Download Integrals of vector fields. (Sect. 16.2) Vector fields on a plane and in

Integrals of vector fields. (Sect. 16.2)
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Vector fields on a plane and in space.
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The line integral of a vector field along a curve.
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The gradient field of a scalar-valued function.
Work done by a force on a particle.
The flow of a fluid along a curve.
The flux across a plane curve.
Vector fields on a plane and in space
Definition
A vector field on a plane or in space is a vector-valued function
F : D ⊂ Rn → Rn , with n = 2, 3, respectively.
Examples from physics:
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Electric and magnetic fields.
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The gravitational field of the Earth.
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The velocity field in a fluid or gas.
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The variation of temperature in a
room. (Gradient field.)
Magnetic field of a small magnet.
Integrals of vector fields. (Sect. 16.2)
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Vector fields on a plane and in space.
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The line integral of a vector field along a curve.
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The gradient field of a scalar-valued function.
Work done by a force on a particle.
The flow of a fluid along a curve.
The flux across a plane curve.
The gradient field of a scalar-valued function
Remark:
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Given a scalar-valued function f : D ⊂ Rn → R, with n = 2, 3,
its gradient vector, ∇f = h∂x f , ∂y f i or ∇f = h∂x f , ∂y f , ∂z f i,
respectively, is a vector field in a plane or in space.
Example
Find and sketch a graph of the gradient field of the function
f (x, y ) = x 2 + y 2 .
Solution: We know the graph of f is a paraboloid. The gradient
field is ∇f = h2x, 2y i.
z
2
f(x,y) = x + y
2
y
f
x
y
x
D uf
Integrals of vector fields. (Sect. 16.2)
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Vector fields on a plane and in space.
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The line integral of a vector field along a curve.
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The gradient field of a scalar-valued function.
Work done by a force on a particle.
The flow of a fluid along a curve.
The flux across a plane curve.
The line integral of a vector field along a curve
Definition
The line integral of a vector-valued function F : D ⊂ Rn → Rn ,
with n = 2, 3, along the curve associated with the function
r : [t0 , t1 ] ⊂ R → D ⊂ R3 is given by
Z
Z s1
F · dr =
F(r̂(s)) · r̂0 (s) ds
s0
C
where r̂(s) is the arc length parametrization of the function r, and
s(t0 ) = s0 , s(t1 ) = s1 are the arc lengths at the points t0 , t1 .
Example
Remark: It is common the notation
y
r’
F
x
r̂0 = T,
since T is tangent to the curve and
unit, since s is the curve arc-length
parameter.
Line integrals in space
Theorem (General parametrization formula)
The line integral of a continuous function F : D ⊂ R3 → R3 along
a differentiable curve r : [t0 , t1 ] ⊂ R → D ⊂ R3 can be written as
Z s1
Z t1
0
F(r̂(s)) · r̂ (s) ds =
F(r(t)) · r0 (t) dt,
s0
t0
where r̂(s) is the arc length parametrization of the function r, and
s(t0 ) = s0 , s(t1 ) = s1 are the arc lengths at the points t0 , t1 .
Z
t
Proof: Recall the curve arc-length function s(t) =
Then ds =
r̂0 (s) =
|r0 (t)| dt.
|r0 (τ )| dτ .
t0
Also, r̂(s(t)) = r(t). And finally
dr̂
dr
dt
r0 (t)
(s) = (t)
= 0
⇒ r̂0 (s) ds = r0 (t) dt.
ds
dt
ds
|r (t)|
This substitution provides the equation in the Theorem.
Integrals of vector fields. (Sect. 16.2)
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Vector fields on a plane and in space.
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The line integral of a vector field along a curve.
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The gradient field of a scalar-valued function.
Work done by a force on a particle.
The flow of a fluid along a curve.
The flux across a plane curve.
Work done by a force on a particle
Definition
If the vector valued function F : D ⊂ Rn → Rn , with n = 2, 3,
represents a force acting on a particle with position function
r : [t0 , t1 ] ⊂ R → D ⊂ R3 , then the line integral
Z
W = F · dr,
C
is called the work done by the force on the particle.
A mass m projectile near the Earth surface.
Example
y
r’
F
x
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The movement takes place on a plane,
and F = h0, −mg i.
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W 6 0 in the first half of the
trajectory, and W > 0 on the second
half.
Work done by a force on a particle
Example
Find the work done by the force F(x, y , z) = h(3x 2 − 3x), 3z, 1i on
a particle moving along the curve with r(t) = ht, t 2 , t 4 i, t ∈ [0, 1].
Solution:
First: Evaluate F along r. This is: F(t) = h(3t 2 − 3t), 3t 4 , 1i.
Second: Compute r0 (t). This is: r0 (t) = h1, 2t, 4t 3 i.
Third: Integrate the dot product F(t) · r0 (t).
Z
1
(3t 2 − 3t) + (6t 5 ) + (4t 3 ) dt
0
1
3 2
3
3
6
4 = t − t + t + t = 1 − + 1 + 1.
2
2
0
W =
3
3
So, W = 3 − . We conclude: The work done is W = .
2
2
C
Integrals of vector fields. (Sect. 16.2)
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Vector fields on a plane and in space.
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The line integral of a vector field along a curve.
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The gradient field of a scalar-valued function.
Work done by a force on a particle.
The flow of a fluid along a curve.
The flux across a plane curve.
The flow of a fluid along a curve
Definition
In the case that the vector field v : D ⊂ Rn → Rn , with n = 2, 3, is
the velocity field of a flow and r : [t0 , t1 ] ⊂ R → D ⊂ R3 is any
smooth curve, then the line integral
Z
F = v · dr,
C
is called a flow integral. If the curve is a closed loop, the flow
integral is called the circulation of the fluid around the loop.
Example
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The flow of a viscous fluid in a
pipe is maximal along a line
through the center of the pipe.
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The flow vanishes on any curve
perpendicular to the section of the
pipe.
Viscous fluid in a pipe.
v
y
x
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The flow of a fluid along a curve
Example
Find the circulation of a fluid with velocity field v = h−y , xi along
the closed loop given by r1 = ha cos(t), a sin(t)i for t ∈ [0, π], and
r2 = ht, 0i for t ∈ [−a, a].
Z
Z
Solution: The circulation is: F =
v · dr2 .
v · dr1 +
C2
C1
y
The first term is given by:
Z
Z π
v(t) · r01 (t) dt.
v · dr1 =
C1
0
C1
v(t) = h−a sin(t), a cos(t)i,
−a
C2
a
x
r01 (t) = h−a sin(t), a cos(t)i.
Z
π
Z
a2 sin2 (t) + cos2 (t) dt
v · dr1 =
Z
0
C1
v · dr1 = πa2 .
⇒
C1
The flow of a fluid along a curve
Example
Find the circulation of a fluid with velocity field v = h−y , xi along
the closed loop given by r1 = ha cos(t), a sin(t)i for t ∈ [0, π], and
r2 = ht, 0i for t ∈ [−a, a].
Z
Z
Solution: The circulation is: F =
v · dr1 +
v · dr2 .
C1
y
The second term is given by:
Z
Z a
v · dr2 =
v(t) · r02 (t) dt,
C1
C2
v(t) = h0, ti,
−a
C2
C2
a
−a
r02 (t) = h1, 0i.
x
v(t) · r02 (t) = 0
Z
⇒
v · dr2 = 0.
C2
Since
R
C1
v · dr1 = πa2 , we conclude: F = πa2 .
C
Integrals of vector fields. (Sect. 16.2)
Vector fields on a plane and in space.
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The gradient field of a scalar-valued function.
The line integral of a vector field along a curve.
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Work done by a force on a particle.
The flow of a fluid along a curve.
The flux across a plane curve.
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The flux across a plane curve
Definition
The flux of a vector field F : {z = 0} ⊂ R3 → {z = 0} ⊂ R3 along
a closed plane loop r : [t0 , t1 ] ⊂ R → {z = 0} ⊂ R3 is given by
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F = F · n ds,
C
where n is the curve outer unit normal vector in the plane {z = 0}.
Remarks:
Example
z
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F is defined on {z = 0}.
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The loop C lies on {z = 0}.
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Simple formula for n? Yes.
y
C
x
n
{z=0}
n=
1
hy 0 (t), −x 0 (t), 0i.
0
|r |
The flux across a plane curve
Theorem (Counterclockwise loops.)
The flux of a vector field F = hFx (x, y ), Fy (x, y ), 0i along a closed,
counterclockwise plane loop r(t) = hx(t), y (t), 0i for t ∈ [t0 , t1 ] is
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Z t1
given by
F · n ds =
Fx y 0 (t) − Fy x 0 (t) dt.
t0
C
Proof:
Remarks: Since C is counterclockwise
traversed, n = u × k, where u = r0 /|r0 |.
z
k
u
x
C
{z=0}
y
u(t) =
n=uxk
i j k
1 n = 0 x 0 y 0 0
|r | 0 0 1
1
hx 0 (t), y 0 (t), 0i,
0
|r (t)|
⇒
n=
k = h0, 0, 1i.
1
hy 0 (t), −x 0 (t), 0i.
0
|r |
The flux across a plane curve
Theorem (Counterclockwise loops.)
The flux of a vector field F = hFx (x, y ), Fy (x, y ), 0i along a closed,
counterclockwise plane loop r(t) = hx(t), y (t), 0i for t ∈ [t0 , t1 ] is
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Z t1
given by
F · n ds =
Fx y 0 (t) − Fy x 0 (t) dt.
t0
C
Proof: Recall: n =
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Z
t1
F · n ds =
1
hy 0 (t), −x 0 (t), 0i.
0
|r |
hFx , Fy , 0i · hy 0 (t), −x 0 (t), 0i
t0
C
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Z
F · n ds =
C
t1 t0
1
|r0 (t)| dt
0
|r (t)|
Fx y 0 (t) − Fy x 0 (t) dt.
The flux across a plane curve
Example
Find the flux of a field F = h−y , x, 0i across the plane closed loop
given by r1 = ha cos(t), a sin(t), 0i for t ∈ [0, π], and r2 = ht, 0, 0i
for t ∈ [−a, a].
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Z
Z
Solution: Recall:
F · n ds =
F2 · n2 ds
F1 · n1 ds +
C
C2
C1
Along C1 we have: F1 (t) = h−a sin(t), a cos(t), 0i and
x 0 (t) = −a sin(t),
y 0 (t) = a cos(t).
Therefore,
F1x (t) y 0 (t)−F1y (t) x 0 (t) = −a2 sin(t) cos(t)+a2 sin(t) cos(t) = 0.
Z
F · n ds = 0.
Hence:
C1
The flux across a plane curve
Example
Find the flux of a field F = h−y , x, 0i across the plane closed loop
given by r1 = ha cos(t), a sin(t), 0i for t ∈ [0, π], and r2 = ht, 0, 0i
for t ∈ [−a, a].
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Z
Z
Solution: Recall:
F · n ds =
F1 · n1 ds +
F2 · n2 ds
C
C1
C2
Along C2 we have: F2 (t) = h0, t, 0i and x 0 (t) = 1, y 0 (t) = 0. So,
Z
Z a
F · n ds =
−t dt,
F2x (t) y 0 (t) − F2y (t) x 0 (t) = 0 − t ⇒
C2
t 2 a F · n ds = −
2
−a
C2
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We conclude:
F · n ds = 0.
Z
Z
C
−a
⇒
F · n ds = 0.
C2
C