Answer
$$\approx 20.420$$
Work Step by Step
$f(x,y) =x^3y^4+xy^2$
The volume under the surface $z=f(x,y)$ and above the region $D$ in the $xy$-plane can be expressed as follows: $$\ Volume =\iint_{D} f(x,y) \ dA$$
and the region $D$ using the point of intersection can be expressed as follows:
$D=\left\{ (x, y) | -\sqrt {1-x^2} \leq y \leq \sqrt {1-x^2}, -1 \leq x \leq 1 \right\}$
$\ Volume ; V =\iint_{D} f(x,y) \ dA\\=\iint_{D} (3x^2+3y^2-8) \ dy \ dx $
or, $= \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (3x^2+3y^2-8) \ dy \ dx$
or, $ = 2\times \int_0^1 (2) \int_{0}^{\sqrt {1-x^2}} (3x^2+3y^2-8) dy dx $
or, $ = 4 \times \int_0^1 [3x^2 y +y^3 -8y]_{0}^{\sqrt {1-x^2}} $
or, $= 4 \times \int_0^1 (2x^2 \sqrt {1-x^2} -7 \sqrt {1-x^2} ) dx $
The integral can be computed by using a calculator as follows:
$\ Volume; V = 4 \int_0^1 (2x^2 \sqrt {1-x^2} -7 \sqrt {1-x^2} dx ) \\ \approx 20.420$
