Answer
$\approx 0.713$
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:
$$D=\left\{ (x, y) | x^4 \leq y \leq 3x-x^2, \ 0 \leq x \leq 1.213 \right\}
$$
Now, $V =\iint_{D} f(x,y) \ dA \\= \int_{0}^{1.213} \int_{x^4}^{3x-x^2}x dy \ dx \\ = \int_{0}^{1.213} [xy]_{x^4}^{3x-x^2} \ dx \\= \int_{0}^{1.213} 3x^2-x^3-x^5 \ dx \\=[x^3 -\dfrac{x^4}{4}-\dfrac{x^6}{6}]_0^{1.213} \\ = [(1.213)^3 -0]-[\dfrac{(1.213)^4}{4}-0]-[\dfrac{(1.213)^6}{6} -0] \\ \approx 0.713$
