Answer
$\dfrac{16}{3}$
Work Step by Step
The domain $D$ for given region can be expressed as: $-1 \leq y \leq 1$ and $1-y^2 \leq y \leq y^2-1$
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A=\int_{-1}^1 \int_{-1+y^2}^{-y^2+1} dz dy\\=\int_{-1}^1 (-2y^2+2) \ dy \\=[-\dfrac{2y^3}{3}+2y]_{-1}^1 \\=\dfrac{-2}{3}+2-\dfrac{2}{3}-2 \\=\dfrac{8}{3}$
Thus, the volume of a given region is:
$2 \times \dfrac{8}{3}=\dfrac{16}{3}$
