integraldisplay
r
′′
0
r dr
parenrightBigg
d
Ω =
−
1
2
G
ρ
(
r
′
2
+
r
′′
2
)
d
Ω
.
(E.3)
The net potential at
P
is obtained by integrating over all solid angle, and dividing the result by two to adjust for double
counting. This yields
Ψ
=
−
1
4
G
ρ
contintegraldisplay
(
r
′
2
+
r
′′
2
)
d
Ω
.
(E.4)
From Figure E.1, the position vector of point
Q
, relative to the origin,
O
, is
x
′
=
x
+
r
′
n
,
(E.5)
where
x
=
(
x
1
,
x
2
,
x
3
) is the position vector of point
P
, and
n
a unit vector pointing from
P
to
Q
. Likewise, the
position vector of point
R
is
x
′′
=
x
+
r
′′
n
.
(E.6)
However,
Q
and
R
both lie on the body’s outer boundary. It follows, from (E.1), that
r
′
and
r
′′
are the two roots of
summationdisplay
i
=
1
,
3
parenleftBigg
x
i
+
r n
i
a
i
parenrightBigg
2
=
1
,
(E.7)
which reduces to the quadratic
A r
2
+
B r
+
C
=
0
,
(E.8)
where
A
=
summationdisplay
i
=
1
,
3
n
2
i
a
2
i
,
(E.9)
B
=
2
summationdisplay
i
=
1
,
3
x
i
n
i
a
2
i
,
(E.10)
C
=
summationdisplay
i
=
1
,
3
x
2
i
a
2
i
−
1
.
(E.11)

280
FLUID MECHANICS
P
O
x
r
Q
R
d
Ω
Figure E.1:
Calculation of ellipsoidal gravitational potential.
Now, according to standard polynomial equation theory,
r
′
+
r
′′
=
−
B
/
A
, and
r
′
r
′′
=
C
/
A
. Thus,
r
′
2
+
r
′′
2
=
(
r
′
+
r
′′
)
2
−
2
r
′
r
′′
=
B
2
A
2
−
2
C
A
,
(E.12)
and (E.4) becomes
Ψ
=
−
1
2
G
ρ
contintegraldisplay
2
parenleftBig
∑
i
=
1
,
3
x
i
n
i
/
a
2
i
parenrightBig
2
parenleftBig
∑
i
=
1
,
3
n
2
i
/
a
2
i
parenrightBig
2
+
1
−
∑
i
=
1
,
3
x
2
i
/
a
2
i
∑
i
=
1
,
3
n
2
i
/
a
2
i
d
Ω
.
(E.13)
The above expression can also be written
Ψ
=
−
1
2
G
ρ
contintegraldisplay
2
∑
i
,
j
=
1
,
3
x
i
x
j
n
i
n
j
/
(
a
2
i
a
2
j
)
parenleftBig
∑
i
=
1
,
3
n
2
i
/
a
2
i
parenrightBig
2
+
1
−
∑
i
=
1
,
3
x
2
i
/
a
2
i
∑
i
=
1
,
3
n
2
i
/
a
2
i
d
Ω
.
(E.14)
However, the cross terms (
i.e.
,
i
nequal
j
) integrate to zero by symmetry, and we are left with
Ψ
=
−
1
2
G
ρ
contintegraldisplay
2
∑
i
=
1
,
3
x
2
i
n
2
i
/
a
4
i
parenleftBig
∑
i
=
1
,
3
n
2
i
/
a
2
i
parenrightBig
2
+
1
−
∑
i
=
1
,
3
x
2
i
/
a
2
i
∑
i
=
1
,
3
n
2
i
/
a
2
i
d
Ω
.
(E.15)
Let
J
=
contintegraldisplay
d
Ω
∑
i
=
1
,
3
n
2
i
/
a
2
i
.
(E.16)
It follows that
1
a
i
∂
J
∂
a
i
=
contintegraldisplay
2
n
2
i
/
a
4
i
parenleftBig
∑
i
=
1
,
3
n
2
i
/
a
2
i
parenrightBig
2
d
Ω
.
(E.17)
Thus, (E.15) can be written
Ψ
=
−
1
2
G
ρ
J
−
summationdisplay
i
=
1
,
3
A
i
x
2
i
,
(E.18)
where
A
i
=
J
a
2
i
−
1
a
i
∂
J
∂
a
i
.
(E.19)

Ellipsoidal Potential Theory
281
At this stage, it is convenient to adopt the spherical angular coordinates,
θ
and
φ
(see Section C.4), in terms of
which
n
=
(sin
θ
cos
φ,
sin
θ
sin
φ,
cos
θ
)
,
(E.20)
and
d
Ω =
sin
θ
d
θ
d
φ
. We find, from (E.16), that
J
=
8
integraldisplay
π/
2
0
sin
θ
d
θ
integraldisplay
π/
2
0
sin
2
θ
cos
2
φ
a
2
1
+
sin
2
θ
sin
2
φ
a
2
2
+
cos
2
θ
a
2
3
−
1
d
φ.
(E.21)
Let
t
=
tan
φ
. It follows that
J
=
8
integraldisplay
π/
2
0

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- Fluid Dynamics, Fluid Mechanics, stress tensor, Fluid Motion