question_answer

The area of the region bounded by \[{{x}^{2}}+{{y}^{2}}-2x-3=0\] and \[y=\left| x \right|+1\]is

A) \[\frac{\pi }{2}-1\] sq. units

B) \[2\pi \]sq. units

C) \[4\pi \]sq. units 

D) \[\pi /2\]sq. units

Correct Answer: A

Solution :

[a] \[{{x}^{2}}+{{y}^{2}}-2x-3=0\] or \[{{(x-1)}^{2}}+{{y}^{2}}=4\] \[A=\int\limits_{1-\sqrt{2}}^{0}{(\sqrt{4-{{(x-1)}^{2}}}-(-x+1))\,}dx\] \[+\int\limits_{0}^{1}{(\sqrt{4-{{(x-1)}^{2}}}-(x+1))}dx\]

\[=\frac{x-1}{2}\sqrt{4-{{(x-1)}^{2}}}\left. +\frac{4}{2}{{\sin }^{-1}}\frac{x-1}{2}+\frac{{{x}^{2}}}{2}-x \right|_{1-\sqrt{2}}^{0}\] \[+\frac{x-1}{2}\sqrt{4-{{(x-1)}^{2}}}+\left. \frac{4}{2}{{\sin }^{-1}}\frac{x-1}{2}-\frac{{{x}^{2}}}{2}-x \right|\]

\[=\left( -\frac{\sqrt{3}}{2}-\frac{\pi }{3} \right)-\left( \frac{-\sqrt{2}}{2}\sqrt{2}-\frac{\pi }{2}+\frac{3-2\sqrt{2}}{2}-1+\sqrt{2} \right)\] \[+\left( -\frac{1}{2}-1 \right)-\left( -\frac{\sqrt{3}}{2}-\frac{\pi }{3} \right)\]

\[=-\left( -1-\frac{\pi }{2}+\frac{3}{2}-\sqrt{2}-1+\sqrt{2} \right)-\frac{3}{2}\]

\[=\frac{\pi }{2}-1\,\,sq.units.\]