Answer
$2$
Work Step by Step
Graphing the given equations, we find the intersections
or we solve the equation $f(y)=g(y), $ where
$ f(y)=|y|\sqrt{1-y^{2}} \quad$ (the curve on the right side) and
$ g(y)=y^{2}-1\quad$ (the curve on the left).
When $y\in[c,d]=[-1,1]$, the area between the graphs is
$A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{-1}^{1}[|y|\sqrt{1-y^{2}}-y^{2}+1]dy $
Replacing $y$ with $-y$, the integrand remains the same$\Rightarrow$
integrand is symmetric about the x-axis.
So we calculate the area for $y\in[0,1]$ and double it.
(On this interval, $|y|=y$)
$A=2\displaystyle \int_{0}^{1}[y\sqrt{1-y^{2}}-y^{2}+1]dy $
$=2\displaystyle \left[\int_{0}^{1}[y\sqrt{1-y^{2}}-y^{2}+1]dy \right]$
$=2\displaystyle \int_{0}^{1}[y\sqrt{1-y^{2}}dy+2\left[-\frac{y^{3}}{3}+y \right]_{0}^{1}$
... substitute: $\left[\begin{array}{ll}
u=1-y^{2} & du=-2ydy\\
y=0\Rightarrow u=1 & \\
y=1\Rightarrow u=0 &
\end{array}\right]$
$=2\displaystyle \int_{1}^{0}u^{1/2}(-\frac{du}{2})+2\left[-\frac{1-0}{3}+(1-0) \right]$
$=-\displaystyle \left[\frac{u^{3/2}}{3/2}\right]_{1}^{0}+\frac{4}{3}$
$=-(0-\displaystyle \frac{2}{3})+\frac{4}{3}$
$=2$
