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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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