Given circles are $x^{2}+y^{2}=1 \ldots$ (i)
and $(x-1)^{2}+y^{2}=1 \ldots$ (ii)
Centre of (i) is $O (0,0)$ and radius $=1$
Both these circle are symmetrical about $x$-axis
solving (i) and (ii), we get,
$-2 x+1=0$
$\Rightarrow x=\frac{1}{2}$
then $y^{2}=1-\left(\frac{1}{2}\right)^{2}=34$
$\Rightarrow y=\frac{\sqrt{3}}{2}$
$\therefore $ The points of intersection are
$P\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $Q\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
It is clear from the figure that the shaded portion in region whose area is required.
$\therefore $ Required area = area $ O Q A P O$
$=2 \times $ area of the region $ O L A P$
$=2 \times($ area of the region $O L P O+$ area of $ L A P L)$
$=2\left[\int\limits_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} d x+\int\limits_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} dx \right] $
$=2\left[\frac{(x-1) \sqrt{1-(x-1)^{2}}}{2}+\frac{1}{2} \sin ^{-1}(x-1)\right]_{0}^{\frac{1}{2}}+2\left[\frac{x \sqrt{1-x^{2}}}{2}+\frac{1}{2} \sin ^{-1} x\right]_{\frac{1}{2}}^{1}$
$=-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{-1}{2}\right)-\sin ^{-1}(-1)+0+\sin ^{-1}(1)-\left(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{1}{2}\right)\right)$
$=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$ sq.units .
