Answer
$ \approx 0.713$
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:
$D=\left\{ (x, y) | x^4 \leq y \leq 3x-x^2, \ 0 \leq x \leq 1.213 \right\}$
Now, $\ V =\iint_{D} f(x,y) \ dA \\= \int_{0}^{1.213} \int_{x^4}^{3x-x^2}x dy \ dx \\ = \int_{0}^{1.213} [xy]_{x^4}^{3x-x^2} \space dx \\= \int_{0}^{1.213} 3x^2-x^3-x^5 \space dx \\=[x^3 -\dfrac{1}{4} x^4-\dfrac{1}{6} x^6]_0^{1.213} \\ = (1.213)^3 -\dfrac{(1.213)^4}{4}-\dfrac{(1.213)^6}{6} \\ \approx 0.713$
