Answer
$\dfrac{2336}{27}$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $$\ Volume =\iint_{D} f(x,y) \space dA$$
The region $D$ can be defined as:
$D=\left\{ (x, y) | y^2 \leq x \leq 4, \ -2 \leq y \leq 2 \right\}
$
Now, $V =\iint_{D} f(x,y) \ dA \\=\int_{-2}^{2} \int_{y^2}^{4} 1+x^2y^2 \ dx \ dy \\ =\int_{-2}^{2} [x+\dfrac{x^3y^2}{3}]_{y^2}^{4} \ dx \\= \int_{-2}^{2} [ 4+\dfrac{64y^2}{3}]-[y+\dfrac{y^8}{3}] \ dy \\ = [4y+\dfrac{61 y^3}{9}-\dfrac{y^9}{27}]_{-2}^{2} \\= \dfrac{2336}{27}$
