EXAMPLE IX. Let it be proposed to find the Area of a Semicircle
AREH.
Here, putting the Diameter AH $=a$, AB $=x$, and BR $=y~$ &c.
(as usual) we have $y^2$ (BR$^2$) $=axxx$ (AB$\times$BH),
and consequently $\dot{u}$ ($y\dot{x}$) $=\dot{x}\sqrt{axxx}$
$=\displaystyle{ a^{\frac{1}{2}} x^{\frac{1}{2}} \dot{x}\times
\overline{1\frac{x}{a}}^{\frac{1}{2}}}:$ Which Expression not
being of the Kind [described previously], that admit of Fluents in
finite Terms, let it be therefore resolved into an Infinite Series
and you will have
$\displaystyle{\dot{u} = a^{\frac{1}{2}} x^{\frac{1}{2}} \dot{x}\times
\overline{1\frac{x}{2a}\frac{x^2}{8a^2}\frac{x^3}{16a^3}
\frac{5x^4}{128a^4}}}~$ &c.
$\displaystyle{
=a^{\frac{1}{2}}\times \overline{
x^{\frac{1}{2}}\dot{x}
\frac{x^{\frac{3}{2}}}{2a}\dot{x}
\frac{x^{\frac{5}{2}}}{8a^2}\dot{x}
\frac{x^{\frac{7}{2}}}{16a^3}\dot{x}}}~$
&c.
From whence, the Fluent of every Term being taken, according to
the common Method, there will come out
$u =\displaystyle{=a^{\frac{1}{2}}\times \overline{ \frac{2x^{\frac{3}{2}}}{3}
\frac{x^{\frac{5}{2}}}{5a} \frac{x^{\frac{7}{2}}}{28a^2}
\frac{x^{\frac{9}{2}}}{72a^3} \frac{5x^{\frac{11}{2}}}{704a^4}}}~$&c.
$=\displaystyle{x\sqrt{ax}\times
\overline{ \frac{2}{3}
\frac{x}{5a} \frac{x^2}{28a^2}
\frac{x^3}{72a^3} \frac{5x^4}{704a^4}}}~$&c.
= the Area ABR. Now, when $x=\frac{1}{2}a$, the Ordinate BR
will coincide with the Radius OE; in which Case the Area becomes
$=\frac{1}{2}a\sqrt{\frac{1}{2}aa}\times
\overline{ \frac{2}{3}
\frac{1}{10} \frac{1}{112}
\frac{1}{576} \frac{5}{11264}
}~$&c.
$=\displaystyle{\frac{a^2\sqrt{\frac{1}{2}}}{2}\times
\overline{0,66660,10,00890,00170,0004
}}~$&c. $=0,1964a^2$; which, when
multiplied by 2, gives $0,3928a^2$ for the Area
of the Semicircle AEH, nearly.
