4, 1, 2d 18. yC sy 1 zd dx 1 sx 1 zd dy 1 sx 1 yd dz,

C consists of line segments from s0, 0, 0d to s1, 0, 1d and from s1, 0, 1d to s0, 1, 2d

Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

1618 ¤ CHAPTER 16 VECTOR CALCULUS

6. Choosing  as the parameter, we have  = ³,  = , −1 ≤  ≤ 1. Then



^ =1

−1^³· 3² = ^³1

−1= ¹− ⁻¹=  −¹.

7.  = 1+ 2

On 1:  = ,  = ¹₂ ⇒  = ¹2, 0 ≤  ≤ 2.

On 2:  = ,  = 3 −  ⇒  = −, 2 ≤  ≤ 3.

Then



( + 2)  + ² =

₁( + 2)  + ² +

₂( + 2)  + ²

=2 0

 + 2₁

2

+ ²₁

2

 +3 2

 + 2(3 − ) + ²(−1)



=2 0

2 +¹₂²

 +3 2

6 −  − ²



=

²+¹₆³2 0+

6 −¹2²− ¹3³3

2= ¹⁶₃ − 0 +⁹2 −²²3 = ⁵₂

8.  = 1+ 2

On 1:  = 2 cos  ⇒  = −2 sin  ,

 = 2 sin  ⇒  = 2 cos  , 0 ≤  ≤ ^2. On 2:  = − ⇒  = −,

 = 2 −  ⇒  = −, 0 ≤  ≤ 1.

Then 

² + ² =

₁² + ² +

₂² + ²

=2

0 (2 cos )²(−2 sin  ) + (2 sin )²(2 cos  ) +1

0(−)²(−) + (2 − )²(−)

=2

0 (−8 cos² sin  + 8 sin² cos )  − 21

0(²− 2 + 2) 

= 81

3cos³ +¹₃sin³^2

0 − 21

3³− ²+ 2¹

0= 81 3− ¹3

− 21

3− 1 + 2

= −⁸3

9.  = cos ,  = sin ,  = , 0 ≤  ≤ 2. Then by Formula 9,



²  =2

0 (cos )²(sin )_



² +



² +



²



=2

0 cos² sin 

(− sin )²+ (cos )²+ (1)² =2

sin² + cos² + 1 

=√ 22

0 cos² sin   =√ 2

−¹3cos³2

0 =√

2 0 +¹₃

= ^√₃²

10. Parametric equations for the line segment  from (3 1 2) to (1 2 5) are  = 3 − 2,  = 1 + ,  = 2 + 3, 0 ≤  ≤ 1.

Then by Formula 9,



 ²  =1

0 (1 + )²(2 + 3)

(−2)²+ 1²+ 3² =√ 141

0 (3³+ 8²+ 7 + 2) 

=√ 143

4⁴+⁸₃³+⁷₂²+ 21 0=√

143

4+ ⁸₃+⁷₂+ 2

= ¹⁰⁷₁₂ √ 14

12. Evaluate the line integral, where C is the given space curve. R

C(x²+ y²+ z²)ds, C : x = t, y = cos 2t, z = sin 2t, 0 ≤ t ≤ 2π.

Solution:

9.  = cos ,  = sin ,  = , 0 ≤  ≤ 2. Then by Formula 9,



²  =2

0 (cos )²(sin )_



2

+



² +



2



=2

(− sin )²+ (cos )²+ (1)² =2

sin² + cos² + 1 

=√ 22

0 cos² sin   =√ 2

−¹3cos³2

0 =√

2 0 +¹₃

= ^√₃² 10. Parametric equations for  are  = 3 − 2,  = 1 + ,  = 2 + 3, 0 ≤  ≤ 1. Then



 ²  =1

0 (1 + )²(2 + 3)

(−2)²+ 1²+ 3² =√ 14 1

0 (3³+ 8²+ 7 + 2) 

=√ 14₃

4⁴+⁸₃³+⁷₂²+ 21 0=√

14₃

4+⁸₃+ ⁷₂+ 2

= ¹⁰⁷₁₂√ 14 11. Parametric equations for  are  = ,  = 2,  = 3, 0 ≤  ≤ 1. Then



 ^ =1

0 ^(2)(3)√

1²+ 2²+ 3² =√ 141

0 ^6² =√ 14

1 12^6²1

0= ^√₁₂¹⁴(⁶− 1).

12. Parametric equations for  are  = −1 + 2,  = 5 + ,  = 4, 0 ≤  ≤ 1. Then



² =1

0 (−1 + 2)(5 + )(4)²√

2²+ 1²+ 4² =√ 211

0 (32⁴+ 144³− 80²) 

=√ 21

 32 ·⁵

5 + 144 ·⁴

4 − 80 ·³ 3

1 0

=√ 21₃₂

5 + 36 −⁸⁰3

= ²³⁶₁₅√ 21

13. 

 ^ =1

0()(²)^(²^)(³⁾· 2  =1

0 2⁴^⁵ = ²₅^⁵1

0= ²₅(¹− ⁰) = ²₅( − 1) 14. 

  +   +   =1

0 ²· 2  + ²· 3² + ³· 2  =1

0(2³+ 5⁴)  =1

2⁴+ ⁵¹

0= ¹₂+ 1 = ³₂ 15. Parametric equations for  are  = 1 + 3,  = ,  = 2, 0 ≤  ≤ 1. Then



 ² + ² + ² =1

0(2)²· 3  + (1 + 3)² + ²· 2  =1 0

23²+ 6 + 1



=₂₃

3³+ 3²+ 1

0= ²³₃ + 3 + 1 = ³⁵₃

16. On 1:  =  ⇒  =   = 0 ⇒

 = 0   =  ⇒  =  0 ≤  ≤ 1.

On 2:  = 1 −  ⇒  = −  =  ⇒

 =   = 1 +  ⇒  =  0 ≤  ≤ 1.

Then

( + )  + ( + )  + ( + ) 

=

₁( + )  + ( + )  + ( + )  +

₂( + )  + ( + )  + ( + ) 

=1

0(0 + )  + ( + ) · 0  + ( + 0)  +1

0( + 1 + )(−) + (1 −  + 1 + )  + (1 −  + ) 

=1

0 2  +1

0(−2 + 2)  =

²1 0+

−²+ 21

0= 1 + 1 = 2

20. The figure shows a vector field F and two curves C1 and C2. Are the line integrals of F over C1 and C2 positive, negative, or zero? Explain.

SeCtion16.2 Line Integrals 1085

24. yC F dr, where Fsx, y, zd − yze^xi 1 zxe^y j 1 xye^z k and rstd − sin t i 1 cos t j 1 tan t k, 0 < t < y4

25. yC xy arctan z ds, where C has parametric equations x − t², y − t³, z −st, 1 < t < 2

26. yC z lnsx 1 yd ds, where C has parametric equations x − 1 1 3t, y − 2 1 t², z − t⁴, 21 < t < 1

27–28 Use a graph of the vector field F and the curve C to guess whether the line integral of F over C is positive, negative, or zero. Then evaluate the line integral.

27. Fsx, yd − sx 2 yd i 1 xy j,

C is the arc of the circle x²1y²− 4 traversed counter

clockwise from (2, 0) to s0, 22d 28. Fsx, yd − x

sx²1y² i 1 y sx²1y² j,

C is the parabola y − 1 1 x² from s21, 2d to (1, 2)

29. (a) Evaluate the line integral yC F dr, where Fsx, yd − e^x21 i 1 xy j and C is given by rstd − t²i 1 t³j, 0 < t < 1.

(b) Illustrate part (a) by using a graphing calculator or com

puter to graph C and the vectors from the vector field corresponding to t − 0, 1ys2, and 1 (as in Figure 13).

30. (a) Evaluate the line integral yC F dr, where Fsx, y, zd − x i 2 z j 1 y k and C is given by rstd − 2t i 1 3t j 2 t²k, 21 < t < 1.

(b) Illustrate part (a) by using a computer to graph C and the vectors from the vector field corresponding to t − 61 and 6¹₂ (as in Figure 13).

31. Find the exact value of yC x³y²zds, where C is the curve with parametric equations x − e^2t cos 4t, y − e^2t sin 4t, z − e^2t, 0 < t < 2.

32. (a) Find the work done by the force field

Fsx, yd − x²i 1 xy j on a particle that moves once around the circle x²1y²− 4 oriented in the counter

clockwise direction.

(b) Use a computer algebra system to graph the force field and circle on the same screen. Use the graph to explain your answer to part (a).

33. A thin wire is bent into the shape of a semicircle x²1y²− 4, x > 0. If the linear density is a constant k, find the mass and center of mass of the wire.

34. A thin wire has the shape of the firstquadrant part of the circle with center the origin and radius a. If the density function is sx, yd − kxy, find the mass and center of mass of the wire.

35. (a) Write the formulas similar to Equations 4 for the center of mass sx, y, zd of a thin wire in the shape of a space curve C if the wire has density function sx, y, zd.

CAS

;

CAS

17. Let F be the vector field shown in the figure.

(a) If C1 is the vertical line segment from s23, 23d to s23, 3d, determine whether yC1 F dr is positive, nega

tive, or zero.

(b) If C2 is the counterclockwiseoriented circle with radius 3 and center the origin, determine whether yC2 F dr is positive, negative, or zero.

y

x

0 1

1

2 3

_3 _2 _1

18. The figure shows a vector field F and two curves C1 and C2. Are the line integrals of F over C1 and C2 positive, negative, or zero? Explain.

y

x C¡

C™

19–22 Evaluate the line integral yC F dr, where C is given by the vector function rstd.

19. Fsx, yd − xy²i 2 x² j, rstd − t³ i 1 t² j, 0 < t < 1

20. Fsx, y, zd − sx 1 y²di 1 xz j 1sy 1 zd k, rstd − t² i 1 t³ j 2 2t k, 0 < t < 2

21. Fsx, y, zd − sin x i 1 cos y j 1 xz k, rstd − t³i 2 t²j 1 t k, 0 < t < 1 22. Fsx, y, zd − x i 1 y j 1 xyk,

rstd − cos t i 1 sin t j 1 tk, 0 < t < 

23–26 Use a calculator to evaluate the line integral correct to four decimal places.

23. yC F dr, where Fsx, yd −sx 1 y i 1syyxd j and rstd − sin²t i 1 sin t cos t j, y6 < t < y3

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

1

(2)

Solution:

SECTION 16.2 LINE INTEGRALS ¤ 641

17. (a) Along the line  = −3, the vectors of F have positive -components, so since the path goes upward, the integrand F · T is always positive. Therefore

₁F· r =

₁F· T  is positive.

(b) All of the (nonzero) ﬁeld vectors along the circle with radius 3 are pointed in the clockwise direction, that is, opposite the direction to the path. So F · T is negative, and therefore

₂F· r =

₂F· T  is negative.

18. Vectors starting on 1point in roughly the same direction as 1, so the tangential component F · T is positive. Then



₁F· r =

₁F· T  is positive. On the other hand, no vectors starting on ²point in the same direction as 2, while some vectors point in roughly the opposite direction, so we would expect

₂F· r =

₂F· T  to be negative.

19. r() = ³i+ ²j, so F(r()) = (³)(²)²i− (³)²j= ⁷i− ⁶j and r⁰() = 3²i+ 2 j. Then



 F· r =1

0 F(r()) · r⁰()  =1

0(⁷· 3²− ⁶· 2)  =1

0(3⁹− 2⁷)  =3

10¹⁰−¹4⁸1

0= ₁₀³ −¹4 = ₂₀¹. 20. F(r()) =

²+ (³)²

i+ (²)(−2) j + (³− 2) k = (²+ ⁶) i − 2³j+ (³− 2) k, r⁰() = 2 i + 3²j− 2 k. Then



 F· r =2

0 F(r()) · r⁰()  =2

0(2³+ 2⁷− 6⁵− 2³+ 4)  =2

0(2⁷− 6⁵+ 4) 

=₁

4⁸− ⁶+ 2²2

0= 64 − 64 + 8 = 8 21. 

F· r =1 0

sin ³ cos(−²) ⁴

·

3² −2 1



=1

0(3²sin ³− 2 cos ²+ ⁴)  =

− cos ³− sin ²+¹₅⁵1

0= ⁶₅− cos 1 − sin 1

22. F(r()) = (²+ ³) i + (³− ²) j + (²)²k= (²+ ³) i + (³− ²) j + ⁴k, r⁰() = 2 i + 3²j+ 2 k. Then



F· r =1

0 F(r()) · r⁰()  =1

0(2³+ 2⁴+ 3⁵− 3⁴+ 2⁵)  =1

0(5⁵− ⁴+ 2³) 

=5

6⁶− ¹5⁵+¹₂⁴1

0= ⁵₆ −¹5+¹₂ = ¹⁷₁₅ 23. F(r()) =

sin² + sin  cos  i +

(sin  cos ) sin² j=

sin² + sin  cos  i + cot  j, r⁰() = 2 sin  cos  i + (cos² − sin²) j. Then



 F· r =3

6 F(r()) · r⁰()  =3

6



2 sin  cos 

sin² + sin  cos  + (cot )(cos² − sin²)



≈ 05424

24. F(r()) = (cos  tan )^{sin }i+ (tan  sin )^{cos }j+ (sin  cos )^{tan }k

= (sin )^{sin }i+ (tan  sin )^{cos }j+ (sin  cos )^{tan }k, r⁰() = cos  i − sin  j + sec² k. Then



 F· r =4

0 F(r()) · r⁰()  =4 0

(sin  cos )^{sin }− (tan  sin²)^{cos }+ (tan )^{tan }

 ≈ 08527

36. A thin wire has the shape of the first-quadrant part of the circle with center the origin and radius a. If the density function is ρ(x, y) = kxy, find the mass and center of mass of the wire.

Solution:

Therefore 

³²  =2

0 (^−cos 4)³(^−sin 4)²(^−) (3√

2 ^−) 

=2

0 3√

2 ^−7cos³4 sin²4  = _5,632,705^172,704 √

2 (1 − ^−14)

32. (a) We parametrize the circle  as r() = 2 cos  i + 2 sin  j, 0 ≤  ≤ 2. So F(r()) =

4 cos² 4 cos  sin  , r⁰() = h−2 sin  2 cos i, and  =

F· r =2

0 (−8 cos² sin  + 8 cos² sin )  = 0.

(b) From the graph, we see that all of the vectors in the field are perpendicular to the path. This indicates that the field does no work on the particle, since the field never pulls the particle in the direction in which it is going. In other words, at any point along , F · T = 0, and so certainly

F· r = 0 .

33. We use the parametrization  = 2 cos ,  = 2 sin , −^2 ≤  ≤ ^2. Then

 =_



2

+



²

 =

(−2 sin )²+ (2 cos )² = 2 , so  =

  = 22

−2  = 2(),

 = _2¹ 

  = _2¹ 2

−2(2 cos )2  = _2¹ 

4 sin 2

−2 = _⁴,  = _2¹ 

  = _2¹ 2

−2(2 sin )2  = 0.

Hence ( ) =₄

 0.

34. We use the parametrization  =  cos ,  =  sin , 0 ≤  ≤ ^2. Then

 =_



2

+



²

 =

(− sin )²+ ( cos )² =  , so

 =

( )  =

  =2

0 ( cos )( sin )   = ³2

0 cos  sin   = ³1

2sin²2

0 = ¹₂³,

 = 1

³2





()  = 2

³

 2 0

( cos )²( sin )  = 2

³ · ⁴

 2 0

cos² sin  

= 2

−¹3cos³2 0 = 2

0 +¹₃

= ²₃, and

 = 1

³2



()  = 2

³

 2 0

( cos )( sin )²  = 2

³ · ⁴

 2 0

sin² cos  

= 21

3sin³2

0 = 21 3− 0

= ²₃.

Therefore the mass is¹₂³and the center of mass is ( ) =2 3²₃.

35. (a)  = 1







(  ) ,  = 1







(  ) ,  = 1







(  ) where  =

(  ) .

49. (a) Show that a constant force field does zero work on a particle that moves once uniformly around the circle x²+ y²= 1.

(b) Is this also true for a force field F(x) = kx, where k is a constant and x =< x, y >?

Solution:

40. Choosing  as the parameter, the curve  is parametrized by  = ²+ 1,  = , 0 ≤  ≤ 1. Then

 =

 F· r =1 0

²+ 12

 ^²⁺¹

· h2 1i  =1 0

2

²+ 12

+ ^²⁺¹



=₁

3

²+ 1³+¹₂^²⁺¹¹

0= ⁸₃+¹₂²−¹3−¹2 = ¹₂²−¹2 + ⁷₃ 41. r() = h2  1 − i, 0 ≤  ≤ 1.

 =

F· r =1 0

2 − ²  − (1 − )² 1 −  − (2)²

· h2 1 −1i 

=1

0 (4 − 2²+  − 1 + 2 − ²− 1 +  + 4²)  =1

0 (²+ 8 − 2)  =1

3³+ 4²− 21 0= ⁷₃ 42. r() = 2 i +  j + 5 k, 0 ≤  ≤ 1. Therefore

 =

F· r =

 1 0

h2  5i

(4 + 26²)^32 · h0 1 5i  = 

 1 0

26

(4 + 26²)^32 = 

−(4 + 26²)^−121 0= 

1

2 −^√¹₃₀ .

43. (a) r() = ²i+ ³j ⇒ v() = r⁰() = 2 i + 3²j ⇒ a() = v⁰() = 2 i + 6 j, and force is mass times acceleration: F() =  a() = 2 i + 6 j.

(b)  =

 F· r =1

0(2 i + 6 j) · (2 i + 3²j)  =1

0 (4² + 18²³) 

=

2²²+⁹₂²⁴1

0= 2²+⁹₂²

44. r() =  sin  i +  cos  j +  k ⇒ v() = r⁰() =  cos  i −  sin  j +  k ⇒ a() = v⁰() = − sin  i −  cos  j and F() =  a() = − sin  i −  cos  j. Thus

 =

 F· r =2

0 (− sin  i −  cos  j) · ( cos  i −  sin  j +  k) 

=2

0 (−²sin  cos  + ²sin  cos )  = (²− ²)1

2sin²2

0 = ¹₂(²− ²)

45. The combined weight of the man and the paint is 185 lb, so the force exerted (equal and opposite to that exerted by gravity) is F= 185 k. To parametrize the staircase, let  = 20 cos ,  = 20 sin ,  = _6⁹⁰ =¹⁵_, 0 ≤  ≤ 6. Then the work done is

 =

 F· r =6

0 h0 0 185i ·

−20 sin  20 cos ¹⁵

 = (185)¹⁵_ 6

0  = (185)₁₅



(6) ≈ 167 × 10⁴ft-lb

46. This time  is a function of :  = 185 −_6⁹  = 185 −2³ . So let F =

185 −2³ k. To parametrize the staircase, let  = 20 cos ,  = 20 sin ,  =_6⁹⁰ = ¹⁵_, 0 ≤  ≤ 6. Therefore

 =

 F· r =6

0

0 0 185 −2³ 

·

−20 sin  20 cos ¹⁵

 = ¹⁵_ 6

0

185 − 2³ 



= ¹⁵_

185 −4³ ²6

0 = 90 185 − ⁹2

≈ 162 × 10⁴ft-lb

47. (a) r() = hcos  sin i, 0 ≤  ≤ 2, and let F = h i. Then

 =

 F·  r =2

0 h i · h− sin  cos i  =2

0 (− sin  +  cos )  =

 cos  +  sin 2

0

=  + 0 −  + 0 = 0 (b) Yes. F ( ) =  x = h i and

 =

 F·  r =2

0 h cos   sin i · h− sin  cos i  =2

0 (− sin  cos  +  sin  cos )  =2

0 0  = 0.

2

But this last integral is precisely the line integral in (10). Therefore we have

Section 16.2 Line Integrals

But this last integral is precisely the line integral in (10). Therefore we have

y

F  dr − y

P dx 1 Q dy 1 R dz where F − P i 1 Q j 1 R k

For example, the integral y

y dx 1 z dy 1 x dz in Example 6 could be expressed as y

F  dr, where

sx, y, zd − y i 1 z j 1 x k A similar result holds for vector fields F on R

:

y

F  dr − y

P dx 1 Q dy

where F − P i 1 Q j.

16.2 Exercises

F dr − y

F dr, where

F dr − y