Answer
$\approx 20.420$
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:
$$D=\left\{ (x, y) | -\sqrt {1-x^2} \leq y \leq \sqrt {1-x^2}, -1 \leq x \leq 1 \right\}$$
Therefore, $\ Volume =\iint_{D} f(x,y) \ dA\\=\iint_{D} (3x^2+3y^2-8) \ dy \ dx \\= \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (3x^2+3y^2-8) \ dy \ dx \\ = 2 \int_0^1 (2) \int_{0}^{\sqrt {1-x^2}} (3x^2+3y^2-8) dy dx \\ = 4 \int_0^1 [3x^2 y +y^3 -8y]_{0}^{\sqrt {1-x^2}} \\= 4 \int_0^1 (2x^2 \sqrt {1-x^2} -7 \sqrt {1-x^2} ) dx $
Now, we will use a calculator to compute the integral:
$\ Volume = 4 \int_0^1 (2x^2 \sqrt {1-x^2} -7 \sqrt {1-x^2} dx ) \approx 20.420$
