Chapter 18

Section 18.1Line Integrals

Vector Fields (or Force Fields)A vector field is function with several variables as inputs and several variables as outputs.

The function and the function . Since we cannot represent 4 or 6 dimensions on 3 dimensional coordinate system we need a different way to visualize these types of functions.

Graphing Vector FieldsTo graph a vector field at the point draw the vector with its tail located at the point . This represents how energy (gravitation, electrical, thermal, etc.) is being directed though the plane.

A similar type of diagram can also be done for a vector field to show how energy is directed through space.

ExampleGraph the constant vector field given by:

At every single point the vector field produces the vector .

4 3 2 1 1 2 3 4

4

3

2

1

1

2

3

4

This type of a vector field is sometimes called a river because all force “flows” in the same direction at the same rate.

ExampleGraph the vector given by:

All vectors will end up pointing away from the y-axis. The farther away from the y-axis the smaller the vector.

4 3 2 1 1 2 3 4

4

3

2

1

1

2

3

4

This type of a vector field is sometimes called the red sea because all force “flows” away from the y-axis and gets smaller.⟨1,0 ⟩

⟨ 12 ,0 ⟩

⟨−1,0 ⟩

⟨− 12 ,0 ⟩

ExampleGraph the vector given by:

All vectors will end up pointing away from the origin. The farther away from the origin the larger the vector.

4 3 2 1 1 2 3 4

4

3

2

1

1

2

3

4

ExampleGraph the vector given by:

All vectors will end up pointing away from the origin. All of the vectors are of unit length.

4 2 0 2 4

4

2

0

2

4

ExampleGraph the vector given by:

All vectors will end up pointing toward the origin. All of the vectors are of unit length.

4 2 0 2 4

4

2

0

2

4

ExampleGraph the vector given by:

All vectors will end up pointing away from the origin. All of the vectors increase in length as you move toward the origin. (Explosion)

ExampleGraph the vector given by:

All vectors will end up pointing toward the origin. All of the vectors increase in length as you move toward the origin. (Black Hole)

2 1 0 1 2

2

1

0

1

2

2 1 0 1 2

2

1

0

1

2

Work and EnergyPreviously you saw that which is a measurement of energy is found by:

This is when objects or fluid was lifted or pumped straight up or down.

10 lbs

𝑑=12 𝑓𝑡𝑤𝑜𝑟𝑘= ⟨0,10 ⟩ ∙ ⟨0,12 ⟩

𝑤𝑜𝑟𝑘=⟨ 𝑓𝑜𝑟𝑐𝑒𝑣𝑒𝑐𝑡𝑜𝑟 ⟩ ∙ ⟨𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑣𝑒𝑐𝑡𝑜𝑟 ⟩

2 1 0 1 2

2

1

0

1

2Work Moving on a Path . If is a path from an initial point A to a terminal point B in a vector field F is found by partitioning a parameterization of , say into small vectors, finding the dot product of each one with F (i.e. ), summing them up and taking the limit as the number of partition goes to infinity.

Parametric Curve for :

𝐴=r (𝑎 )

𝜸 : r (𝒕 )𝐵=r (𝑏)

𝐸𝑛𝑒𝑟𝑔𝑦𝑀𝑜𝑣𝑖𝑛𝑔𝑜𝑛 h𝑃𝑎𝑡 𝛾

= lim𝑛→∞

∑𝑖=1

𝑛

F ∙∆ r𝑖=∫𝛾

❑

F∙𝑑 rThis is the limit of a Riemann sum which is an integral.

∫𝛾

❑

F ∙𝑑 r=∫𝑎

𝑏

F (r (𝑡 ) )∙ r ′ (𝑡 )𝑑𝑡The way this formula is implemented with a parametric curve :

ExampleFind the work done moving along the curve , given by from the point to the point ,in the vector field: .

Use two different parameterizations of the curve .

0 .5 1 .0 1 .5 2 .0

1

2

3

4

𝛾

𝛾 : r1 (𝑡 )= ⟨𝑡 , 𝑡 2 ⟩ ,0≤ 𝑡≤2

∫𝛾

❑

F ∙𝑑 r1=∫0

2

F (𝑡 , 𝑡2 ) ∙ ⟨1,2 𝑡 ⟩𝑑𝑡

r 1′=⟨1,2𝑡 ⟩ 𝑑𝑡

¿∫0

2

⟨2 𝑡2+𝑡 2 ,6 (𝑡 2 )2 ⟩ ∙ ⟨1,2 𝑡 ⟩𝑑𝑡

¿∫0

2

⟨3 𝑡 2 ,6 𝑡 4 ⟩ ∙ ⟨1,2 𝑡 ⟩𝑑𝑡

¿∫0

2

3 𝑡 2+12𝑡 5𝑑𝑡

¿ 𝑡3+2𝑡 6|02=8+128=136

𝛾 : r2 (𝑡 )= ⟨√𝑡 , 𝑡 ⟩ ,0≤ 𝑡≤4r 2′=⟨ 1

2√𝑡,1⟩𝑑𝑡

∫𝛾

❑

F ∙𝑑 r2=∫0

4

F (√𝑡 , 𝑡 ) ∙ ⟨ 12√𝑡 ,1⟩𝑑𝑡¿∫0

4

⟨2 (√𝑡 )2+𝑡 ,6 𝑡 2 ⟩ ∙⟨ 12√𝑡

,1⟩𝑑𝑡¿∫0

4

⟨3 𝑡 ,6 𝑡 2 ⟩ ∙ ⟨ 12√𝑡

,1⟩𝑑𝑡¿∫0

432

√𝑡+6 𝑡 2𝑑𝑡

¿ 𝑡3 /2+2 𝑡3|04=8+128=136

Notice that different parameterizations of the same curve give equal values for the integral. ∫

𝛾

❑

F ∙𝑑 r1=∫𝛾

❑

F∙𝑑 r=∫𝛾

❑

F∙𝑑 r 2

ExampleFind the work done moving through the 3 dimensional vector field along the path parameterized to the right.

∫𝛾

❑

F ∙𝑑 r=∫0

1

F (𝑡−1,2 𝑡 , 𝑡2 ) ∙ ⟨1,2,2 𝑡 ⟩𝑑𝑡

Before we can set up the integral we have to compute . 𝑑 r=⟨1,2,2𝑡 ⟩ 𝑑𝑡

¿∫0

1

⟨2 (2𝑡 )𝑡 2 ,6 (𝑡−1+1 )2 ,3 (𝑡2 )2 ⟩ ∙ ⟨1,2,2𝑡 ⟩ 𝑑𝑡

¿∫0

1

⟨4 𝑡 3 ,6 𝑡 2 ,3 𝑡 4 ⟩ ∙ ⟨1,2,2 𝑡 ⟩𝑑𝑡

¿∫0

1

4 𝑡 3+12𝑡 2+6 𝑡 5𝑑𝑡

¿ 𝑡 4+4 𝑡3+𝑡6|01=1+4+1=6

Notice that this formula can also be interpreted for vector fields for dimensions other than 2!

The Path For a path with initial point A and terminal point B the path is the exact same curve but in the opposite direction with initial point B and terminal point A. The energy returning from B is the negative of the energy required to get to B. x

y

𝛾A: Initial Point

B: Terminal Point

x

y

−𝛾A: Terminal Point

B: Initial Point

∫𝛾

❑

F ∙𝑑 r+∫−𝛾

❑

F ∙𝑑 r=0 ∫−𝛾

❑

F ∙𝑑 r=−∫𝛾

❑

F∙𝑑 r

A Sequence of Paths If a path is made up a sequence of paths , ,, with the terminal point of one being the initial point of the next then the work on is the sum of the work on each of the curves .

⋰

𝛾1 𝛾 2 𝛾 3𝛾𝑛

x

y

∫𝛾1⋃𝛾2⋃⋯⋃𝛾 𝑛

❑

F ∙𝑑 r=∫𝛾 1

❑

F ∙𝑑 r+∫𝛾2

❑

F ∙𝑑 r+⋯+∫𝛾𝑛

❑

F ∙𝑑 r=∑𝑖=1

𝑛

∫𝛾 𝑖

❑

F∙𝑑 r

ExampleFor the vector field and the path that goes from to along , and from to along , and from to along the y-axis, compute .

0 .5 1 .0 1 .5 2 .0

1

2

3

4

Γ𝛾1

𝛾 2

𝛾 3

We already found .

The curve is made up of the curves .

∫𝛾1

❑

F ∙𝑑 r=∫0

2

F (𝑡 , 𝑡2 ) ∙ ⟨1,2 𝑡 ⟩𝑑𝑡

¿∫0

2

⟨2 𝑡2+𝑡 2 ,6 (𝑡 2 )2 ⟩ ∙ ⟨1,2 𝑡 ⟩𝑑𝑡

¿∫0

2

⟨3 𝑡 2 ,6 𝑡 4 ⟩ ∙ ⟨1,2 𝑡 ⟩𝑑𝑡

¿∫0

2

3 𝑡 2+12𝑡 5𝑑𝑡

¿ 𝑡3+2𝑡 6|02=8+128=136

𝛾1 : r (𝑡 )= ⟨𝑡 , 𝑡 2 ⟩r ′ (𝑡 )=⟨1,2𝑡 ⟩

Find and take negative.

∫𝛾2

❑

F ∙𝑑 r=−∫0

2

F (𝑡 ,4 ) ∙ ⟨1,0 ⟩𝑑𝑡

−𝛾 2: r (𝑡 )= ⟨𝑡 ,4 ⟩r ′ (𝑡 )=⟨1,0 ⟩

¿−∫0

2

⟨2𝑡 2+4,96 ⟩ ∙ ⟨1,0 ⟩𝑑𝑡

¿−∫0

2

2𝑡 2+4𝑑𝑡

¿− ( 23 𝑡 3+4 𝑡)|02

¿−( 163 +8)=−8( 53 )=−403

0≤ 𝑡≤2 0≤ 𝑡≤2

Find and take negative.

∫𝛾3

❑

F ∙𝑑 r=−∫0

4

F (0 , 𝑡 ) ∙ ⟨0,1 ⟩ 𝑑𝑡

−𝛾 3: r (𝑡 )=⟨0 , 𝑡 ⟩r ′ (𝑡 )=⟨0,1 ⟩ 𝑑𝑡

¿−∫0

4

⟨ 𝑡 ,6 𝑡2 ⟩ ∙ ⟨0,1 ⟩𝑑𝑡

¿−∫0

4

6 𝑡2𝑑𝑡

¿− (2𝑡 3 )|04

¿−128

0 .5 1 .0 1 .5 2 .0

1

2

3

4

Γ𝛾1

𝛾 2

𝛾 30≤ 𝑡≤ 4 F (𝑥 , 𝑦 )= ⟨2 𝑥2+𝑦 ,6 𝑦 2 ⟩

∫Γ

❑

F ∙𝑑 r=∫𝛾1

❑

F ∙𝑑 r+∫𝛾2

❑

F ∙𝑑 r+∫𝛾3

❑

F ∙𝑑 r=136− 403−128=8− 40

3=−16

3

Adding up the previous values from the other paths we obtain the answer for the integral along the entire path .

Parameterizing curvesIn order to do these integrals you will need to parameterize a curve. Lines, functions, circles, and ellipses all have a standard method for parameterization.

x

yy is a function of x

𝛾

a bx

yx is a function of y

𝛾c

d

y

z

A line segment from to

𝛾

𝑃 (𝑥0 , 𝑦 0 ,𝑧 0 )

x

𝑄 (𝑥1 , 𝑦1 , 𝑧1 )

x

y

A circle with center and radius r

𝛾

(h ,𝑘 )r