2 66 Mr. Knight on the Attraction of such Solids 
C == jcip ( 1 , m, r) + i' j L . ( l/ i -j- m ' 1 -f- m ') ~ L . 
A/ i + m‘ z + + »i' 7 
Vi-j-r 1 ^ * 
These are, perhaps, as simple expressions as can be had in 
this case ; we may however find others ; nor is it necessary 
to conceive the solid divided into pyramids. 
Thus we may obtain the value of the force A, by the me- 
thod of the preceding propositions : a triangular section of the 
prism, parallel to the end rpv', and at a distance x from that end, 
will have its action on p expressed by <p (x, x ', r) — arc (tang. 
= y), using the same notation as before ; therefore, the action 
of the pyramid, in the direction pm, is 
A = fxcp ( x , x', r) — x arc (tang. = y) = F ( x' } r, x) — x arc 
(tang. = yj -f- con% 
Another Way of finding the force B. 
Draw ks parallel to rV; call pk, x'\ then ks = r x x\ Con- 
ceive a plane to pass through ks, parallel to the back rvv , r / of 
the prism ; the section made by it will be a rectangle whose 
sides are x and rx'. The action of this rectangle on p, in the 
direction pr', is arc (tang. 
VJC \ 
===,) by Prop. 4 , where- 
V' x tz + r 
fore the action of the prism, in the same direction, is 
B = fx' arc (tang. = ■ -===^ ^======.) where x' is the variable 
J & Vx z +(i + r z )x' z t 
quantity, we have then 
B = x 1 arc (tang. — — 
rx 
— fx' arc (tang. 
rx 
V X 1 -J- ( I -J- 7 
=). 
l )x z ‘ 
Vx z -{- (i -j- r 1 ) x lz 
