Answer
$\dfrac{341\sqrt 2}{60}+\dfrac{81}{20} \arcsin (\dfrac{\sqrt 3}{3})$
Work Step by Step
$div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}=\dfrac{\partial e^y \tan z}{\partial x}+\dfrac{\partial y \sqrt{3-x^2}}{\partial y}+\dfrac{\partial (x \sin y)}{\partial z}=\sqrt{3-x^2}$
$I=\iiint_E \sqrt{3-x^2} dV=\int_{-1}^{1}\int_{-1}^{1} \int_{0}^{2-x^4-y^4} \sqrt{3-x^2} dz dxdy$
and $I=\int_{-1}^{1}\int_{-1}^{1} (2-x^4-y^4) \times \sqrt{3-x^2} dxdy$
We need to use a calculating tool:
$\iint_S F \cdot dS=\dfrac{341\sqrt 2}{60}+\dfrac{81}{20} \arcsin (\dfrac{\sqrt 3}{3})$
