Answer
$\dfrac{3a^2b^2c^2}{4}$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
This implies that $div F=\dfrac{\partial (x^2yz)}{\partial x}+\dfrac{\partial (xy^2z)}{\partial y}+\dfrac{\partial (xyz^2)}{\partial z}=6xyz$
$=\int_{0}^a\int_0^b \int_0^c (6xyz)dzdydx$
$=\int_{0}^a\int_0^b [6xy(z^2/2)]_0^cdydx$
$=\int_{0}^a\int_0^b [3xy \times (c^2) ]dydx$
$=\dfrac{3a^2b^2c^2}{4}$
