Question

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

Answer 1

$x^2+y^2=1$ $(x-1{)}^2+(y-0{)}^2=1$

centre $(0,0)$ centre $(1,0)$

radius =$1$ radius=$1$

$x2+y^2=1\Rightarrow y^2=1-x^2$

$\therefore (x-1{)}^2+(1-x^2)=1$

$\Rightarrow x^2+1-2x+1-X^2=1$

$x=\frac{1}{2}$

If $x=\frac{1}{2}$,then $y=\pm \frac{\sqrt{3}}{2}$

Hence points of intersection are $(\frac{1}{2},\frac{\sqrt{3}}{2}),(\frac{1}{2},\frac{-\sqrt{3}}{2})$

The curve $y=\sqrt{1-(x-1{)}^2}$ moves from $0$ to $\frac{1}{2}$

and $y=\sqrt{1-x^2}$ moves from $\frac{1}{2}$ to $1$ $(x-0.005)$

Area $=[2{\int }_0^{\frac{1}{2}}\sqrt{1-(x-1{)}^2}dx+{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}dx]$

$2\left[\frac{1}{2}\right(x-1)\sqrt{1-(x-1{)}^2}+\frac{1}{2}{\sin }^{-1}(\frac{x-1}{1}){]}_0^{\frac{1}{2}}$

$+[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}{\sin }^{-1}(x){]}_{\frac{1}{2}}^1$

On applying limits , we get

$Area=[\frac{-\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}+\frac{\pi }{2}-\frac{\sqrt{3}}{4}-\frac{\pi }{6}]=[\frac{2\pi }{3}-\frac{\sqrt{3}}{2}]$