Answer
A = $\frac{8}{3}$
Work Step by Step
Find the area of the region enclosed by the given curves : x = $ y^{2}$-1, x = 1-$ y^2$
First find where the functions intersect:
$ y^{2}$-1 = 1-$ y^2$
$2y^2$-2 = 0
y = $\sqrt ($2/2$)$
y = 1, y = -1
Then, with both graph's aspects calculated, the graphs can be drawn.
By the formula, A =$\int(f(y)-g(y))dy $ we consider, f(y) = 1-$ y^2$ and g(y) = $ y^{2}$-1 (based on the graph).
Then,
A = $\int $( (1-$ y^2$) - ($ y^{2}$ -1 ) )dy, [-1,1]
= ( y - $\frac{y^2}{3}$ - $\frac{y^2}{3}$ + y), [-1,1]
= ( -$\frac{2}{3}$ + 2) - ( $\frac{2}{3}$ - 2 ) = $\frac{8}{3}$