## Calculus 8th Edition

$\dfrac{9 \pi}{2}$
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$ Here, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial (3xy^2)}{\partial x}+\dfrac{\partial (xe^z)}{\partial y}+\dfrac{\partial (z^3)}{\partial z}=3(y^2+z^2)$ $\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\iiint_E 3 \times (y^2+z^2)dV=\int_{-1}^{2}\int_0^{2 \pi} \int_0^{1} \times 3 (r^2 \cos^2 \theta+r^2 \sin^2 \theta) \times r dr d\theta dx$ $=[\int_{-1}^{2} dx] \times [\int_0^{2 \pi} d\theta] \times [\int_0^{1} 3 r^3 dr]$ $=[x]_{-1}^{2} \times [\theta]_0^{2 \pi} \times [3\dfrac{r^4}{4}]_0^{1}$ or, $=(2+1) \times [{2 \pi}-0] \times \dfrac{3}{4}$ Hence, we have $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\dfrac{9 \pi}{2}$