Line, surface, and volume integrals | Engineering Mathematics for Electrical Engineering - Electrical Engineering (EE) PDF Download

 Page 1


Line, Surface and Volume
Integrals
1
Page 2


Line, Surface and Volume
Integrals
1
Line integrals
Z
C
Ádr;
Z
C
a¢dr;
Z
C
a£dr (1)
(Á is a scalar ¯eld and a is a vector ¯eld)
We divide the path C joining the points A and B
into N small line elements ¢r
p
, p=1;:::;N. If
(x
p
;y
p
;z
p
) is any point on the line element ¢r
p
,
then the second type of line integral in Eq. (1) is
de¯ned as
Z
C
a¢dr= lim
N!1
N
X
p=1
a(x
p
;y
p
;z
p
)¢r
p
where it is assumed that allj¢r
p
j!0 as N !1.
2
Page 3


Line, Surface and Volume
Integrals
1
Line integrals
Z
C
Ádr;
Z
C
a¢dr;
Z
C
a£dr (1)
(Á is a scalar ¯eld and a is a vector ¯eld)
We divide the path C joining the points A and B
into N small line elements ¢r
p
, p=1;:::;N. If
(x
p
;y
p
;z
p
) is any point on the line element ¢r
p
,
then the second type of line integral in Eq. (1) is
de¯ned as
Z
C
a¢dr= lim
N!1
N
X
p=1
a(x
p
;y
p
;z
p
)¢r
p
where it is assumed that allj¢r
p
j!0 as N !1.
2
Evaluating line integrals
The ¯rst type of line integral in Eq. (1) can be
written as
Z
C
Ádr = i
Z
C
Á(x;y;z)dx+j
Z
C
Á(x;y;z)dy
+k
Z
C
Á(x;y;z)dz
The three integrals on the RHS are ordinary scalar
integrals.
The second and third line integrals in Eq. (1) can
also be reduced to a set of scalar integrals by
writing the vector ¯eld a in terms of its Cartesian
components as a=a
x
i+a
y
j+a
z
k. Thus,
Z
C
a¢dr =
Z
C
(a
x
i+a
y
j+a
z
k)¢(dxi+dyj+dzk)
=
Z
C
(a
x
dx+a
y
dy+a
z
dz)
=
Z
C
a
x
dx+
Z
C
a
y
dy+
Z
C
a
z
dz
3
Page 4


Line, Surface and Volume
Integrals
1
Line integrals
Z
C
Ádr;
Z
C
a¢dr;
Z
C
a£dr (1)
(Á is a scalar ¯eld and a is a vector ¯eld)
We divide the path C joining the points A and B
into N small line elements ¢r
p
, p=1;:::;N. If
(x
p
;y
p
;z
p
) is any point on the line element ¢r
p
,
then the second type of line integral in Eq. (1) is
de¯ned as
Z
C
a¢dr= lim
N!1
N
X
p=1
a(x
p
;y
p
;z
p
)¢r
p
where it is assumed that allj¢r
p
j!0 as N !1.
2
Evaluating line integrals
The ¯rst type of line integral in Eq. (1) can be
written as
Z
C
Ádr = i
Z
C
Á(x;y;z)dx+j
Z
C
Á(x;y;z)dy
+k
Z
C
Á(x;y;z)dz
The three integrals on the RHS are ordinary scalar
integrals.
The second and third line integrals in Eq. (1) can
also be reduced to a set of scalar integrals by
writing the vector ¯eld a in terms of its Cartesian
components as a=a
x
i+a
y
j+a
z
k. Thus,
Z
C
a¢dr =
Z
C
(a
x
i+a
y
j+a
z
k)¢(dxi+dyj+dzk)
=
Z
C
(a
x
dx+a
y
dy+a
z
dz)
=
Z
C
a
x
dx+
Z
C
a
y
dy+
Z
C
a
z
dz
3
Some useful properties about line integrals:
1. Reversing the path of integration changes the
sign of the integral. That is,
Z
B
A
a¢dr=¡
Z
A
B
a¢dr
2. If the path of integration is subdivided into
smaller segments, then the sum of the separate
line integrals along each segment is equal to the
line integral along the whole path. That is,
Z
B
A
a¢dr=
Z
P
A
a¢dr+
Z
B
P
a¢dr
4
Page 5


Line, Surface and Volume
Integrals
1
Line integrals
Z
C
Ádr;
Z
C
a¢dr;
Z
C
a£dr (1)
(Á is a scalar ¯eld and a is a vector ¯eld)
We divide the path C joining the points A and B
into N small line elements ¢r
p
, p=1;:::;N. If
(x
p
;y
p
;z
p
) is any point on the line element ¢r
p
,
then the second type of line integral in Eq. (1) is
de¯ned as
Z
C
a¢dr= lim
N!1
N
X
p=1
a(x
p
;y
p
;z
p
)¢r
p
where it is assumed that allj¢r
p
j!0 as N !1.
2
Evaluating line integrals
The ¯rst type of line integral in Eq. (1) can be
written as
Z
C
Ádr = i
Z
C
Á(x;y;z)dx+j
Z
C
Á(x;y;z)dy
+k
Z
C
Á(x;y;z)dz
The three integrals on the RHS are ordinary scalar
integrals.
The second and third line integrals in Eq. (1) can
also be reduced to a set of scalar integrals by
writing the vector ¯eld a in terms of its Cartesian
components as a=a
x
i+a
y
j+a
z
k. Thus,
Z
C
a¢dr =
Z
C
(a
x
i+a
y
j+a
z
k)¢(dxi+dyj+dzk)
=
Z
C
(a
x
dx+a
y
dy+a
z
dz)
=
Z
C
a
x
dx+
Z
C
a
y
dy+
Z
C
a
z
dz
3
Some useful properties about line integrals:
1. Reversing the path of integration changes the
sign of the integral. That is,
Z
B
A
a¢dr=¡
Z
A
B
a¢dr
2. If the path of integration is subdivided into
smaller segments, then the sum of the separate
line integrals along each segment is equal to the
line integral along the whole path. That is,
Z
B
A
a¢dr=
Z
P
A
a¢dr+
Z
B
P
a¢dr
4
Example
Evaluate the line integral I =
R
C
a¢dr, where
a=(x+y)i+(y¡x)j, along each of the paths in
the xy-plane shown in the ¯gure below, namely,
1. the parabola y
2
=x from (1;1) to (4;2),
2. the curve x=2u
2
+u+1, y =1+u
2
from
(1;1) to (4;2),
3. the line y =1 from (1;1) to (4;1), followed by
the line y =x from (4;1) to (4;2).
FIG. 1: Di®erent possible paths between points (1;1)
and (4;2).
5