Answer
$\dfrac{128}{15}$
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:
$$D=\left\{ (x, y) | -2 \leq x \leq 2, \ x^2 \leq y \leq 4 \right\}
$$
Now, $V =\iint_{D} f(x,y) \ dA \\=\int_{-2}^{2} \int_{x}^{4} x^2 \ dy \ dx \\ =2 \int_{0}^{2} \int_{x^2}^{4} x^2 dy dx \\=2 \times \int_{0}^{2}[x^2(4-x^2) \\=2 \times \int_0^2 (4x^2-x^4) \\=2[\dfrac{4x^3}{3}-\dfrac{x^5}{5}]_0^2 \\=\dfrac{128}{15}$
