Answer
$-12 \pi$
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 $div F=\dfrac{\partial (xy+2xz)}{\partial x}+\dfrac{\partial (x^2+y^2)}{\partial y}+\dfrac{\partial (xy-z^2)}{\partial z}$
$div F=(y+2z)+2y-2z=3y$
Now, we have
$\iiint_E 3y dV=\int_{0}^{2 \pi}\int_0^{2} \int_{r \sin \theta-2}^{0} 3 r \sin \theta r dz dr d\theta$
$=-\int_{0}^{2 \pi} \int_0^{2} 3r^3 \sin^2 \theta-6r^2 \sin \theta dr d\theta$
$=-\int_{0}^{2 \pi} [\dfrac{3r^4 \sin^2 \theta}{4}-2r^3 \sin \theta]_0^2 d\theta$
$=-\int_{0}^{2 \pi} 12 \sin^2 \theta-16 \sin \theta d\theta$
$=-\int_{0}^{2 \pi} 12 (\dfrac{1-\cos 2 \theta}{2}) -16 \sin \theta d\theta$
$=-[6 \theta -3 \sin 2 \theta +16 \cos \theta]_0^{2\pi}$
$=-12 \pi$
