Question

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

Answer 1

Given, curve $(x-1{)}^2+y^2=1$ ...(i)

Which represents a circle with centre (1, 0) and radius 1

and curve $x^2+y^2=1$ ...(ii)

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

Which represents a circle with centre (0, 0) and radius 1.

Both the c urves are circle and meet where $(x-1{)}^2=x^2$ i.e., where 2x = 1 or $x=\frac{1}{2}$

Required area (shown in shaded region).

$=2[{\int }_0^{\frac{1}{2}}{y}_{1}dx+{\int }_{\frac{1}{2}}^1{y}_{2}dx]=2\left\{{\int }_0^{\frac{1}{2}}\right\}\sqrt{1-(x-1{)}^2}dx+{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}dx]=2[\frac{x-1}{2}\sqrt{1-(x-1{)}^2}+\frac{1}{2}si{n}^{-1}\frac{x-1}{1}{]}_0^{\frac{1}{2}}+2[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}si{n}^{-1}x{]}_{\frac{1}{2}}^1=2[\frac{\frac{1}{2}-1}{2}\sqrt{1-\frac{1}{4}}+\frac{1}{2}si{n}^{-1}(-\frac{1}{2})-\left(\frac{-1}{2}\right)0-\frac{1}{2}si{n}^{-1}(-1)]+2[0+\frac{1}{2}si{n}^{-1}\left(1\right)-\frac{1}{4}\sqrt{1-\frac{1}{4}}-\frac{1}{2}si{n}^{-1}\frac{1}{2}]=2[-\frac{1}{4}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{\pi }{6}+0+\frac{1}{2}.\frac{\pi }{2}]+\frac{\pi }{2}-\frac{1}{2}.\frac{\sqrt{3}}{2}-\frac{\pi }{6}=-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}+\frac{\pi }{2}-\frac{\sqrt{3}}{4}-\frac{\pi }{6}=(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})squnit$