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 defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$
Thus, we can express the region $D$ using the point of intersection as follows:
$$D=\left\{ (x, y) | x^4 \leq y \leq 3x-x^2, \ 0 \leq x \leq 1.213 \right\}
$$
Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA \\= \int_{0}^{1.213} \int_{x^4}^{3x-x^2}x dy \ dx \\ = \int_{0}^{1.213} [xy]_{x^4}^{3x-x^2} \ dx \\= \int_{0}^{1.213} 3x^2-x^3-x^5 \ dx \\=[x^3 -\dfrac{x^4}{4}-\dfrac{x^6}{6}]_0^{1.213} \\ = (1.213)^3 -\dfrac{(1.213)^4}{4}-\dfrac{(1.213)^6}{6} \\ \approx 0.713$$
