Chapter 13. Vector Analysis
Le Cong Nhan
Faculty of Applied Sciences
HCMC University of Technology and Education
January 8, 2021
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Content
1
Properties of a Vector Field: Divergence and Curl
2
Line Integrals
Line Integrals in R2 and R3
Line Integrals of Vector Fields
Applications of Line Integrals
3
The Fundamental Theorem and and Path Independence
4
Green’s Theorem
5
Surface Integrals
Applications of Surface Integrals
Flux Integral
6
Stokes’ Theorem and Applications
7
Divergence Theorem and Applications
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13.1 Properties of a Vector Field: Divergence and Curl
Definition 1 (Vector field)
Let D be a set in R2 (a plane region). A vector field on D is a function
that assigns to each point (x, y ) in D a two-dimensional vector F(x, y ).
F(x, y ) can be written in terms of its
component functions P and Q as follows:
F(x, y ) =P(x, y )i + Q(x, y )j
= hP(x, y ), Q(x, y )i ,
Figure: Vector filed on R2
where P(x, y ) and Q(x, y ) are scalar
functions of two variables.
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Figure: Velocity vector fields showing San Francisco Bay wind patterns. Source:
James Stewart’s book
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Figure: Map of global surface currents. Source: NOC
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Definition 2 (Vector field on R3 )
Let E be a set in R3 . A vector field on E is a function that assigns to each
point (x, y , z) in E a three-dimensional vector F(x, y , z)
F(x, y , z) = P(x, y , z)i + Q(x, y , z)j + R(x, y , z)k,
where P, Q and R are component functions of F.
Figure: Vector filed on R3
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Example 3
Sketch the graph of the vector field F(x, y ) = y i − xj
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Divergence
Definition 4 (Divergence)
The divergence of a differentiable vector field
F(x, y , z) = P(x, y , z)i + Q(x, y , z)j + R(x, y , z)k
is denoted by div F and is given by
div F =
∂P
∂Q
∂R
+
+
∂x
∂y
∂z
(1)
Denote the del operator ∇ by
∇=
∂
∂
∂
i+
j+
k
∂x
∂y
∂z
then
div F = ∇ · F
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Example 5
Find the divergence for each of the following vector field
a. F(x, y ) = x 2 y i + xy 3 j
b. F(x, y , z) = xi + y 3 z 2 j + xz 3 k
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Curl
Definition 6 (Curl)
The curl of a differentiable vector field
F(x, y , z) = P(x, y , z)i + Q(x, y , z)j + R(x, y , z)k
is denoted by curl F and is given by
∂Q
∂R
∂P
∂Q
∂P
∂R
curl F =
−
i−
−
j+
−
k
∂y
∂z
∂x
∂z
∂x
∂y
curl F = ∇ × F =
i
j
k
∂
∂x
∂
∂y
∂
∂z
P
Q
R
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Example 7
Find the curl of each of the following vector fields
F = x 2 yzi + xy 2 zj + xyz 2 k and G = (x cos y )i + xy 2 j
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Divergence and Curl in the Context of Fluid Flow
Let V be the velocity field of a fluid with constant density ρ. Then the flux
density
F = ρV
is the rate of flow per unit area and used to measure the tendency of the
fluid to diverge from the point P.
If div F(P) > 0, the net flow is outward
near P and P is called a source.
If div F(P) < 0, the net flow is inward
near P and P is called a sink.
If div F(P) = 0, then fluid is said to be
incompressible.
Figure: The vector field
F = x 2i + y 2j
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Assume that F represents the velocity field
of a fluid flow.
If curl F = 0 at a point P, then the fluid
is free from rotations at P and F is called
irrotational at P
Figure: Curl vector is associated
with rotations
If curl F(P) 6= 0, then particles near
P(x, y , z) in the fluid tend to rotate
about the axis that points in the direction of curl F, and the length of this curl
vector is a measure of how quickly the
particles move around the axis.
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Definition 8 (Laplacian operator)
Let f (x, y , z) define a function with continuous first and second partial
derivatives. Then the Laplacian of f is
∇2 f = ∇ · ∇f =
∂2f
∂2f
∂2f
+
+
∂x 2 ∂y 2 ∂z 2
Example 9
Show that f (x, y ) = e x cos y is harmonic, that is,
∇2 f (x, y ) = 0.
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13.2 Line Integrals
A plane curve C given by the parametric
equations
x = x(t)
a≤t≤b
y = y (t),
or by the vector function
R(t) = x(t)i + y (t)j.
Then the line integral of f along C is given by
s 2
Z
Z b
dx 2
dy
f (x, y )ds =
f (x(t), y (t))
+
dt
dt
dt
C
a
|
{z
}
ds
Z
=
b
f (R(t)) R0 (t) dt
(5)
a
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In particular case where C is the line segment that joins (a, 0) to (b, 0)
using x as the parameter x = x, y = 0 for a ≤ x ≤ b, then
Z
Z
f (x, y )ds =
C
b
f (x, 0)dx
(6)
a
Example 10
R
Evaluate C (2 + x 2 y )ds, where C is the upper half of the unit circle
x 2 + y 2 = 1.
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Properties of line integrals
Let f , f1 , and f2 be a scalar functions defined on a smooth curve, orientable
curve. Then for any constant k
Z
Z
kfds = k
fds
C
C
Z
Z
Z
(f1 + f2 )ds =
f1 ds +
f2 ds
C
C
C
Z
Z
fds =
fds
−C
C
where −C denotes the curve C traversed in the opposite site
direction.
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Properties of line integrals
Suppose now that C is a piecewise-smooth curve; that is, C is a
union of a finite number of smooth curves C1 , C2 , ..., Cn . Then the
line integral of f along C is
Z
Z
Z
Z
f (x, y )ds =
f (x, y )ds +
f (x, y )ds + · · · +
f (x, y )ds
C
C1
C2
Cn
(7)
Figure: A piecewise-smooth curve
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Example 11
R
Evaluate C 2xds, where C consists of the arc C1 of the parabola y = x 2
from (0, 0) to (1, 1) followed by the vertical line segment from (1, 1) to
(1, 2).
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Line Integrals of f along C with respect to x and y
A plane curve C given by the parametric equations
x = x(t)
y = y (t),
a≤t≤b
or by the vector function R(t) = x(t)i + y (t)j. The line integrals of f along
C with respect to x and y are given by
Z
Z b
f (x, y )dx =
f (x(t), y (t))x 0 (t)dt
(8)
C
a
Z
Z b
f (x, y )dy =
f (x(t), y (t))y 0 (t)dt
(9)
C
a
It frequently happens that line integrals with respect to x and y occur
together, we obtain
Z
Z
Z
P(x, y )dx +
Q(x, y )dy =
[P(x, y )dx + Q(x, y )dy ]
(10)
C
C
C
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Example 12
Evaluate
R
C
y 2 dx + xdy , where
(a) C = C1 is the line segment from (−5, −3) to (0, 2);
(b) C = C2 is the arc of the parabola x = 4 − y 2 from (−5, −3) to (0, 2).
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Remark
Given parametrization
x = x(t)
y = y (t) a ≤ t ≤ b
determines an orientation of a curve C , with the positive direction corresponding to increasing values of the parameter t and denote −C the curve
C traversed in the opposite site direction. Then
Z
Z
Z
Z
f (x, y )dx = −
f (x, y )dx,
f (x, y )dy = −
f (x, y )dy
−C
C
−C
C
But if we integrate with respect to arc length
Z
Z
f (x, y )ds =
f (x, y )ds
−C
C
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Line Integrals in Space
Suppose that C is a smooth space curve given by the parametric equations
x = x(t)
y = y (t)
z = z(t),
a≤t≤b
or by the vector function R(t) = x(t)i + y (t)j + z(t)k.
Line integral with respect to arc length
Then the line integral of f along C
Z
s
b
Z
f (x, y , z)ds =
f (x(t), y (t), z(t))
C
a
dx
dt
2
+
2
dy
dt
+
dz
dt
2
dt
which can be written in the more compact vector notation
Z
Z
f (x, y , z)ds =
C
b
f (R(t)) R0 (t) dt
(11)
a
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Line integrals along C with respect to x, y , and z
Line integrals along C with respect to x is defined by
Z
Z
f (x, y , z)dx =
C
b
f (x(t), y (t), z(t)) x 0 (t)dt
(12)
a
Similarly, line integrals with respect to x, y , and z occur together
Z
[P(x, y , z)dx + Q(x, y , z)dy + R(x, y , z)dz]
(13)
C
can be evaluated by expressing everything (x, y , z, dx, dy , dz) in terms of
the parameter t.
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Example 13
Evaluate
R
C
y sin zds, where C is the circular helix given by the equations
x = sin t
y = cos t
z = t,
0 ≤ t ≤ 2π.
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Example 14
R
Evaluate C ydx + zdy + xdz, where C is the straight line segment from
(2, 0, 0) to (3, 4, 5)
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Line Integrals of Vector Fields
Let F(x, y , z) = P(x, y , z)i + Q(x, y , z)j + R(x, y , z)k be a vector field, and
let C be a piecewise smooth orientable curve with parametric representation
R(t) = x(t)i + y (t)j + z(t)k a ≤ t ≤ b
Denote dR = dxi + dy j + dzk, we can define the line integral of F along
C by
Z
Z b
F · dR =
F (R(t)) · R0 (t)dt
(14)
C
a
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Example 15
Let F(x, y ) = xy 2 i + x 2 y j and evaluate
and (2, 4) along the following paths:
R
C
F · dR between the point (0, 0)
a. the line segment connecting the points
b. the parabolic arc y = x 2 connecting the points.
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Example 16
R
Evaluate C F · dR, where F(x, y , z) = xy i + yzj + zxk and C is the
twisted cubic given by
x =t
y = t2
z = t 3,
0 ≤ t ≤ 1.
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Applications of Line Integrals: Mass
Consider a thin wire with the shape of a curve C and let ρ (x, y , z) be the
density at each point P(x, y , z) on the wire. Suppose the curve is
described by the parametric equation
x = x(t), y = y (t), z = z(t),
a ≤ t ≤ b.
The mass of the wire is given by
Z
m=
(15)
ρ (x, y , z) ds
C
The center of mass of the wire is then the point (x̄, ȳ , z̄), where
Z
Z
1
1
x̄ =
xρ (x, y , z) ds, ȳ =
y ρ (x, y , z) ds,
m C
m C
Z
1
z̄ =
zρ (x, y , z) ds
m C
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Example 17
A wire has a shape of the curve
√
x = 2 sin t, y = cos t,
z = cos t,
0 ≤ t ≤ π.
If the wire has density ρ(x, y , z) = xyz at each point (x, y , z), what is its
mass?
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Applications of Line Integrals: Work
Suppose that F = Pi + Qj + Rk is a continuous force field on R3 . Then
the work done by this force in moving a particle along a smooth curve C is
Z
Z
F (x, y , z) · T (x, y , z) ds =
W =
C
F · Tds
(16)
C
where T (x, y , z) is the unit tangent vector at the point (x, y , z) on C .
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If the curve C is represented by arc length parameter R(s), then
Z
dR
T=
and W =
F · dR.
(17)
ds
C
If the curve C is parameterized by vector function R(t) for a ≤ t ≤ b,
then
R0 (t)
R0 (t)
T(t) =
and ds = R0 (t) dt
and the work W
Z
Z
F · dR =
W =
C
b
F (R(t)) · R0 (t)dt
(18)
a
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Example 18
Find the work done by the force field F(x, y ) = x 2 i − xy j in moving a
particle along the quarter-circle
R(t) = cos ti + sin tj,
0≤t≤
π
.
2
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Example 19
An object moving in the force field F(x, y ) = y 2 i + 2xy j. How much work
is performed as the object moves from the point (2, 0) counterclockwise
along the elliptical path x 2 + 4y 2 = 1 to (0, 1), then back to (2, 0) along
the line segment joining the two points, as shown in the figure.
Figure: The curve C
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13.3 The Fundamental Theorem and Path Independence
Theorem 20 (Fundamental Theorem for line integrals)
Let C be a smooth curve given by the vector function R(t), a ≤ t ≤ b.
Let f be a differentiable function of two or three variables whose gradient
vector is continuous on C . Then
Z
∇f · dR = f (R(b)) − f (R(a))
(19)
C
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Example 21
Evaluate the line integral
R
C
F · dR, where
F = ∇ 3x − x 2 y − y 3
and C is the path described by R(t) = (sin t) i + (cos t) j, 0 ≤ t ≤ π.
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Conservative Vector Fields
Definition 22 (Conservative Vector Fields)
A vector field F is said to be conservative in a region D if F = ∇f for
some scalar function f in D. The function f is called a scalar potential of
F in D.
Test for Consevative vector fields:
Cross-partials test for a conservative vector field in R2
The curl criterion for a conservative vector field in R3
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Conservative Vector Fields
Test for Conservative Vector Fields
Theorem 23 (Cross-partials test for a conservative vector field in R2 )
Let F(x, y ) = P(x, y )i + Q(x, y )j be a vector field on an open simply connected region D. Then F is conservative in D if and only if
∂P
∂Q
=
∂y
∂x
throughout D.
(20)
Example 24
a. If F(x, y ) = (3 + 2xy ) i + x 2 − 3y 2 j, find a function f such that
F = ∇f .
R
b. Evaluate the line integral C F · dR, where C is the curve given by
R(t) = e t sin ti + e t cos tj,
0 ≤ t ≤ π.
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Theorem 25 (The curl criterion for a conservative vector field in R3 )
Suppose the vector field F and curl F are continuous in the simply
connected region D of R3 . Then F is conservative in D if and only if
curl F = 0.
Example 26
Show that the vector field
F = 20x 3 z + 2y 2 , 4xy , 5x 4 + 3z 2
is conservative in R3 and find a scalar potential function of F.
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Independence of Path
Definition 27
If F is a continuous vector field with domain D. The line integral
is independent of path if
Z
C
F · dR
Z
F · dR =
C1
R
F · dR
C2
for any two paths C1 and C2 in D that have
the same initial and terminal points.
Figure: Independence of Path
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Theorem 28 (Equivalent conditions for path independence)
If F is a continuous vector field on the open connected set D, then the
following three conditions are either all true or false:
a. F is conservative on D.
H
b. C F · dR = 0 for every piecewise smooth closed curve C in D.
R
c. C F · dR is independent of path within D.
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Example 29
Evaluate the line integral
R
C
F · dR, where
F(x, y ) = (3 + 2xy ) i + x 2 − 3y 2 j
for each of following curves:
x2 y2
+
= 1.
a. the ellipse
4
9
b. the curve with parametric equations
x = t 2 cos πt
y = e −t sin πt,
0 ≤ t ≤ 1.
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13.4 Green’s Theorem
Theorem 30 (Green’s Theorem)
Let D be a simply connected region that is bounded by a positively oriented,
piecewise-smooth, simple closed curve C . Then if the vector field F(x, y ) =
P(x, y )i + Q(x, y )j is continuous differentiable on D, we have
Z
I
ZZ ∂Q
∂P
F · dR =
(Pdx + Qdy ) =
−
dA
(21)
∂x
∂y
C
C
D
Figure: A simply connected region D
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Example 31
A closed path C in the plane is defined by figure below. Find the work
done on an object moving along C in the force field
F(x, y ) = x + xy 2 i + 2 x 2 y − y 2 sin y j
Figure: Closed path C
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Applications of Green’s Theorem
In the Green’s Theorem, choose P and Q such that
∂P
∂Q
−
=1
∂x
∂y
we can use the line integral to compute the area A of the region D.
ZZ I
∂P
∂Q
A=
−
dA =
(Pdx + Qdy ) .
∂x
∂y
D
C
(22)
Theorem 32 (Area as line integral)
Let D be a simply connected region in the plane with piecewise smooth,
positively oriented closed boundary C . Then the area A of the region D is
given by
I
I
I
1
A=
xdy = −
ydx =
(xdy − ydx)
(23)
2 C
C
C
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Example 33
Find the area enclosed by the ellipse
x2 y2
+ 2 = 1.
a2
b
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Theorem 34 (Green’s Theorem for Multiply-Connected Regions)
Assume that the boundary C of the region D (as in figure) consists of two
simple closed curves C1 and C2 . We assume that these boundary curves
are oriented so that the region D is always on the left as the curve C is
traversed. Then if the vector field F(x, y ) = P(x, y )i+Q(x, y )j is continuous
differentiable on D, we have
I
I
ZZ ∂Q
∂P
Green’s theorem
F · dR =
(Pdx + Qdy )
=
−
dA (24)
∂x
∂y
C
C
D
Figure: A doubly-connected region with
oriented boundary curve
Figure: D = D 0 ∪ D 00
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Example 35
Show that
R
C
F · dR = 2π, where
F(x, y ) =
−y i + xj
x2 + y2
and C is any positively oriented simple closed path that encloses the origin.
Figure: The region D for doubly connected regions
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Alternative Form of Green’s Theorem
Let D be a simply connected region with a positive oriented boundary C .
Then if vector field F(x, y ) = P(x, y )i+Q(x, y )j is continuous differentiable
on D, we have
I
I
F · dR =
(Pdx + Qdy )
C
C
ZZ ZZ
∂P
∂Q
Green’s theorem
−
dA =
=
(curl F · k) dA (25)
∂x
∂y
D
D
and
I
I
dR
F · dR =
F·
ds =
ds
C
C
I
F · Tds
(26)
C
And therefore, we obtain
ZZ
I
F · Tds =
C
(curl F · k) dA
(27)
D
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Theorem 36 (Divergence theorem)
Suppose that F(x, y ) = P(x, y )i + Q(x, y )j is continuous differentiable on
D with a piecewise smooth boundary C . Then
I
ZZ
F · Nds =
div FdA
(28)
C
D
Figure: The outward unit normal and tangential vector to a point on C
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Normal Derivative
Definition 37 (Normal Derivative)
The normal derivative of f , denoted by ∂f /∂n, is the directional
derivative of f in the direction of the normal vector pointing to the
exterior of the domain of f . In other words
∂f
= ∇f · N
∂n
(29)
where N is the outward unit normal vector.
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Example 38 (Green’s formula for the integral of the Laplacian)
Suppose f is a scalar function with continuous first and second partial
derivatives in the simply connected region D. If the piece-wise smooth
positively oriented closed curve C bounds D, then we have
ZZ
I
∂f
2
ds
∇ fdA =
D
C ∂n
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13.5 Surface Integrals
Definition 39 (Surface integral)
The surface integral of g over the surface S is defined by
ZZ
g (x, y , z)dS =
S
lim
m,n→∞
Figure: Projection D of the surface S
m X
n
X
g Pij∗ ∆Sij
i=1 j=1
Figure: Surface S
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Compute Surface Integral
If a surface S is defined by z = f (x, y ), then the surface integral of g
over S is given by
ZZ
ZZ
q
(30)
g (x, y , z)dS =
g (x, y , f (x, y )) fx2 + fy2 + 1dA
S
D
where D is the projection of the surface S onto the xy -plane.
If a surface S is defined parametrically by vector field
R(u, v ) = x(u, v )i + y (u, v )j + z(u, v )k
over a region D in the uv -plane, then the surface integral of g over S
is given by
ZZ
ZZ
(31)
g (x, y , z)dS =
g (R(u, v )) kRu × Rv k dudv
S
D
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Example 40
Evaluate
RR
S
ydS, where S is the surface
z = x + y 2,
0 ≤ x ≤ 1,
0 ≤ y ≤ 2.
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Example 41
Compute the surface integral
RR
S
x 2 dS, where S is the unit sphere
x 2 + y 2 + z 2 = 1.
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Example 42
Evaluate the surface integral
ZZ
g (x, y , z)dS
S
where g (x, y , z) = xz + 2x 2 − 3xy and the surface S is the portion of the
plane 2x − 3y + z = 6 that lies over the unit square R: 2 ≤ x ≤ 3,
2 ≤ y ≤ 3.
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Applications of Surface Integrals: Surface Area Formula
If S is a piecewise smooth surface, its area is given by
ZZ
A=
dS
(32)
S
If a surface S is given by the function z = f (x, y ) then we have
ZZ q
A=
fx2 + fy2 + 1dA
(33)
D
where D is the projection of S onto the xy -plane.
If S is given by a vector function R(u, v ), then we have
ZZ
A=
kRu × Rv k dA
(34)
D
where D is the parameter domain.
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Applications of Surface Integrals: Mass
Consider a thin curved lamina whose shape is part of a surface S and let
ρ (x, y , z) be the density at each point P(x, y , z) on the lamina.
The total mass of the lamina is given by
ZZ
m=
(35)
ρ (x, y , z) dS
S
The center of mass of the lamina is then the point (x̄, ȳ , z̄), where
ZZ
ZZ
1
1
xρ (x, y , z) dS, ȳ =
y ρ (x, y , z) dS,
x̄ =
m
m
S
S
ZZ
1
z̄ =
zρ (x, y , z) dS
m
S
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Example 43
Find the mass of a lamina
p of density ρ(x, y , z) = z in the shape of the
upper hemisphere z = a2 − x 2 − y 2 .
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Flux Integral
Oriented Surfaces
Suppose a surface S has a tangent plane at every point (x, y , z) on S (except
at any boundary point).
Figure: Two unit normal vector N1 and N2 = −N1
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Definition 44 (Oriented Surfaces)
If it is possible to choose a unit normal vector n at every such point
(x, y , z) so that n varies continuously over S, then is called an oriented
surface and the given choice of n provides S with an orientation.
Figure: The two orientations of an orientable surface
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For a surface z = f (x, y ) given as the graph
of f , by setting F (x, y , z) = z − f (x, y ) the
unit normal vector is
N1 =
Figure: Upward and downward
unit normal vectors
−fx i − fy j + k
∇F
=q
k∇F k
fx2 + fy2 + 1
(36)
where ∇F = h−fx , −fy , 1i. Since the
k-component is positive, this gives the
upward orientation of the surface.
If S is a smooth orientable surface given in parametric form by a vector
function R(u, v ), then it is automatically supplied with the orientation of
the unit normal vector
N1 =
Ru × Rv
kRu × Rv k
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For a closed surface, that is, a surface that is the boundary of a solid
region E , the convention is that the positive orientation is the one for
which the normal vectors point outward from E , and inward-pointing
normals give the negative orientation.
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Flux Integral
Definition 45 (Flux Integral)
If F is a continuous vector field defined on an oriented surface S with unit
normal vector N, then the flux of F across S is given by the surface
integral
ZZ
F · NdS
(38)
Flux =
S
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Flux Integral
Compute the flux of F across S
If a surface S is given by the function z = f (x, y ) then we have
ZZ
ZZ
F · NdS =
F (x, y , f (x, y )) · h−fx , −fy , 1i dA
(39)
S
D
where D is the projection of S onto the xy -plane.
If S is given by a vector function R(u, v ), then we have
ZZ
ZZ
F · NdS =
F · (Ru × Rv ) dA
S
(40)
D
where D is the parameter domain.
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Example 46
RR
Compute the flux integral S F · NdS, where F = y i + xj + zk and S is
the triangular surface cut off from the plane x + y + z = 1 by the
coordinate planes.
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Example 47
Find the flux of the vector field F = y i + xj + zk through the surface S
parameterized by
R(u, v ) = (uv )i + (u − v )j + (u + v )k
over the triangular region D in the uv -plane that is bounded by u = 0,
v = 0, and u + v = 1.
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Example 48
Let R be the region that is bounded above by paraboloid z = 9 − x 2 − y 2
and below by the xy -plane. Experiments indicate that the velocity of heat
flow is given by the vector field H = −K ∇T , where
T (x, y , z) = 2x + 3y − 3z 2
is the temperature at each point P(x, y , z) in the region R and K is a
constant (the heat conductivity, which is obtained
RR experimentally for each
different substance). Find the total heat flow R H · NdS out of the
region (that is, N is the outward unit normal).
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13.6 Stokes’ Theorem and Applications
Definition 49 (Compatible orientation)
The orientation of the closed path C on the surface S is compatible with
the orientation on S if you walk in the positive direction around C with
your head pointing in the direction of N, then the surface will be on your
left.
Figure: Compatible orientation
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Theorem 50 (Stokes’ Theorem)
Let S be an oriented surface with unit normal vector field N, and assume
that S is bounded by a a simple, closed, piecewise smooth boundary curve
C whose orientation is compatible with that of S. If F is a vector field that
is continuously differentiable on S, then
I
ZZ
F · dR =
(curl F · N) dS
(41)
C
S
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Example 51
H
Evaluate C F · R, where F(x, y , z) = −y 2 i + xj + z 2 k and C is the curve
of intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1.
(Orient C to be counterclockwise when viewed from above.)
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Example 52
RR
Use Stokes’ Theorem to compute the integral S curl F · NdS, where
F(x, y , z) = xzi + yzj + zxk and S is the part of the sphere
x 2 + y 2 + z 2 = 1 that lies inside the cylinder x 2 + y 2 = 1 and above the
xy -plane.
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Physical Interpretation of Stokes’ Theorem
Assume that the fluid flows across the
surface S with velocity field V. Then the
cumulative rotational tendency over the
surface S is
ZZ
(curl V · N) dS
(42)
S
Figure: The tendency of a fluid
to swirl across the surface S is
measured by curl V · N
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Notice that the line integral
Z
Z
V · dR =
C
V · Tds
(43)
C
where V·T is the component of V in the direction of the unit tangent vector
R
T. Thus C V · dR is a measure of the tendency of the fluid to move around
C and is called the circulation of V around C .
R
Figure: C V · dR > 0: positive
circulation
R
Figure: C V · dR < 0: negative
circulation
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The Stokes’ Theorem
I
ZZ
(curl V · N) dS
S
{z
}
|
The cumulative tendency
of a fluid to swirl across
the surface S
=
V · Tds
| {z }
C
The circulation of
fluid around the
boundary curve C
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13.7 Divergence Theorem and Applications
Theorem 53 (The Divergence Theorem)
Let S be a smooth, orientable surface that encloses a solid region E in R3 .
If F is a continuous vector field whose components have continuous partial
derivatives in an open set containing E , then
ZZ
ZZZ
F · NdS =
div FdV
(45)
S
E
where N is the outward unit normal filed for the surface.
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Example 54
Find the flux of the vector field F(x, y , z) = zi + y j + xk over the unit
sphere x 2 + y 2 + z 2 = 1.
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Example 55
RR
Evaluate S F · NdS, where F(x, y , z) = x 2 i + xy j + x 3 y 3 k and S is the
surface of the tetrahedron bounded by the plane x + y + z = 1 and the
coordinate planes, with outward unit normal vector N.
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Example 56
RR
Find S F · NdS, where F(x, y , z) = 2xi − 3y j + 5xk and S is the
p
hemisphere z = 9 − x 2 − y 2 together with the disk x 2 + y 2 ≤ 9 in the
xy -plane.
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Example 57
RR
Find S F · NdS, where F(x, y , z) = xy i − z 2 k and S is the surface of the
upper five faces of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1.
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Physical Interpretation of Divergence
Let F = ρV be the flux density associated with a fluid of density ρ flowing
with velocity V and let P0 be a point inside a solid region where the conditions of the divergence theorem are satisfied. Then we have
ZZ
1
div F(P0 ) = lim
F · NdS
(46)
r →0 V (Br )
Sr
where Sr is a sphere of the ball Br with center at P0 and radius r .
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