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:
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).
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
C
F dr
where T is the tangent vector of ( )r t .
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
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)
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.
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).
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
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
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)
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.
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”
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.
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.
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
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).
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.
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.
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).
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 )
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
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
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 .
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
.
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 .
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 .
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,
( , , ) ( , )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
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
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
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.
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 .
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.
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.
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.
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.
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