## Calculus 8th Edition

$\dfrac{9}{2}$
In order to verify the Divergence Theorem which is true for for the vector field over the region $E$, we will have to add all the surface integrals and should be make sure that all are equal to such as: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV$ Here, we have $S$ shows a closed surface. The region $E$ is inside that surface. We have $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}$ and $div F=\dfrac{\partial (xye^z)}{\partial x}+\dfrac{\partial (xy^2z^3)}{\partial y}+\dfrac{\partial (-ye^z)}{\partial z}=ye^z+2xyz^3-ye^z=2xyz^3$ $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV =\int_{0}^3\int_0^2 \int_0^1 2xyz^3 \cdot dzdydx=\int_{0}^3\int_0^2 [\dfrac{xy}{2}] \cdot dydx$ or, $=\int_{0}^3 x dx$ or, $=[\dfrac{x^2}{2}]_0^3$ or, $=\dfrac{(3)^2}{2}-0$ Hence, $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV =\dfrac{9}{2}$