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FIGURE 9.2
Hemispherical surface with radius
.
Thus moving
to the boundary leads to no contribution from the second integral on the
right-hand side of Eq. (9.23).
In the case of the third integral of Eq. (9.23),
ξ
uq
∗
dS
uq
∗
dS
uq
∗
dS
=
lim
→
+
lim
→
(9.26)
0
0
S
S
−
S
S
and
uq
∗
dS
uq
∗
dS
lim
→
=
0
S
−
S
S
r
r
. Then, with
r
i
,q
∗
can be written
For a hemispherical surface of Fig. 9.2,
a
i
=
=
,r
i
=
as
1
u
∗
∂
1
2
3
q
∗
=
∂
n
=
∂
=−
i
a
i
3
=−
3
=−
+
+
i
i
∂
n
4
From the expressions for
1
,
2
,
and
3
of Fig. 9.2, it can be seen that
2
1
2
2
2
3
2
+
+
=
Thus, for a hemispherical surface,
u
∗
∂
q
∗
=
∂
1
n
=−
2
Then the second integral on the right-hand side of Eq. (9.26) becomes
dS
dS
u
1
1
uq
∗
dS
=
−
=
−
lim
→
lim
→
u
lim
→
2
2
0
0
0
S
S
S
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