Answer
$12 \pi$
Work Step by Step
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial (xy+2xz)}{\partial x}+\dfrac{\partial (x^2+y^2)}{\partial y}+\dfrac{\partial (xy-z^2)}{\partial z}=(y+2z)+2y-2z=3y$
$\iiint_E 3y dV=\int_{0}^{2 \pi}\int_0^{2}[3r^2z \sin \theta]_{r \sin \theta-2}^{0} dr d\theta$
$=\int_{0}^{2 \pi} \int_0^{2} [3r^3 \sin^2 \theta-6r^2 \sin \theta] dr d\theta$
$=\int_{0}^{2 \pi} [\dfrac{3r^4 \sin^2 \theta}{4}-2r^3 \sin \theta]_0^2 d\theta$
$=(12) \times \int_{0}^{2 \pi} 12 \sin^2 \theta-16 \sin \theta d\theta$
$=(12) \times \int_{0}^{2 \pi} (\dfrac{1-\cos 2 \theta}{2}) -16 \sin \theta d\theta$
Hence, we have $\iiint_E 3y dV=[6 \theta -3 \sin 2 \theta +16 \cos \theta]_0^{2\pi}=12 \pi$
