Explanation

(b)
The equation of circle is $x^{2} + y^{2} = 9$
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$\therefore$ Area of the smaller segment cut off from the circle
$x^{2} + y^{2} = 9\ by\ x = 1$, given by
$\therefore$ Required area, $A = 2\int_{1}^{3}{\sqrt{9 - x^{2}}dx}$
$= 2 \bullet \frac{1}{2}\left\lbrack x\sqrt{9 - x^{2}} + 9\sin^{- 1}\frac{x}{3}\ \right\rbrack_{1}^{3}$
$= \left\lbrack \frac{9\pi}{2} - \sqrt{8} - 9\sin^{- 1}\left( \frac{1}{3} \right) \right\rbrack$
$= \left\lbrack 9\left( \frac{\pi}{2} - \sin^{- 1}\left( \frac{1}{3} \right) \right) - \sqrt{8} \right\rbrack$
$= \left\lbrack 9\cos^{- 1}{\left( \frac{1}{3} \right) - \sqrt{8}} \right\rbrack$
$= \lbrack 9\sec^{- 1}{(3) - \sqrt{8}\rbrack}\ sq\ unit$