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Then, in Eq. (9.82),
u
ij
dS
lim
→
=
0
(9.84)
0
S
so that moving
to the boundary has no effect on this integral.
Consider the integral on the left-hand side of Eq. (9.80b),
ξ
p
ij
u
j
dS
p
ij
u
j
dS
p
ij
u
j
dS
=
lim
→
+
lim
→
(9.85)
0
0
S
S
−
S
S
In the three-dimensional space, with
r
=
, the final integral of Eq. (9.85) can be developed
as
u
j
[
,i
,j
]
∂
∂
p
ij
u
j
dS
lim
→
=
lim
→
(
1
−
2
ν)δ
ij
+
3
n
−
(
1
−
2
ν)(
,i
a
j
−
,j
a
i
)
0
0
S
S
−
1
×
dS
(9.86)
8
π(
1
−
ν)
2
The expression for
p
ij
of Eq. (9.78) has been used here. Let the boundary be smooth at
ξ
,
so
that
S
is a hemispherical surface of Fig. 9.1, where
θ
varies from 0 to 2
π
and
ϕ
varies from
0to
π/
2. For this hemispherical surface, from the relationships of Eq. (9.71) and Fig. 9.2,
∂
∂
=
∂
∂
x
i
=
i
2
sin
n
=
1
,
i
,
,i
a
j
−
,j
a
i
=
0
,
S
=
ϕ
d
ϕ
d
θ
and (Problem 9.11)
∂
∂
∂
∂
dS
=
0
for
i
=
j
x
i
x
j
S
Then, from Eq. (9.86),
−
1
1
p
ij
u
j
dS
2
]
2
sin
lim
→
=
lim
→
u
j
[
(
1
−
2
ν)
+
3
(
)
2
ϕ
d
ϕ
d
θ
,j
8
π(
1
−
ν)
0
0
S
S
−
1
2
] sin
=
lim
→
u
i
[
(
1
−
2
ν)
+
3
(
)
ϕ
d
ϕ
d
θ
,i
8
π(
1
−
ν)
0
S
It follows from the relationship
2
π
π/
2
sin
ϕ
d
ϕ
d
θ
=
d
θ
sin
ϕ
d
ϕ
=
2
π
S
0
0
and from Fig. 9.2, that
2
π
0
θ
π/
2
0
cos
2
sin
3
θ
d
ϕ
d
ϕ
i
=
1
2
π
0
θ
π/
2
0
2
sin
S
(
,i
)
ϕ
d
ϕ
d
θ
=
sin
2
sin
3
θ
d
ϕ
d
ϕ
i
=
2
2
π
0
θ
π/
2
0
cos
2
d
ϕ
sin
ϕ
d
ϕ
i
=
3
2
3
=
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