question_answer

The area (in sq. units) of the smaller portion enclosed between the curves, \[{{x}^{2}}+{{y}^{2}}=4\]and\[{{y}^{2}}=3x,\]is: [JEE Online 08-04-2017]

A) \[\frac{1}{\sqrt{3}}+\frac{4\pi }{3}\]                       

B) \[\frac{1}{\sqrt{3}}+\frac{2\pi }{3}\]

C) \[\frac{1}{2\sqrt{3}}-\frac{\pi }{3}\]                        

D) \[\frac{1}{2\sqrt{3}}+\frac{2\pi }{3}\]

Correct Answer: A

Solution :

                \[{{x}^{2}}+3x-4=0\]                 \[(x+4)(x-1)=0\]                 \[x=-4,x=1\] Area\[=\left( \int\limits_{0}^{1}{\sqrt{3}.\sqrt{x}dx} \right)+\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\sqrt{3}.\sqrt{x}dx}+\int\limits_{1}^{2}{\sqrt{4-{{x}^{2}}}dx} \right)\times 2}\]\[=\left( \sqrt{3}\left( \frac{{{x}^{3/2}}}{3/2} \right)_{0}^{1}+\left( \frac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\frac{x}{2} \right)_{1}^{2} \right)\times 2\] \[=\left( \sqrt{3}\left( \frac{2}{3} \right)+\left\{ 2.\frac{\pi }{2}-\frac{\sqrt{3}}{2}+\frac{\pi }{3} \right\} \right)\times 2\] \[\left( \frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\frac{2\pi }{3} \right)\times 2\]\[=\left( \frac{2}{2\sqrt{3}}+\frac{2\pi }{3} \right)\times 2=\frac{1}{\sqrt{3}}+\frac{4\pi }{3}\]d