Given equation of the circle is $x^2+ y^2 = 9$.
$\therefore $ Area of the smaller segment cut off from the circle
$x^2 + y^2 = 9$ by $x = 1$, is given by
$A=2\int\limits_{1}^{3}\,\sqrt{9-x^{2}}dx$
$=2\cdot\frac{1}{2}\left[x \sqrt{9-x^{2}}+9\,\sin^{-1} \frac{x}{3}\right]^{3}_{1}$
$=\left[3\cdot\sqrt{9-9}+9\,sin^{-1}\left(\frac{3}{3}\right)-1\cdot\sqrt{9-1}-9\,\sin^{-1}\left(\frac{1}{3}\right)\right]$
$=\left[9\,\sin^{-1}\left(\sin \frac{\pi}{2}\right)-\sqrt{8}-9\,\sin^{-1}\left(\frac{1}{3}\right)\right]$
$=\left[9\left(\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{3}\right)\right)-\sqrt{8}\right]$
$=\left[9\left(\cos^{-1}\left(\frac{1}{3}\right)-\sqrt{8}\right)\right]$
$=\left[9\,\sec^{-1}\,\left(3\right)-\sqrt{8}\right]$ sq unit
