Explanation

(b)
$Required\ area = 2\int_{1}^{3}{\sqrt{9 - x^{2}\ }dx}$
$= 2 \bullet \frac{1}{2}\left\lbrack x\sqrt{9 - x^{2}} + 9\sin^{- 1}\frac{x}{3} \right\rbrack_{1}^{3}$
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$= \left\lbrack 9\ \sin^{- 1}(1) - \sqrt{8} - 9\ \sin^{- 1}\left( \frac{1}{3} \right) \right\rbrack$
$= \left\lbrack 9\left\{ \cos^{- 1}\left( \frac{1}{3} \right) \right\} - \sqrt{8} \right\rbrack\left\lbrack \because\ \cos^{- 1}\theta = \frac{\pi}{2} - \sin^{- 1}\theta \right\rbrack$
$= \left\lbrack 9\sec^{- 1}(3) - \sqrt{8} \right\rbrack sq\ unit$