Question

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

Answer 1

The Equations represent the circles with Center at $(0,0),(1,0)$ Radius $1$ unit

Point of intersection $:(\frac{1}{2},\frac{\sqrt{3}}{2}),$$(\frac{1}{2},\frac{-\sqrt{3}}{2})$

The region is symmetric about $\text{x-axis}$

$(x-1{)}^2+y^2=1\Rightarrow y=\sqrt{1-(x-1{)}^2}$

$x^2+y^2=1\Rightarrow y=\sqrt{1-x^2}$

So, the area bounded is given by

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

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

$\Rightarrow \frac{\sqrt{3}}{4}-\frac{5\pi }{6}+\frac{3\pi }{2}-\frac{\pi }{2}+\frac{\sqrt{3}}{4}+\frac{\pi }{6}=\frac{\sqrt{3}}{2}+\frac{\pi }{3}$