304 Mr. Knight on the Attraction of such Solids 
he taken from 0 to — x, instead of from 0 to x. A being a 
function of r, x, and u, may be represented by 2 <p ( r,x,u ). To 
correct the fluent, let sm (fig. 23) = X, sm' = x, then, the 
attraction of the solid, whose base is the quadrilateral figure 
prr'd, will be 2<p (r, x, u ) — 29 (r, X, u ). 
In figure 24, call ps, u ; sm, X ; sm', x. The action of the 
solid, whose base is prr'p', is expressed by 2$ (r,— x, u) — 2© 
(r,_ X, u). 
In fig. 25, put ps = u , ps' = id, sm = s'm = x, tang, of 
rsm = r: the attraction of the solid, whose base is the rhom- 
bus srs'p, on a point p in the produced diameter of the section, 
is 2 q> ( [r , x,u) — 2<p ( r,o,u ) -{- 2<p (r, — x, id) — 2<p (r, — 0, id). 
Prop. C. 
Let fig. 2 6 represent the base or section of an infinitely 
long prism, and let this base be any right lined figure what- 
ever, regular or irregular : from p, a point in the same plane, 
draw r any line pq, cutting the base at s and m'". It is required 
to find the action of the solid on the point p, in the direc- 
tion pq. 
From the angles r, r', r", r'", &c. of the base, let fall the 
perpendiculars rm, r'm', r"m ,r , r /// m ,// , &c. on the line pq. 
Prolong the sides of the polygon till they meet pq at the 
points s, s', s", s'", &c. 
Put u — ps, id = ps', u" = ps", id" = ps'", &c. ; and 
x' = s'm' | J x" = s"m" 1 f x'" = s"'m" | ^ 
X' = s'm r L X y ' = s"m' ] ’ | X'" = s^m"' J ’ CCs 
Also, let r = tang, rsm, r'= tang, r's'm', r" = tang. r'V'm", 
r"'= tang. &c. then it appears from the last pro- 
position, that the attraction, of the upper half of the solid, is 
