Question
The area of the region described by $$A = \left\{ {\left( {x,\,y} \right):{x^2} + {y^2} \leqslant 1\,{\text{and}}\,{y^2} \leqslant 1 - x} \right\}$$ is:
A.
$$\frac{\pi }{2} - \frac{2}{3}$$
B.
$$\frac{\pi }{2} + \frac{2}{3}$$
C.
$$\frac{\pi }{2} + \frac{4}{3}$$
D.
$$\frac{\pi }{2} - \frac{4}{3}$$
Answer :
$$\frac{\pi }{2} + \frac{4}{3}$$
Solution :
Given curves are $${x^2} + {y^2} = 1$$ and $${y^2} = 1 - x.$$
Intersecting points are $$x=0,\,1$$
Area of shaded portion is the required area.
So, Required Area $$=$$ Area of semi-circle $$+$$ Area bounded by parabola
$$\eqalign{
& = \frac{{\pi {r^2}}}{2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} \cr
& = \frac{\pi }{2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} \,\,\,\,\,\left( {\because {\text{radius of circle}} = 1} \right) \cr
& = \frac{\pi }{2} + 2\left[ {\frac{{{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}{{ - \frac{3}{2}}}} \right]_0^1 \cr
& = \frac{\pi }{2} - \frac{4}{3}\left( { - 1} \right) \cr
& = \frac{\pi }{2} + \frac{4}{3}{\text{ sq}}{\text{. unit}} \cr} $$