Graphs too need Personal space.

The enclosed area is symmetric about the $x$ -axis, and the two curves intersect at $(0,1)$ and $(0, -1).$ As $|y|*\sqrt{1 - y^{2}} \ge y^{2} - 1$ for $-1 \le y \le 1$ , it is probably easiest to integrate with respect to $y$ from $y = 0$ to $y = 1$ and then multiply the result by $2$ . This results in an area of

$2*\displaystyle\int_{0}^{1} (y*\sqrt{1 - y^{2}} - (y^{2} - 1)) dy = 2*[-\dfrac{(1 - y^{2})^{\frac{3}{2}}}{3} - \dfrac{y^{3}}{3} + y]$ ,

which evaluated from $y = 0$ to $y = 1$ is $2*[-\dfrac{1}{3} + 1 - (-\dfrac{1}{3})] = \boxed{2}.$

Note that a substitution of $u = 1 - y^{2}$ was used to evaluate the first term of the integral.

Why did you say decimal upto 3 decimal places????????? I wasted so much time rechecking the sum.highly overrated problem.