Mr. Bootes on a certain Multiple Definite Integral. 147 
Assuming in (12) n = 3, 71 = 3, we have 
to) 
Be dsf(c) ' : 
= — 2h h. hn [GER +h GER (23) 
In this expression a is only found in the function o, vide (16); hence 
dy do dy Qa dv 
a dads — s+h?do 
a > 
,dv_—s- 4h, h hyxa ae (24) 
SS 3 s(s--h2)[(s-Eh,) (8h, (s-Fh2)]* 
Suppose, first, the attracted point external, and the density uniform and equal 
to unity; then, as before, determining A by the equation 
a & Ol 
A+ h? Ath? © A$h2” ” 
we see, that in integrating relatively to s, we must, when s is less than A, regard 
- + (25) 
ed : 
J (c) as 0, and therefore epee) as Oalso; that when sis greater than A, f(c) =1, 
o 
and = J(c) is again equal to 0. It appears, therefore, that ae f (ce) vanishes be- 
Lo o 
fore and after the break in the discontinuous function f(o). To find its value at 
the break we must proceed thus. Since o is a function of s, we have 
$ f(=F + Fe) 
ds d 
SG) = Siem 
Now £ F(o) ds is the differential of f(c) relative to s, i.e. the difference of the 
values orf «) preceding and succeeding the break, and this difference is unity. 
With the value of f(c) at the break we have nothing to do, as it does not extend 
to any differential element of s. pei) at the break, 
ys 
