Find the area bounded by curves $(x-1{)}^2+y^2=1$ and $x^2+y^2=1$.

The area bounded by the curves, $(x-1{)}^2+y^2=1$ and $x^2+y^2=1$, is represented by the shaded area as
On solving the equations, $(x-1{)}^2+y^2=1$ and $x^2+y^2=1$, we obtain the point of intersection as $A$ $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ and $B\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$.
It can be observed that the required area is symmetrical about $x$-axis.
$\therefore$ Area $OBCAO$ $=2\times AreaOCAO$
We join $AB$, which intersects $OC$ at $M$, such that $AM$ is perpendicular to $OC$.
The coordinates of $M$ are $\left(\frac{1}{2},0\right)$.
$\Rightarrow AreaOCAO=AreaOMAO+AreaMCAM$
$=\left[{\int }_0^{\frac{1}{2}}\sqrt{1-(x-1{)}^2}dx+{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}dx\right]$
$={\left[\frac{x-1}{2}\sqrt{1-(x-1{)}^2}+\frac{1}{2}{\sin }^{-1}(x-1)\right]}_0^{\frac{1}{2}}+{\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]}_{\frac{1}{2}}^1$
$=\left[-\frac{1}{4}\sqrt{1-{\left(-\frac{1}{2}\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}-1\right)-\frac{1}{2}{\sin }^{-1}(-1)\right]+\left[\frac{1}{2}{\sin }^{-1}(1)-\frac{1}{4}\sqrt{1-{\left(\frac{1}{2}\right)}^2}-\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)\right]$
$=\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi }{6}\right)-\frac{1}{2}\left(-\frac{\pi }{2}\right)\right]+\left[\frac{1}{2}\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi }{6}\right)\right]$
$=\left[-\frac{\sqrt{3}}{4}-\frac{\pi }{12}+\frac{\pi }{4}+\frac{\pi }{4}-\frac{\pi }{12}\right]$
$=\left[-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}\right]$
$=\left[\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\right]$
Therefore, required area $OBCAO=2\times \left(\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\right)=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$ units