An introduction to the mathematical theory of attraction ...

Gauss' Theorem. 45
for every element of mass outside S, and therefore the part of
the integral J N'dS due to mass outside S is zero.
If O be inside 8, any radius vector r from it must pass
once out of the surface S without a corresponding entrance,
and if r enters S there must for each entrance be a corresponding exit; hence the whole contribution to the surface
integral J NdS due to the cone having r for axis is midw. In
order to get the whole portion of
the surface integral due to ~m, we
must integrate dw all round 0. 9 7
In this manner we obtain 47rm.
A similar result holds good for
every other element of mass inside S. If O be on the surface S, all radii vectores from it
on one side of the tangent plane pass out of S once without
any corresponding entrance. Any radius vector from O on
the other side of the tangent plane either does not meet S at
all or else accomplishes as many exits as entrances. Hence
in this case the entire contribution of the mass rn at 0 to the
surface integral is n dw taken over a hemisphere, that is
27nm.
Finally, if we add together all the parts of the integral
due to the various elements of mass, we obtain
f NdS = 47r + 27rM'. (2)
The number of entrances and exits of any one radius
vector which meets the surface S depends on the form of this
surface.
If the closed surface S contain two adjacent regions
separated by a single sheet, this sheet must be counted twice
over, once as a boundary
to each region, or, which
comes to the same thing,
not counted at all. As
an example of a surface
such ashas been described, o
we may take a sphere and ~
the portion of another
sphere terminated by the
curve of intersection of the two. In this case as r passes