question_answer

The area (in sq. units) of the region \[\{(x,y):{{y}^{2}}\ge 2x\]and\[{{x}^{2}}+{{y}^{2}}\le 4x,x\ge 0,y\ge 0\]\[\}\] is :- [JEE Main Solved Paper-2016 ]

A) \[\frac{\pi }{2}=\frac{2\sqrt{2}}{3}\]                       

B) \[\pi -\frac{4}{3}\]

C) \[\pi -\frac{4}{3}\]                          

D) \[\pi -\frac{4\sqrt{2}}{3}\]

Correct Answer: C

Solution :

\[=\pi \frac{\pi {{(2)}^{2}}}{4}-\sqrt{2}\int\limits_{0}^{2}{\sqrt{x}}dx\] \[=\pi -\sqrt{2}.\frac{2}{3}2\sqrt{2}\]      \[=\pi -8/3\]