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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.