484 
A “ smith’s PRIZE ” PAPER ; SOLUTIONS. 
[534 
+ 7 
3(i — g )2 [K 1 - a ) 2 sin2 + 2of - a2 P - (1 - a) 3 sin 3 fi] 
Let the position of an element of the solid in question be determined by r its 
distance from P, 0 the inclination of r to the axis PQ, and </> the azimuth in regard 
to any fixed plane through the axis; then dm = r 2 sin 6 dr dd d<f>, and the attraction 
in the direction PQ is = j sin 6 cos 6 dr dd d(f>, = 2ir — r ^ sin 0 cos 6 dd, the integral 
in regard to <£ having been taken from <£ = 0 to </> = 2-77, and that in regard to r 
from r = r (value at the face MQ of the lens) to r = (value at the tangent plane 
QN). Taking the radius of the surface QM of the lens to be = 1, we have 
that is, 
(1 — a + r cos 0) 2 + r 2 sin 2 0 = 1, 
r 2 q. 2r cos 0(1 — a) = 2a— a 2 , 
{r + (1 — a) cos 0} 2 = (1 — a) 2 cos 2 0 + 2a — a 2 , 
or 
r = — (1 — a) cos 0 + V{(1 — a) 2 cos 2 0 + 2a — a 2 }, 
which is the value of r to be substituted in the formula 
A = j(a sin 0 — r sin 0 cos 0) 00, 
2rr 
and the integral is to be taken from 0 = 0 to 0 = A; viz. this i 
is 
| [a sin 0 + (1 — a) sin 0 cos 2 0 — sin 0 cos 0 V{(1 — a) 2 cos 2 0 + 2a — a 2 }] dd, 
= - a cos 0 - £ (1 - a) cos 3 0 + ^ ^ {(1 - a) 2 cos 2 0 + 2a - a 2 }^; 
so that taking this between the limits in question, we have 
~ A = a (1 — cos A) + i (1 — a) (1 - cos 3 A) +1—[{(1 - a) 2 cos 2 A + 2a - a 2 } f - 1] 
or writing for greater convenience A = ^ — /¿, (p, = z PNQ), this is 
2^. A = a (1 — sin ft) + i (1 — «) (1 - sin 3 p) + 3 ^3 -, [{(1 - a) 2 sin 2 ^ + 2a - a 2 }^ - 1] 
1 
= a +i 1 -a 
(1 — a ) : 
r — a sm yu. 
+ 3 (1 - af [K 1 “ ^ sin2 ^ + 2a ~ « 2 } f - ~ a) 3 sin 3 p,] 
3 (1 - g)2 (“ 3 “ 2 + 2 “ 3 ) “ 01 Sil1