Line IntegralsMath 212

Brian D. Fitzpatrick

Duke University

March 25, 2020

MATH

Overview

Scalar Line IntegralsDefinitionExamples

Vector Line IntegralsDefinitionExamplesDifferential Form Notation

Scalar Line IntegralsDefinition

ThemeConsider a wire W in Rn connecting endpoints P and Q.

P

Q

f (P)=2 kg/m

f (Q)=13 kg/m

P1

f (P1)=3 kg/m P2

f (P2)=7 kg/m

P3

f (P3)=5 kg/m

P4

f (P4)=9 kg/m

Suppose f ∈ C (Rn) measures density (kg/m) at every point of W.

DefinitionThe scalar line integral of f along W is∫

Wf ds = mass of W (in kg)

Scalar Line IntegralsDefinition

DefinitionThe scalar line integral of f ∈ C (Rn) along x : [a, b]→ Rn is∫

xf ds =

∫ b

af (x(t)) · ‖x ′(t)‖ dt

Here, we adopt the notation ds = ‖x ′(t)‖ dt.

Scalar Line IntegralsDefinition

ObservationTracking units allows us to interpret scalar line integrals.∫

xf ds =

∫ b

af (x(t))

mass units

distance unit

· ‖x ′(t)‖

distance units

time unit

dt

time unit

= mass of the curve

Scalar Line IntegralsExamples

Example

Suppose f measures density (in kg/m) along x(t) where

f (x , y) = x2y x(t) = 〈cos (t) , sin (t)〉

and 0 ≤ t ≤ π. The mass of the curve is∫xf ds =

∫ π

0f (x(t)) · ‖x ′(t)‖ dt

=

∫ π

0x2y ‖〈− sin (t) , cos (t)〉 ‖ dt

=

∫ π

0cos2(t) sin(t) dt

u = cos(t) u(π) = −1du = − sin(t) dt u(0) = 1

=

∫ −11−u2 du

= 2/3 kg

Scalar Line IntegralsExamples

Example

The segment connecting P (−2,−2, 1) to Q (−1,−2,−1) is

x(t) = 〈−2, −2, 1〉P

+ t · 〈1, 0, −2〉

# ‰

PQ

= 〈t − 2, −2, −2 t + 1〉 P

Q

for 0 ≤ t ≤ 1. Suppose density is f (x , y , z) = (x + 2 z) · ex−y .∫xf ds =

∫ 1

0(x + 2 z) · ex−y · ‖〈1, 0, −2〉 ‖ dt

=√

5

∫ 1

0−3 t · et dt u = t du = dt

dv = et dt v = et

= − 3√

5 ·{t et∣∣10−∫ 1

0et dt

}= − 3

√5

Vector Line IntegralsDefinition

DefinitionThe work of a force F

“Newton” N

acting on a particle with displacement d“metre” m

is

work = F · d “joule” J = N ·m

The power of F along a velocity vm/s

is

power = F · v “watt” W = N ·m/s = J/s

Power is also called work density.

Vector Line IntegralsDefinition

ThemeConsider a particle’s path P in Rn from P to Q.

x′

F

x ′

F

x ′

F x′

F

P

Q

Suppose F ∈ X(Rn) exerts a force (in N) at every point of P.

DefinitionThe vector line integral of F along P is∫

PF · ds = work (in J)

Vector Line IntegralsDefinition

DefinitionThe vector line integral of F ∈ X(Rn) along x : [a, b]→ Rn is∫

xF · ds =

∫ b

aF (x(t)) · x ′(t) dt

Here, we adopt the notation ds = x ′(t) dt.

Vector Line IntegralsDefinition

ObservationTracking units allows us to interpret vector line integrals.∫

xF · ds =

∫ b

aF (x(t))

force units

· x ′(t)

distance units

time unit

dt

time unit

= work done

Vector Line IntegralsExamples

Example

Consider the data

F (x , y) =

⟨1

xy,

1

x + y

⟩x(t) =

⟨t, t2

⟩1 ≤ t ≤ 4

The work is∫xF · ds =

∫ 4

1F (x(t)) · x ′(t) dt =

∫ 4

1

1

t3+

2

t + 1dt

=

∫ 4

1

⟨1

xy,

1

x + y

⟩· 〈1, 2 t〉 dt =

{− 1

2 t2+ 2 log(t + 1)

}4

1

=

∫ 4

1

1

xy+

2 t

x + ydt =

15

32+ 2 log(5)− 2 log(2)

=

∫ 4

1

1

t3+

2 t

t2 + tdt

Vector Line IntegralsExamples

Example

Consider the data

F (x , y , z) = 〈yz , xz , xy〉 x(t) =⟨t, t2, t3

⟩0 ≤ t ≤ 1

The work is∫xF · ds =

∫ 1

0F (x(t)) · x ′(t) dt =

∫ 1

06 t5 dt

=

∫ 1

0〈yz , xz , xy〉 ·

⟨1, 2 t, 3 t2

⟩dt = t6

∣∣10

=

∫ 1

0

⟨t5, t4, t3

⟩·⟨1, 2 t, 3 t2

⟩dt = 1

=

∫ 1

0t5 + 2 t5 + 3 t5 dt

Vector Line IntegralsDifferential Form Notation

DefinitionFor F = 〈F1,F2,F3〉 it is common to write∫

xF · ds =

∫xF1 dx

x ′1 dt

+ F2 dy

x ′2 dt

+ F3 dz

x ′3 dt

The expression F1 dx + F2 dy + F3 dz is called a differential form.

Vector Line IntegralsDifferential Form Notation

Example

Consider x(t) =⟨t, t2, 1

⟩for 0 ≤ t ≤ 1. Then∫

xx2 dx + xy dy + 1 dz =

∫ 1

0

⟨x2, xy , 1

⟩· ds

=

∫ 1

0

⟨x2, xy , 1

⟩· 〈1, 2 t, 0〉 dt

=

∫ 1

02 txy + x2 dt

=

∫ 1

02 t4 + t2 dt

=11

15