156' Mr. Woodhouse's Demonstration of a Theorem 
f, %' xy is transformed into 
and, since %' == V r * — x 
r 
y' = V 
£££!£=££ ; 
3 /“• 
e'; fence * = =*=//? giXr'- f ==/(- 
Let the integral commence from the circumference of the 
circle in the plane of which x andy are situated; then f = g = r, 
* 3. 
0, in which expression > r » 0 is the ele- 
(j r -e 
and s 
^ 3 * 3 
ment of the solidity insisting on that part of the plane of x which 
is included between the circumference of the circle whose ra- 
dius is r , and the curve whose nature is determined by the 
relation existing between x and y, or f and 9. 
In the present problem, this latter curve is given to be a semi- 
circle described on rasa diameter; hence, as ^ is a line drawn 
from an extremity of this diameter to the circumference, ma- 
king an angle 9 with the diameter, we have, by similar figures, 
tang. 6 (0 = but 6 = ~ 
* Hence, 5 = -/ * 8 = =fc fV=£i j = Cl _ iL + C 
when ^ — o = — , the value of the portion 
* s — —— 3 r * g * g3 
9 9 7 * 9 
of the solid insisting on a base in the plane of x and y, and the 
bounding lines of which are a quadrantal arc, (rad. r) a semi- 
circle, (diam. r ) and a radius (r) perpendicular to the diameter 
• If s is expressed in terms of t and i, the integration Is less simple ; for 
i = . — II til -llxPt (i + P)~ * = — ± fluxion (f 1 (i + f 1 ) - *) 
V 3 9 * ' 
a (*+<•) 
2 f 1 . 3 7*3 f* 2 r* * . 2 f s 
+ T lt ( l + t% ) * s = — T 7 “ ~ X 7 + correction = — , 
9 9 9 (*+»■)* 9 
when the integral is taken from t to t — oo. 
