Answer
$\pi$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
This implies that $div F=\dfrac{\partial (2x^3+y^3)}{\partial x}+\dfrac{\partial (y^3+z^3)}{\partial y}+\dfrac{\partial (3y^2z)}{\partial z}$
$div F=6x^2+3y^2+3y^2=6(x^2+y^2)$
Now, we have
$\iiint_E 6(x^2+y^2) dV=\int_{0}^{2 \pi}\int_0^{1} \int_0^{1-r^2} 6r^2 r dz dr d\theta$
$=\int_{0}^{2 \pi}\int_0^{1} [6r^3 z]_0^{1-r^2} dr d\theta$
$=\int_{0}^{2 \pi}[\dfrac{3r^4}{2}-r^6]_{0}^{1} d\theta$
$=\int_{0}^{2 \pi}(\dfrac{3}{2}-1) d\theta$
$=[\dfrac{\theta}{2}]_0^{2 \pi}$
$=\pi$
