An introduction to the mathematical theory of attraction ...

158 Surfaces and Curves of the Second Degree.
Again, suppose each semi-axis of the ellipsoid E to receive a small increment,
then the thickness ap of the shell comprised between the ellipsoid thus generated
and E is given by the equation
pop = a2asa + /32bb + y2cc;
whence
aP x2 y2 z2
-a - + - ab + -.
p 3 b3 c3
If we now superpose the two shells, for the total thickness Sp we get
P =P 1a + (a b) xy - &c.;
hence pap can be identified with the given form for o.
The components of the force due to E at an internal point x, y, z are
- Ax, - By, - Cz, and those due to E' are - A (x + bx) + Bya, - Cz&p, &c.,
where bx = y83 - zp. i Hence the component X, parallel to the primary axis
of E, due to the attraction of the superposed shells, is given by the equation
da, db do
x = - (A - B) y + (A - ) - (. a + d b d+ ' ) ~ x,
and similar equations hold good for Y and Z. The quantities p5a, pSO, &c.
being already known in terms of the coefficients of (x, y, z), the forces X, Y, Z
are determined as linear functions of the coordinates.
10. If a concentric ellipsoidal cavity be cut out of a homogeneous sphere,
find the equipotential surfaces in the interior of the cavity.
The force at any point inside the cavity is the resultant of that due to an
attracting sphere and that due to a repelling ellipsoid of equal density. Hence
if X, Y, Z be the components of this force,
X = — 7rpx + Ax
and therefore by (22), Art. 24,
3
X= {A-(A +.B+ C)} x= (2A-B- C) -3
Y= (2B -C-A ) Y, Z= (2C-A- -B);
3 3
and the equipotential surfaces are given by the equation
(2A - B - C) x2 + &c. = constant.
11. If a homogeneous ellipsoid E be divided into two parts by any plane P,
show that the mutual action between the parts is reducible to a single force, and
find its amount.
If E1 and E2 denote the portions into which E is divided by P, the force and
couple produced by the attraction of E2 on Ei, are the same as those produced
by the attraction of E on E1, since the resultant force and couple due to the
attraction of E1 on itself are each zero. Let X, Y, Z be the components of the
resultant force, and L, 1M, N those of the resultant couple, then
X=-Af xdn, Y= -Bfydm, Z=- Czdm,
L = (B-C) S yzdm, M = J (C- A) zxdmn, N = (A - B) J xydm,