Chapter 18

Section 18.4Another Notation for Line Integrals

The Differential Form of Line Integrals (2 dimensional)Let the functions and be the x component and y component of the 2 dimensional vector field F (i.e. ).

,

𝑑 r=⟨ 𝑥′ (𝑡 ) , 𝑦 ′ (𝑡 ) ⟩𝑑𝑡𝑑 r=⟨𝑑𝑥 ,𝑑𝑦 ⟩

The curve is parameterized by .

The Differential Form of Line Integrals (3 dimensional)Let the functions , and be the x, y , z components of the 3 dimensional vector field F (i.e. ).

,

𝑑 r=⟨ 𝑥′ (𝑡 ) , 𝑦 ′ (𝑡 ) ,𝑧 ′ (𝑡 ) ⟩𝑑𝑡𝑑 r=⟨𝑑𝑥 ,𝑑𝑦 ,𝑑𝑧 ⟩

The curve is parameterized by .

∫𝛾

❑

F ∙𝑑 r=∫𝛾

❑

𝑀 (𝑥 , 𝑦 , 𝑧 )𝑑𝑥+𝑁 (𝑥 , 𝑦 ,𝑧 )𝑑𝑦+𝑃 (𝑥 , 𝑦 , 𝑧 ) 𝑑𝑧

¿∫𝑎

𝑏

𝑀 (𝑥 (𝑡 ) , 𝑦 (𝑡 ) , 𝑧 (𝑡 ) ) 𝑥 ′ (𝑡 )+𝑁 (𝑥 (𝑡 ) , 𝑦 (𝑡 ) , 𝑧 (𝑡 ) ) 𝑦 ′ (𝑡 )+𝑃 (𝑥 (𝑡 ) , 𝑦 (𝑡 ) , 𝑧 (𝑡 ) ) 𝑧′ (𝑡 )𝑑𝑡

ExampleFind the value of the integral to the right for the given vector field F where goes from the point to the point along the curve , and also along the curve given by .

𝑥=𝑡

¿∫−1

2

1+2𝑡 3+6 𝑡 3𝑑𝑡

¿∫−1

2

1+8 𝑡3 𝑑𝑡

¿ 𝑡+2𝑡 4|−12

=34−1=33

Notice that and the answers are the same!

For let: ,𝑦=𝑡 2

𝑑𝑥=1 𝑑𝑦=2𝑡

∫−1

2

(1+2𝑡 𝑡 2 )1+ (𝑡 2+2𝑡 2 )2𝑡 𝑑𝑡

𝑥=𝑡

¿ 𝑡3+3 𝑡2+5 𝑡|−12

= (30 )− (−3 )=33

For let: ,

𝑦=𝑡+2𝑑𝑥=1 𝑑𝑦=1

∫−1

2

(1+2𝑡 (𝑡+2 ) )1+(𝑡2+2 (𝑡+2 ) )1𝑑𝑡

∫𝛾𝑖

❑

(1+2𝑥𝑦 )𝑑𝑥+ (𝑥2+2 𝑦 )𝑑𝑦

¿∫−1

2

(2 𝑡2+4 𝑡+1 )+(𝑡 2+2 𝑡+4 )𝑑𝑡

¿∫−1

2

3 𝑡2+6 𝑡+5𝑑𝑡

If a vector field F has an antigradient (i.e. ) and and have the same initial and terminal points then:

Path IndependenceAn integral is said to be path independent (or independent of path) if any two paths with the same initial and terminal points have the value of the integral over those paths equal.

𝛾1

𝛾 2∫𝛾1

❑

𝑀𝑑𝑥+𝑁𝑑𝑦=∫𝛾2

❑

𝑀𝑑𝑥+𝑁𝑑𝑦

1

2

A

B

2 1 1 2 3

1

1

2

3

4

5

In the previous example we saw that if the vector field has an antigradient (potential function) the integral is independent of path since the value is the difference of the end points.

Conservative Vector FieldsA vector field is conservative if energy is conserved along any path in the vector field. That is the energy expended moving to any position along any path is equal to the energy required to return to the same position along any other path possibly different path. The function that has the vector field as its gradient represents the potential energy and is called the potential function.

The Following statements are equivalent:

1. The vector field is conservative.2. We have the follow equality: .3. An antigradient (potential function) exists for . (i.e. )4. The integral is independent of the path .5. If is any closed curve . (All loop integrals are zero!)

All of these have analogues statements for 3 dimensions.

The Following statements are equivalent:

1. The vector field is conservative.2. We have the follow equality: .3. An antigradient (potential function) exists for . (i.e. )4. The integral is independent of the path .5. If is any closed curve . (All loop integrals are zero!)

ExampleFind the value of the integral to the right along the parabolic path .

For let:

𝑥=𝑡 𝑧=2 𝑡2𝑦=𝑡𝑑𝑥=1 𝑑𝑦=1 𝑑𝑧=4 𝑡

∫0

2

(𝑡 ) (𝑡 )1+2 (2𝑡 2 )1+(𝑡+2 𝑡2 )4 𝑡 𝑑𝑡

¿∫0

2

𝑡 2+4 𝑡2+4 𝑡 2+8 𝑡3𝑑𝑡

¿∫0

2

9𝑡 2+8 𝑡 3𝑑𝑡

¿ 3 𝑡 3+2𝑡 4|02

¿24+32=56

Notice that this integral is not independent of path since:

and

This means that we needed to find the value using the parameterization.

ExampleThe triangular path goes on straight lines from the point to then to and back to . Find the value of the following integral:

Γ

𝛾 3𝛾 2

𝛾1x

y(1,2 )

(0,0 )(1 ,0 )

Γ is𝛾 1⋃𝛾2⋃𝛾 3

is on for

is on for

is on for

∫𝛾3

❑

(2𝑥+𝑦 ) 𝑑𝑥+3𝑥𝑦𝑑𝑦=−∫0

1

(2𝑡+2 𝑡 )1+3 𝑡 (2 𝑡 )2𝑑𝑡

¿−∫0

1

4 𝑡+12 𝑡2𝑑𝑡=− (2 𝑡2+4 𝑡 3 )|01=−6

∫𝛾2

❑

(2𝑥+𝑦 ) 𝑑𝑥+3𝑥𝑦𝑑𝑦=∫0

2

3𝑡 𝑑𝑡= 32𝑡 2|

0

2

=6

∫𝛾1

❑

(2𝑥+𝑦 ) 𝑑𝑥+3𝑥𝑦𝑑𝑦=∫0

1

2𝑡 𝑑𝑡=𝑡2|01=1

∫Γ

❑

(2𝑥+𝑦 ) 𝑑𝑥+3𝑥𝑦𝑑𝑦= ∫𝛾1⋃𝛾2⋃𝛾 3

❑

(2𝑥+𝑦 )𝑑𝑥+3 𝑥𝑦𝑑𝑦=1+6−6=1