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and
=
√
g
22
du
2
=
√
g
33
du
3
ds
2
,
ds
3
,
(1.54)
represent infinitesimal displacements in the
u
2
and
u
3
directions, respectively.
Next, we consider an element of surface area
da
1
, on the surface defined by
u
1
constant, contained by the parallelogram formed by the infinitesimal displacements
d
s
2
,
d
s
3
along the directions
u
2
,
u
3
. The area of the parallelogram is
du
2
du
3
=
|
d
s
2
×
d
s
3
|
=
|
b
2
×
b
3
da
1
|
,
(1.55)
with
(
b
2
b
3
)
(
b
2
b
3
)
.
|
b
2
×
b
3
|=
×
·
×
(1.56)
Using the vector identity (A.3),
(
b
2
×
b
3
)
·
(
b
2
×
b
3
)
=
(
b
2
·
b
2
)(
b
3
·
b
3
)
−
(
b
2
·
b
3
)(
b
3
·
b
2
),
(1.57)
we find that
2
23
du
2
du
3
da
1
=
g
22
g
33
−
g
.
(1.58)
Similarly, for elements on the
u
2
and
u
3
surfaces their areas are
2
31
du
3
du
1
da
2
=
g
33
g
11
−
g
(1.59)
and
12
du
1
du
2
da
3
=
g
11
g
22
−
g
.
(1.60)
An element of volume, bounded by the three co-ordinate surfaces, is given by
(
b
2
×
b
3
)
du
1
du
2
du
3
d
v
=
d
s
1
·
(
d
s
2
×
d
s
3
)
=
b
1
·
.
(1.61)
If we set
V
=
b
2
×
b
3
in the expression (1.17), we get
b
i
b
3
)
b
i
.
b
2
×
b
3
=
·
(
b
2
×
(1.62)
Replacing
b
i
by its expressions (1.6) and taking the scalar product with
b
1
,wehave
(
b
2
×
b
1
b
1
·
(
b
2
×
b
3
)
=
b
3
)
·
b
3
)
·
(
b
2
×
b
3
)
b
1
+
(
b
3
×
b
1
)
·
(
b
2
×
b
3
)
b
2
b
1
·
(
b
2
×
b
3
)
b
3
+
(
b
1
×
b
2
)
·
(
b
2
×
.
(1.63)
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