Answer
$\dfrac{341\sqrt 2}{60}+\dfrac{81}{20} \arcsin (\dfrac{\sqrt 3}{3})$
Work Step by Step
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}=\dfrac{\partial e^y \tan z}{\partial x}+\dfrac{\partial y \sqrt{3-x^2}}{\partial y}+\dfrac{\partial (x \sin y)}{\partial z}=0+\sqrt {3-x^2}+0=\sqrt{3-x^2}$
Now, we have
$\iiint_E div F dV=\iiint_E \sqrt{3-x^2} dV$
$=\int_{-1}^{1}\int_{-1}^{1} \int_{0}^{2-x^4-y^4} \sqrt{3-x^2} dz dxdy$
$=\int_{-1}^{1}\int_{-1}^{1} (2-x^4-y^4) \sqrt{3-x^2} dxdy$
$=\int_{0}^{2 \pi}\int_0^{\pi} \int_{0}^{R} 5\rho^2 \sin \phi d \rho d\phi d \theta$
By using a calculating tool, we have
$\iint_S F \cdot dS=\dfrac{341\sqrt 2}{60}+\dfrac{81}{20} \arcsin (\dfrac{\sqrt 3}{3})$
