Question

Find the area bounded by curves(x-1)²+y²=1 and x²+y²=1

Answer 1

The area bounded by the curves,(x-1)²+y²=1 and x²+y²=1,is represented by the shaded area as

On solving the equations,(x-1)²+y²=1 and x²+y²=1,we obtain the point of

intersection as\(A(\frac{1}{2},\frac{\sqrt3}{2})\)and B\((\frac{1}{2},-\frac{\sqrt3}{2})\)

It can be observed that the required area is symmetrical about x-axis.

∴Area OBCAO=2×Area OCAO

We join AB,which intersect OC at M,such that AM is perpendicular to OC.

The coordinates of M are\((\frac{1}{2},0).\)

\(⇒Area\, OCAO=Area\, OMAO+Area\, MCAM\)

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

\(=\bigg[\frac{x-1}{2}\sqrt{1-(x-1)^2}+\frac{1}{2}sin^{-1}(x-1)\bigg]_0^\frac{1}{2}+\bigg[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}sin^{-1}x\bigg]^1_\frac{1}{2}\)

\(=[-\frac{1}{4}\sqrt{1-(-\frac{1}{2})^2}+\frac{1}{2}sin^{-1}(\frac{1}{2}-1)-\frac{1}{2}sin^{-1}(-1)]\)+\([\frac{1}{2}sin^{-1}(1)-\frac{1}{4}\sqrt{1-(\frac{1}{2})^2}-\frac{1}{2}sin^{-1}(\frac{1}{2})]\)

\(=[-\frac{\sqrt3}{8}+\frac{1}{2}(-\frac{π}{6})-\frac{1}{2}(-\frac{π}{2})]+[\frac{1}{2}(\frac{π}{2})-\frac{\sqrt3}{8}-\frac{1}{2}(\frac{π}{6})]\)

\(=[-\frac{\sqrt3}{4}-\frac{π}{12}+\frac{π}{4}+\frac{π}{4}-\frac{π}{12}]\)

\(=[-\frac{\sqrt3}{4}-\frac{π}{6}+\frac{π}{2}]\)

\(=[\frac{2π}{6}-\frac{\sqrt3}{4}]\)

Therefore,required area OBCAO\(=2\times[\frac{2π}{6}-\frac{\sqrt3}{4}]\)=\(=[\frac{2π}{3}-\frac{\sqrt3}{2}]\) units.