question_answer

The area of the portion of the circle \[{{x}^{2}}+{{y}^{2}}=1,\] which lies inside the parabola\[{{y}^{2}}=1-x\], is-

A) \[\frac{\pi }{2}-\frac{2}{3}\]      

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

C) \[\frac{\pi }{2}+\frac{4}{3}\]    

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

Correct Answer: C

Solution :

[c] Area = semicircle + 2(area of arc OAB) \[=\frac{\pi \times {{1}^{2}}}{1}+2\int\limits_{0}^{1}{\sqrt{1-x}dx=\frac{\pi }{2}-2\left( \frac{{{(1-x)}^{3/2}}}{3/2} \right)_{0}^{1}}\] \[=\frac{\pi }{4}+\frac{4}{3}\]