CHAPTER 5 VECTOR ANALYSIS · CHAPTER 5 VECTOR ANALYSIS 5.2 Line Integrals Definition The line...
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CHAPTER 5 VECTOR ANALYSIS
5.2 Line Integrals
Definition
The line integral of the scalar function ( , )f x y
along a smooth and orientable curve C is given by
* *
1
( , ) lim ( , )n
k k kn
kC
f x y ds f x y s
Evaluation of Line Integrals
Theorem
Let C be a smooth curve with parametric
representation
( ), ( ) ( )x x t y y t a t b
and ( , )f x y is a continuous function over C. The
line integral is given as:
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2 2( , ) ( ( ), ( )) ( ( )) ( ( ))b
C a
f x y ds f x t y t x t y t dt
If C is a curve in 3-space with parametric
representation
( ), ( ), ( )( )x x t y y t z z t a t b
and ( , , )f x y z is continuous on C, then
( , , ) ( ( ), ( ), ( ))b
C a
dsf x y z ds f x t y t z t dtdt
where
2 2 2( ( )) ( ( )) ( ( ))ds x t y t z tdt
Example
Evaluate x
C
ye ds , where C is the segment
joining (1, 2) and (4, 7).
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Example
Evaluate 2 2( )
C
x y ds , along the helix
( ) (cos 4 ) (sin 4 )r t t i t j tk , 0 2t
Line Integrals in a Vector Field
Definition
Let F be a vector field and let C be the curve with
parametric representation
( ) ( ), ( ), ( )r t x t y t z t for a t b
Then the line integral of F over C is
C
F T ds ( )b
a
drF t dtdt
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C
F dr
where T is the tangent vector of ( )r t .
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Connection between the line integrals of vector
fields and line integrals of scalar fields:
Let
( , , ) ( , , ) ( , , )F P x y z i Q x y z j R x y z k
t b
C t a
F dr P dx Q dy Rdz
[ ( ( ), ( ), ( ))
( ( ), ( ), ( ))
( ( ), ( ), ( )) ]
t b
t a
b
a
b
a
dxP x t y t z t dt
dyQ x t y t z t dt
dzR x t y t z t dtdt
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Example
Let ( ) ( )F x y i y x j . Compute
C
F dr where C is the path along the parabola
2y x from (1, 1) to (4, 2).
Example
Compute
C
F dr where F yi x j and C
is the closed path shown in the figure.
(0, 0)
(0, 1)
(1, 0)
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Work as a Line Integral
Definition
The work done by a force F Pi Qj Rk
over a smooth curve ( )r t from t = a to t = b is
C
W F dr
What this say:
If F is a force field and C(t) represents the position
of a mass in the plane at time t, then the dot
product gives the the amount of force exerted on
the mass in its direction of travel, how much of a
push the force field will give the mass at the point.
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Example
If 22 ( ) (3 2 4 )F x y z i x y z j x y z k ,
find the work done by F in moving a particle once
around a circle C in the xy-plane, if the circle has
radius 3 and centre at the origin.
Example
Find the work done by the force
2 2 3F x z i yx j xz k ,
in moving a particle along a straight line segments
from (1,1,0) to (1,1,1) and then to (0,0,0).
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5.2.1 Green’s Theorem
Theorem
Let R be a plane region with a positively oriented
piecewise smooth, simple closed curve boundary C.
If the vector field
( , ) ( , ) ( , )F x y P x y i Q x y j
is continuously differentiable on R, then
C C R
Q PF dr P dx Qdy dAx y
R is a 2D space
enclosed by a simple
closed curve C
C
R
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Note
R is required to be simply connected with a positively oriented boundary C.
The notation
C
P dx Q dy or
C
P dx Q dy
is used to indicate that the line integral is
calculated using the positive (counter-
clockwise) orientation of the close curve C.
Example
Use Green’s Theorem to evaluate the line integral
C
F dr
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along the positively oriented triangular path shown
and that 2( , )F x y x yi x j .
Example
Find the work done by the force field
1 2 2( , ) tan ln( )yF x y i x y jx
when an object moves counterclockwise around
the circular path 2 2 4x y .
(1, 2)
(1, 0) (0, 0)
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5.2.2 Conservative Fields
Definition
A vector field F is said to be conservative if there
exists a differentiable function such that the
gradient of is F . That is
F
The function is called the scalar potential
function for F .
Not all vector fields are conservative.
Terminology
i. A path C is called closed if its initial and terminal points are the same point.
ii. A path C is simple if it doesn’t cross itself.
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iii. A region D is open if it doesn’t contain any of its boundary points.
iv. A region D is connected if any two points in D can be joined by a path that lies entirely within D.
v. D is simply connected if every closed curve in D encloses only points in D.
simply connected region is one with “no holes”
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A simply connected region A region that is not
simply connected
Theorem: Testing for conservative field
Let ( , ) ( , )F P x y i Q x y j be a vector field
where P and Q have continuous first partials in the
open connected region D. Then F is conservative
in D if and only if
Q Px y
throughout D.
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By similar reasoning, it can be shown that
( , , ) ( , , ) ( , , )F P x y z i Q x y z j R x y z k
is conservative if and only if
QRy z
, R Px z
, Q Px y
This is equivalent to
conservative curl 0F F .
Example
Determine whether the vector field
y yF e i xe j
is conservative. If it is, find a potential.
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Example
Show that the vector field
3 2 22 3F xy z i x j xz k
is conservative and then find a scalar potential
function for F .
Theorem: Fundamental theorem for line
integrals
Let F be a conservative vector field continuous on
an open connected region D. That is, there exists a
function such that F . Then, if C is any
smooth curve from A to B lying entirely in D, we
have
( ) ( )B
A
F dr B A
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What this say:
Recall FTC,
( ) ( ) ( )b
a
F x dx F b F a
A version of this for line integrals over certain kinds
of vector fields,
( ( )) ( ( ))C
dr r b r a
Example
Let 2 22 ( 2 )F xyzi x z j x y z k be a
vector field describing a force.
(a) Show that F is conservative.
(b) Find the work done by F in moving an object along the line segment beginning at (1, 1, 1) and ending at (2, 2, 4).
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Theorem: Independence of path
Let F be a vector field continuous on an open
connected region D.
C
F dr is independent of
path in D, if and only if F is a conservative vector
field.
What this says:
CF dr is independent of path if
1 2C CF dr F dr for any two paths 1C
and 1C in D with the same endpoints.
C
dr is independent of path.
If F is conservative then CF dr is
independent of path.
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Theorem: Closed-loop property
C
F dr is independent of path in D if and only if
0C
F dr for every closed path C in D.
This is equivalent to,
conservative 0C
F F dr
The symbol indicate that the curve C must
be closed.
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Example
Show that
C
F dr is independent of path.
Hence find a potential function and evaluate the
line integral along C.
(a) 10 7 7 2F x y i x y j and C is
the curve 21y x from (0,1) to (1,0).
(b) F yz i xz j xy k and C is the
the line segment from (0,0,0) to (1,1,1), and
then to (0,0,1).
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5.3 Surface Integrals
Surface Integrals of Scalar Fields
Definition
Suppose f is defined and continuous on a surface .
The surface integral of f over is denoted by
( , , )f x y z dS
where S is the area of the surface.
When the surface projects onto the region Rxy in
the xy-plane and has the representation
( , )z f x y then
22
1z zdS dAx y
where dA is either dx dy or dy dx (or rdrd )
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The surface integral is
22
( , , )
( , , ( , )) 1R
f x y z dS
z zf x y g x y dAx y
can obtain similar formulas for surfaces given by ( , )y h x z (with R in the xz-plane) and
( , )x k y z (with R in the yz-plane).
Formula for Surface Integrals
1. ( , )z g x y , R in the xy-plane
22
( , , )
( , , ( , )) 1R
f x y z dS
z zf x y g x y dAx y
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2. ( , )y h x z , R in the xz-plane
2 2
( , , )
( , ( , ), ) 1R
f x y z dS
y yf x h x z z dAx z
3. ( , )x k y z , R in the yz-plane
2 2
( , , )
( ( , ), , ) 1R
f x y z dS
x xf k y z y z dAy z
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Example
Evaluate the surface integral
2 2y z dS
where is part of the cone 2 2z x y that
lies between the planes 1z and 2z .
Example
Evaluate
2x z dS
where is the portion of the cylinder 2 2 1x y between 0z and 1z .
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Surface Integrals of Vector Fields
For a general surface in space, each element of
surface dS has a vector area dS such that
dS n dS .
If F is a vector field, the surface integral
F dS F n dS
where n is the outward unit normal to the surface
.
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Definition
If F Pi Qj Rk is a continuous vector
field on an oriented surface with outward unit
normal vector n , then the surface integral of F
over is
F dS F n dS
22
1R
z zF n dAx y
where ( , )z g x y
Note
This integral is also called the flux of F across .
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Example:
Calculating the flux of a vector field outward
through the surface S.
Other forms
1. F dS F n dS
2 2
1R
y yF n dAx z
where ( , )y h x z .
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2. F dS F n dS
2 2
1R
x xF n dAy z
where ( , )x k y z
Note
To work with surface integrals of vector fields we need to be able to write down a formula for the unit normal vector corresponding to the chosen orientation.
depends on how the surface is given
Let’s suppose that the surface is given by
( , )z g x y . We define a new function,
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( , , ) ( , )x y z z g x y
Thus, in term of the new function, the surface is
a level surface for ( , , ) 0x y z .
Recall: will be normal to at ( , )x y . This
means that we have a normal vector to the surface.
We obtain a unit normal vector:
n
To compute the gradient vector:
( , , ) ( , )x y z z g x y
g g z zi j k i j kx y x y
2 2 1
x y
x y
z i z j kn
z z
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Notice that the component of the normal vector
in the z-direction (identified by the k in the normal vector) is always positive and so this normal vector will generally point upwards.
Multiplying by -1 produces the negative orientation.
Likewise for surfaces in the form
( , )y h x z , so ( , , ) ( , )x y z y h x z
y yh hi j k i j kx z x z
Thus, 2 21x z
y yi j kx zny y
For surfaces in the form
( , )x k y z , so ( , , ) ( , )x y z x k y z
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2 21 y z
x xi j ky znx x
Surfaces positive
orientation
negative
orientation
( , )z f x y 2 2 1
x y
x y
z z
z z
i j kn
2 2 1
x y
x y
z z
z z
i j kn
( , )y g x z 2 21x z
x z
y y
y y
i j kn
2 21x z
x z
y y
y y
i j kn
( , )x h y z 2 21
y z
y z
x x
x x
i j kn
2 21
y z
y z
x x
x x
i j kn
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Theorem
Let be a smooth surface of the form
( , )z g x y , ( , )y h x z or ( , )x k y z and
let R be the projection of on the xy-plane, xz-
plane and yz-plane respectively. Suppose that the
surfaces can be rewritten as ( , , ) 0x y z . If is
continuous on R, then
i.
R
F n dS F dA
if has a positive orientation.
ii.
R
F n dS F dA
if has a negative orientation.
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Example
Let be the part of the surface of the cone
2 2z x y from 0z to 1z and let
( , , )F x y z i j k . Evaluate
F n dS
Example
Let is the closed surface of the tetrahedron with
vertices (1,0,0), (0,3,0), (0,0,2) and (0,0,0). Evaluate
the surface integral
F n dS
where 2F x i xy j xzk .
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5.3.1 Stokes’ Theorem
Let C be any closed curve in 3D space, and let be
an oriented surface bounded by C with unit normal
vector n . If F is a vector field that is continuously
differentiable on , then
C
F dr F ndS
can be any surface bounded by the curve.
Stokes Theorem reduces to Green's Theorem when the curve is a 2D curve.
The integral gives the circulation of the vector
field F around C.
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Example
Evaluate
C
F dr where
2 3 24F x i xy j xy k and C is the
close curve on the plane z y with vertices (1,3,3),
(0,3,3) and (0,0,0) with the orientation
counterclockwise when viewed from the positive z-
axis.
Example
Evaluate
C
F dr where F zi x j yk and C
is the intersection of the xy-plane with the paraboloid 2 29y x z with the orientation
counterclockwise direction when viewed from the
positive y-axis.
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5.3.2 Gauss’ Theorem
Gauss’ Theorem (Divergence Theorem)
Let be a smooth orientable surface that encloses
a solid region G in 3
. If F Pi Qj Rk
is a vector field whose components P, Q, and R
have continuous partials derivatives in G, then
divG
F n dS F dV
where n is an outward unit normal vector.
Note
Under appropriate conditions, the flux of a vector field across a closed surface with outward orientation is equal to the triple integral of the divergence of the field over the solid region enclosed by the surface.
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Flux integral Flow out of small cubes
divG
F ndS F dV
Flux integral (left) - measures total fluid flow across
the surface per unit time.
Right integral – measures the fluid flow leaving the
volume dV
For divergence free vector field F , the flux through a closed surface area is zero. Such fields are also called incompressible or source free.
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Example
Use the Divergence Theorem to calculate the
surface integral F n dS
where 3 2 2( , , ) 2 3F x y z x i xz j y zk and
is the surface of the solid bounded by the
paraboloid 2 24z x y and the xy-plane.
Example
Let be the surface of solid enclosed by
2 2 2z a x y and 0z oriented
outward. If 3 3 3F x i y j z k , evaluate
.F n dS