268 
ON THE POTENTIALS OF POLYGONS AND POLYHEDRA. 
[602 
Cone on a plane base, and plane figure. 
6. Suppose that the surface 2 is a plane; the surface 2' is, of course, a parallel 
plane. Taking here mp for the perpendicular distance of the plane 2 from the origin, 
then, if 8 be the infinitesimal distance of the two planes from each other, we have 
8 c 
8 - p dm, that is, dm - -; the potential of the cone is, as before, = \m 2 I r 2 dco, and 
that of the plane figure, thickness 8, is =— J r 2 dco. 
7. Taking, for greater convenience, m= 1, we have 
Potential of cone 
-*/ 
8 f 
Do. of plane figure =-J 
r 2 dco, 
r 2 dco, 
where p is now the perpendicular distance of the plane from the vertex; or if, as 
regards the plane figure, the infinitesimal thickness 8 is taken as unity, then 
Potential of plane figure = -jr 2 dco. 
In each case r is the value of the radius vector corresponding to a point of the plane 
figure which is the base of the cone, and the integration extends over the spherical 
aperture of the cone. 
8. If the position of the radius vector is determined by the usual angular 
coordinates, 6 its inclination to the axis of z, and </> its azimuth from the plane of 
zx—viz. if we have 
x = r sin 0 COS cf), 
y = r sin 6 sin cf), 
z = r cos 6; 
then, as is well-known, dco = sin 6 dd d<f>, and the integral J r 2 dco is =J r 2 sin 0 dd dcp. 
Taking the inclination of p to the axes to be a, /3, 7 respectively, the equation 
of the plane which is the base of the cone is 
x cos a + y cos /3 + z cos 7 = p; 
viz. we have 
r [(cos a cos cf> + cos /3 sin <£) sin 0 + cos 7 cos 0] = p; 
that is, 
(cos a. cos cj) + cos /3 sin cf>) sin 0 + cos 7 cos 0 ’