chapter 18

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