Answer
$\approx 961.181$
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: $$\ Volume =\iint_{D} f(x,y) \ dA$$
$f(x,y) =x^3y^4+xy^2$
Thus, we can express the region $D$ using the point of intersection as follows:
$$D=\left\{ (x, y) | 0 \leq x \leq 2, \ x^2+x \geq y \geq x^3-x \right\}$$
We will use polar co-ordinates as follows: $x= r \cos \theta; y= r \sin \theta$
Therefore, $\ Volume =\iint_{D} f(x,y) \ dA\\=\iint_{D} x^3y^4+xy^2 \ dA \\= \int_{0}^{2} \int_{x^3-x}^{x^2+x} (x^3y^4+xy^2) \ dy \ dx $
Now, we will use a calculator to compute the integral:
$\ Volume =\int_{0}^{2} \int_{x^3-x}^{x^2+x} (x^3y^4+xy^2) \ dy \ dx \approx 961.181$
