Answer
$\dfrac{-40}{3}$
Work Step by Step
As we know that $div F=\dfrac{\partial A}{\partial x}i+\dfrac{\partial B}{\partial y}j+\dfrac{\partial C}{\partial z}k$
Now, we have
$Flux =\iiint_{o} -x dA$
Also,
$Flux =\nabla \cdot F=\int_{0}^{2}\int_{0}^{\sqrt {16-4x^2}}\int_{0}^{4-y} (-x) dz dx dy$
This implies that
$\int_{0}^{2}(\dfrac{x\sqrt {16-4x^2}}{2}-4x\sqrt {16-4x^2} )dx = \dfrac{-40}{3}$
