Answer
$\dfrac{341\sqrt 2}{60}+\dfrac{81}{20} \arcsin (\dfrac{\sqrt 3}{3})$
Work Step by Step
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial e^y \tan z}{\partial x}+\dfrac{\partial y \sqrt{3-x^2}}{\partial y}+\dfrac{\partial (x \sin y)}{\partial z}=\sqrt{3-x^2}$
$\iiint_E div F dV=\iiint_E \sqrt{3-x^2} dV=\int_{-1}^{1}\int_{-1}^{1} \int_{0}^{2-x^4-y^4} \times \sqrt{3-x^2} \times dz dxdy$
$=\int_{-1}^{1}\int_{-1}^{1} (2-x^4-y^4) \times [\sqrt{3-x^2}] dxdy$
$=\int_{0}^{2 \pi}\int_0^{\pi} \int_{0}^{R} 5\rho^2 \times \sin \phi \times d \rho d\phi d \theta$
Use CAS Method.
Hence, we have $\iint_S F \cdot dS=\dfrac{341\sqrt 2}{60}+\dfrac{81}{20} \arcsin (\dfrac{\sqrt 3}{3})$
