question_answer

The area bounded by the circle \[{{x}^{2}}+{{y}^{2}}=4,\] line \[x=\sqrt{3}y\] and \[x-\]axis lying in the first quadrant, is [RPET 1997; Kurukshetra CEE 1998]

A)            \[\frac{\pi }{2}\]                 

B)            \[\frac{\pi }{4}\]

C)            \[\frac{\pi }{3}\]                 

D)            \[\pi \]

Correct Answer: C

Solution :

           Required area \[=\int_{0}^{\sqrt{3}}{\frac{x}{\sqrt{3}}dx+\int_{\sqrt{3}}^{2}{\sqrt{4-{{x}^{2}}}dx}}\]                                    \[=\frac{1}{\sqrt{3}}\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{\sqrt{3}}+\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{\sqrt{3}}^{2}\]                                    \[=\frac{\sqrt{3}}{2}+\left[ \pi -\frac{\sqrt{3}}{2}-\frac{2\pi }{3} \right]=\frac{\pi }{3}\].                    Trick : Area of sector made by an arc = \[\frac{{{\theta }^{c}}{{R}^{2}}}{2}\]                                                                   \[=\frac{\pi }{6}.\frac{4}{2}=\frac{\pi }{3}\].