We have,
$ x=-2 y^{2} $
$\Rightarrow y^{2} =-\frac{1}{2} x \,\,\,\,...(i)$
and $x=1-3 y^{2} $
$\Rightarrow y^{2}=-\frac{1}{3}(x-1)\,\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$-\frac{1}{2} x=-\frac{1}{3}(x-1) $
$\Rightarrow 3 x=2(x-1) $
$\Rightarrow 3 x=2 x-2 $
$\Rightarrow x=-2 $
$\therefore y^{2}=-\frac{1}{2}(-1)=1 $
$\Rightarrow y=\pm 1$
$\therefore $ Point of intersection of two curves is $(-2, \pm 1)$
$\therefore $ Required Area $=2 \int\limits_{0}^{1} \sin x d x$
$=2\left[y-\frac{y^{3}}{3}\right]_{0}^{1}$
$=2\left[1-\frac{1}{3}\right]$
$=\frac{4}{3}$ sq units
