Integrals of vector fields. (Sect. 16.2)
I
Vector fields on a plane and in space. I
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The line integral of a vector field along a curve. I I
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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: I
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)
I
Vector fields on a plane and in space. I
I
The line integral of a vector field along a curve. I I
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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: I
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)
I
Vector fields on a plane and in space. I
I
The line integral of a vector field along a curve. I I
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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(ˆr(s)) · ˆr0 (s) ds s0
C
where ˆr(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
ˆr0 = 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(ˆr(s)) · ˆr (s) ds = F(r(t)) · r0 (t) dt, s0
t0
where ˆr(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 = ˆr0 (s) =
|r0 (t)| dt.
|r0 (τ )| dτ .
t0
Also, ˆr(s(t)) = r(t). And finally
dˆr dr dt r0 (t) (s) = (t) = 0 ⇒ ˆr0 (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. I
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The line integral of a vector field along a curve. I I
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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. I
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The line integral of a vector field along a curve. I I
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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 z
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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I I
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 I 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 I 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 I Z t1 given by F · n ds = Fx y 0 (t) − Fy x 0 (t) dt. t0
C
Proof: Recall: n = I
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
I
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]. I 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]. I 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 I We conclude: F · n ds = 0.
Z
Z
C
−a
⇒
F · n ds = 0. C2
C