Answer: $ \bar {F}=2x ^2yi-y^2j+4xz^2k$

$\therefore \nabla \cdot\bar{F}=\dfrac{\partial }{\partial x }(2x^2y)-\dfrac{\partial }{\partial y}(y^2)+\dfrac{\partial }{\partial z }(4xz^2)$

$\therefore \nabla \cdot\bar{F}=4xy-2y+8xz$

By Guass-Divergence theorem,

$\iint\bar{N}\cdot \bar{F}ds =\iiint \nabla \cdot\bar{F}$

$\iiint \nabla \cdot\bar{F}=\int_{z=0}^3\int _{y=0}^{\sqrt{9-z^2}}\int _{x=0}^2 (4xy-2y+8xz)dxdydz$

$=\int_{z=0}^3\int _{y=0}^{\sqrt{9-z^2}} [2x^2y-2xy+4x^2z8xz]_{x=0}^2dydz$

$=\int_{z=0}^3\int _{y=0}^{\sqrt{9-z^2}} (4y+16z)dydz$

$=\int_{z=0}^3 [2y^2+16yz]_{y=0}^{\sqrt{9-z^2}}dz$

$=\int_{0}^3 [2(9-z^2)+16z\sqrt{9-z^2}]dz$

$= [2(9z-\dfrac{z^3}{3})+16(9-z^2)^\dfrac{3}{2}(\dfrac{2}{3}) (-\dfrac{1}{2})]_0^3$

$=2(18)+16(9)$

$=180$