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Mathematics Question on applications of integrals

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

Answer

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

Area bounded by curvesx-12+y2=1 and x2+y2=1

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

intersection asA(12,32)A(\frac{1}{2},\frac{\sqrt3}{2})and B(12,−32)(\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(12,0).(\frac{1}{2},0).

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

=[∫0121−(x−1)2dx+∫1211−x2dx]=[∫_0^\frac{1}{2}\sqrt{1-(x-1)^2}dx+∫^1_\frac{1}{2}\sqrt{1-x^2}dx]

=[x−121−(x−1)2+12sin−1(x−1)]012+[x21−x2+12sin−1x]121=\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}

=[−141−(−12)2+12sin−1(12−1)−12sin−1(−1)]=[-\frac{1}{4}\sqrt{1-(-\frac{1}{2})^2}+\frac{1}{2}sin^{-1}(\frac{1}{2}-1)-\frac{1}{2}sin^{-1}(-1)]+[12sin−1(1)−141−(12)2−12sin−1(12)][\frac{1}{2}sin^{-1}(1)-\frac{1}{4}\sqrt{1-(\frac{1}{2})^2}-\frac{1}{2}sin^{-1}(\frac{1}{2})]

=[−38+12(−π6)−12(−π2)]+[12(π2)−38−12(π6)]=[-\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})]

=[−34−π12+π4+π4−π12]=[-\frac{\sqrt3}{4}-\frac{π}{12}+\frac{π}{4}+\frac{π}{4}-\frac{π}{12}]

=[−34−π6+π2]=[-\frac{\sqrt3}{4}-\frac{π}{6}+\frac{π}{2}]

=[2π6−34]=[\frac{2π}{6}-\frac{\sqrt3}{4}]

Therefore,required area OBCAO=2×[2π6−34]=2\times[\frac{2π}{6}-\frac{\sqrt3}{4}]==[2π3−32]=[\frac{2π}{3}-\frac{\sqrt3}{2}] units.