Answer
$\dfrac{128}{15}$
Work Step by Step
The volume under the surface is given by :$z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $\ Volume ; V=\iint_{D} f(x,y) \space dA$
Our aim is to calculate the volume of the given surface.
The domain $D$ can be expressed as follows:
$D=\left\{ (x, y) | -2 \leq x \leq 2, \ x^2 \leq y \leq 4 \right\}
$
Now, $\ V =\iint_{D} f(x,y) \ dA \\=\int_{-2}^{2} \int_{x}^{4} x^2 \ dy \ dx \\ =2 \int_{0}^{2} \int_{x^2}^{4} x^2 dy dx \\=2 \int_{0}^{2}[x^2(4-x^2) \\=(2) \int_0^2 (4x^2-x^4) \\= (2) [\dfrac{4}{3} x^3-\dfrac{1}{5} x^5]_0^2 \\= 2 [\dfrac{4(2)^3}{3}-\dfrac{2^5}{5}-0]\\=\dfrac{128}{15}$
