Aea of the region nclosed between the curves `x=y^2-1` and `x=|y|sqrt(1-y^2)` is

Options

A. 1 sq. units
B. `4//3` sq. units
C. `2//3` sq. units
D. 2 sq. units

Answer

D

Explanation

`x=y^(2)-1` is a parabola having vertex at (-1,0) and concave right hand side.
`x=|y|sqrt(1-y^(2))`
Here `xge0` as R.H.S. is positive
When `ygt0`,
`x=ysqrt(1-y^(2))" (1)"`
(1) meets y-axis at (0,0), (0,1), thus,
`(dx)/(dy)=sqrt(1-y^(2))-(y^(2))/(sqrt(1-y^(2)))=(1-2y^(2))/(sqrt(1-y^(2)))`
`"For "(dx)/(dy)=0, y=(1)/(sqrt(2))`
Hence, graph of (1) is as shown in the figure, with `x=-ysqrt(1-y^(2))(ylt0)` as its mirror image.

`A=2overset(1)underset(0)int[ysqrt(1-y^(2))-(y^(2)-1)]dy`
`=2 ` sq. units