Section 16.2 Line Integrals

(1)

Section 16.2 Line Integrals

SECTION 16.2 LINE INTEGRALS ¤ 639 2. = ³and  = ⁴, 1 ≤  ≤ 2, so by Formula 3



()  =2

1(³⁴)

(3²)²+ (4³)² =2

1(1) · ²√

9 + 16² =2 1 √

9 + 16²

= ₃₂¹ ·²3

9 + 16²322

1 =₄₈¹(73^32− 25^32) or ₄₈¹(73√

73 − 125)

3.Parametric equations for  are  = 4 cos ,  = 4 sin , −^2 ≤  ≤^2. Then



 ⁴ =2

−2(4 cos )(4 sin )⁴

(−4 sin )²+ (4 cos )² =2

−24⁵cos  sin⁴

16(sin² + cos²) 

= 4⁵2

−2(sin⁴ cos )(4)  = (4)⁶₁

5sin⁵2

−2= 4⁶·²5 = 16384 4.Parametric equations for  are  = 2 + 3,  = 4, 0 ≤  ≤ 1. Then



 ^ =1

0 (2 + 3) ^4√

3²+ 4² = 51

0 (2 + 3) ^4

Integrating by parts with  = 2 + 3 ⇒  = 3 ,  = ^4 ⇒  = ¹4^4gives



 ^ = 51

4(2 + 3)^4−16³^41 0= 55

4⁴−16³⁴−¹2+₁₆³

=⁸⁵₁₆⁴−²⁵16

5.If we choose  as the parameter, parametric equations for  are  = ,  = ²for 0 ≤  ≤  and by Equations 7





² + sin 

 = 0

²(²) + sin 

· 2  = 2 0

⁵+  sin 



= 2₁

6⁶−  cos  + sin  0

where we integrated by parts in the second term



= 2₁

6⁶+  + 0 − 0

=¹₃⁶+ 2

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

SECTION 16.2 LINE INTEGRALS ¤ 639 2. = ³and  = ⁴, 1 ≤  ≤ 2, so by Formula 3



()  =2

1(³⁴)

(3²)²+ (4³)² =2

1(1) · ²√

9 + 16² =2 1 √

9 + 16²

= ₃₂¹ ·²3

9 + 16²322

1 =₄₈¹(73^32− 25^32) or ₄₈¹(73√

73 − 125)

3.Parametric equations for  are  = 4 cos ,  = 4 sin , −^2 ≤  ≤^2. Then



 ⁴ =2

−2(4 cos )(4 sin )⁴

(−4 sin )²+ (4 cos )² =2

−24⁵cos  sin⁴

16(sin² + cos²) 

= 4⁵2

−2(sin⁴ cos )(4)  = (4)⁶₁

5sin⁵2

−2= 4⁶·²5 = 16384 4.Parametric equations for  are  = 2 + 3,  = 4, 0 ≤  ≤ 1. Then



 ^ =1

0 (2 + 3) ^4√

3²+ 4² = 51

0 (2 + 3) ^4

Integrating by parts with  = 2 + 3 ⇒  = 3 ,  = ^4 ⇒  = ¹4^4gives



 ^ = 51

4(2 + 3)^4−16³^41 0= 55

4⁴−16³⁴−¹2+₁₆³

=⁸⁵₁₆⁴−²⁵16

5.If we choose  as the parameter, parametric equations for  are  = ,  = ²for 0 ≤  ≤  and by Equations 7





² + sin 

 = 0

²(²) + sin 

· 2  = 2 0

⁵+  sin 



= 2₁

6⁶−  cos  + sin  0

where we integrated by parts in the second term



= 21

6⁶+  + 0 − 0

=¹₃⁶+ 2

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

640 ¤ CHAPTER 16 VECTOR CALCULUS

8.  = 1+ 2

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

 = 2 cos  , 0 ≤  ≤ ^2.

On 2:  = 4 ⇒  = 4 ,  = 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(4)²(4 ) + (2 + )²

= 82

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

0(65²+ 4 + 4) 

= 8₁

3cos³ +¹₃sin³2 0 +₆₅

3³+ 2²+ 41 0= 8₁

3−¹3

+⁶⁵₃ + 2 + 4 =⁸³₃

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

0 cos² sin 

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. 

()²+ ()²+ ()²=

1²+ (−2 sin 2)²+ (2 cos 2)²=

1 + 4(sin²2 + cos²2) =√5. Then



(²+ ²+ ²)  =2

0 (²+ cos²2 + sin²2)√

5  =√ 52

0 (²+ 1) 

=√ 51

3³+ 2

0 =√

51

3(8³) + 2

=√ 58

3³+ 2 13. 

 ^ =1

0()(²)^(²^)(³⁾· 2  =1

0 2⁴^⁵ = ²₅^⁵1

0= ²₅(¹− ⁰) = ²₅( − 1) 14. 

   +   +   =4

1  ·¹2^−12 + ²·  +√

 · 2  =4 1

1

2^12+ ²+ 2^32



=

1

3^32+¹₃³+⁴₅^524

1=⁸₃ +⁶⁴₃ +¹²⁸₅ −¹3−¹3−⁴5 =⁷²²₁₅

SECTION 16.2 LINE INTEGRALS ¤ 641

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

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

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

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

SECTION 16.2 LINE INTEGRALS ¤ 645 35. (a)  = 1







(  ) ,  = 1







(  ) ,  = 1







(  ) where  =

(  ) .

(b)  =

  = 2

0

4 sin² + 4 cos² + 9  = √ 132

0  = 2√ 13,

 = 1

2√ 13

 2

0

2√

13 sin   = 0,  = 1 2√

13

2

0

2√

13 cos   = 0,

 = 1

2√ 13

 2

0

√ 13

(3)  = 3 2

2²= 3. Hence (  ) = (0 0 3).

36.  =

(²+ ²+ ²)  =2

0 (²+ 1)

(1)²+ (− sin )²+ (cos )² =2

0 (²+ 1)√

2  =√ 28

3³+ 2,

 = 1

√28

3³+ 2

 2

0

√2 (³+ )  = 4⁴+ 2²

8

3³+ 2 = 3

2²+ 1 4²+ 3 ,

 = 3

2√

2 (4²+ 3)

2

0

√2 cos 

(²+ 1)  = 0, and

 = 3

2√

2 (4²+ 3)

 2

0

√2 sin 

(²+ 1)  = 0. Hence (  ) =

3(2²+ 1) 4²+ 3  0 0

 .

37. From Example 3, ( ) = (1 − ),  = cos ,  = sin , and  = , 0 ≤  ≤  ⇒

=

²( )  =

0 sin² [(1 − sin )]  = 

0(sin² − sin³) 

=¹₂

0(1 − cos 2)  − 

0 (1 − cos²) sin  

Let  = cos ,  = − sin  

in the second integral



= 

 2 +₋₁

1 (1 − ²) 

=  2 −⁴3



=

²( )  = 

0 cos² (1 − sin )  =^2



0(1 + cos 2)  − 

0 cos² sin  

=  2 −²3

, using the same substitution as above.

38. The wire is given as  = 2 sin ,  = 2 cos ,  = 3, 0 ≤  ≤ 2 with (  ) = . Then

 =

(2 cos )²+ (−2 sin )²+ 3² =

4(cos² + sin²) + 9  =√ 13 and

=

(²+ ²)(  )  =2

0 (4 cos² + 9²)()√

13  =√ 13 

41

2 +¹₄sin 2

+ 3³2

0

=√

13 (4 + 24³) = 4√

13 (1 + 6²)

=

(²+ ²)(  )  =2

0

4 sin² + 9² ()√

13  =√ 13 

41

2 −¹4sin 2

+ 3³2

0

=√

13 (4 + 24³) = 4√

13 (1 + 6²)

 =

(²+ ²)(  )  =2

0 (4 sin² + 4 cos²)()√

13  = 4√ 13 2

0  = 8√ 13 

39.  =

F· r =2

0 h − sin  3 − cos i · h1 − cos  sin i 

=2

0 ( −  cos  − sin  + sin  cos  + 3 sin  − sin  cos ) 

=2

0 ( −  cos  + 2 sin )  =₁

2²− ( sin  + cos ) − 2 cos 2

0

integrate by parts in the second term



= 2²

646 ¤ CHAPTER 16 VECTOR CALCULUS

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 ,  = ⁹⁰_6 =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.

1

(2)

48. Consider the base of the fence in the -plane, centered at the origin, with the height given by  = ( ). To graph the fence, observe that the fence is highest when  = 0 (where the height is 5 m) and lowest when  = 0 (a height of 3 m). When

 = ±, the height is 4 m.

Also, the fence can be graphed using parametric equations (see Section 16.6):  = 10 cos ,  = 10 sin ,

 = 

4 + 001((10 cos )²− (10 sin )²)

= (4 + cos² − sin²)

= (4 + cos 2), 0 ≤  ≤ 2, 0 ≤  ≤ 1.

The area of the fence is

( ) where , the base of the fence, is given by  = 10 cos ,  = 10 sin , 0 ≤  ≤ 2.

Then 

 ( )  =2

0

4 + 001((10 cos )²− (10 sin )²) 

(−10 sin )²+ (10 cos )²

=2

0 (4 + cos 2)√

100  = 10

4 +¹₂sin 22

0 = 10(8) = 80m²

If we paint both sides of the fence, the total surface area to cover is 160 m², and since 1 L of paint covers 100 m², we require

160

100 = 16 ≈ 503 L of paint.

49. Let r() = h() () ()i and v = h¹ 2 3i. Then



 v· r =

h¹ 2 3i · h⁰() ⁰() ⁰()i  =

[1⁰() + 2⁰() + 3⁰()] 

=

1() + 2() + 3()

= [1() + 2() + 3()] − [¹() + 2() + 3()]

= 1[() − ()] + ²[() − ()] + ³[() − ()]

= h¹ 2 3i · h() − () () − () () − ()i

= h¹ 2 3i · [h() () ()i − h() () ()i] = v · [r() − r()]

50. If r() = h() () ()i then



 r· r =

h() () ()i · h⁰() ⁰() ⁰()i  =

[() ⁰() + () ⁰() + () ⁰()] 

=1

2[()]²+¹₂[()]²+¹₂[()]²



=¹₂

[()]²+ [()]²+ [()]²

−

[()]²+ [()]²+ [()]²

=¹₂

|r()|²− |r()|² 51. The work done in moving the object is

F· r =

F· T . We can approximate this integral by dividing  into 7segments of equal length ∆ = 2 and approximating F · T, that is, the tangential component of force, at a point (^∗^ ^∗_)on each segment. Since  is composed of straight line segments, F · T is the scalar projection of each force vector onto .

If we choose (^∗ ^∗)to be the point on the segment closest to the origin, then the work done is

48. Consider the base of the fence in the -plane, centered at the origin, with the height given by  = ( ). To graph the fence, observe that the fence is highest when  = 0 (where the height is 5 m) and lowest when  = 0 (a height of 3 m). When

 = ±, the height is 4 m.

Also, the fence can be graphed using parametric equations (see Section 16.6):  = 10 cos ,  = 10 sin ,

 = 

4 + 001((10 cos )²− (10 sin )²)

= (4 + cos² − sin²)

= (4 + cos 2), 0 ≤  ≤ 2, 0 ≤  ≤ 1.

The area of the fence is

( ) where , the base of the fence, is given by  = 10 cos ,  = 10 sin , 0 ≤  ≤ 2.

Then 

 ( )  =2

0

4 + 001((10 cos )²− (10 sin )²) 

(−10 sin )²+ (10 cos )²

=2

0 (4 + cos 2)√

100  = 10

4 +¹₂sin 22

0 = 10(8) = 80m²

If we paint both sides of the fence, the total surface area to cover is 160 m², and since 1 L of paint covers 100 m², we require

160

100 = 16 ≈ 503 L of paint.

49. Let r() = h() () ()i and v = h¹ 2 3i. Then



 v· r =

h¹ 2 3i · h⁰() ⁰() ⁰()i  =

[1⁰() + 2⁰() + 3⁰()] 

=

1() + 2() + 3()

= [1() + 2() + 3()] − [¹() + 2() + 3()]

= 1[() − ()] + ²[() − ()] + ³[() − ()]

= h¹ 2 3i · h() − () () − () () − ()i

= h¹ 2 3i · [h() () ()i − h() () ()i] = v · [r() − r()]

50. If r() = h() () ()i then



 r· r =

h() () ()i · h⁰() ⁰() ⁰()i  =

[() ⁰() + () ⁰() + () ⁰()] 

=₁

2[()]²+¹₂[()]²+¹₂[()]²



=¹₂

[()]²+ [()]²+ [()]²

−

[()]²+ [()]²+ [()]²

=¹₂

|r()|²− |r()|² 51. The work done in moving the object is

F· r =

F· T . We can approximate this integral by dividing  into 7segments of equal length ∆ = 2 and approximating F · T, that is, the tangential component of force, at a point (^∗^ ^∗)on each segment. Since  is composed of straight line segments, F · T is the scalar projection of each force vector onto .

If we choose (^∗_ _^∗)to be the point on the segment closest to the origin, then the work done is