$\displaystyle \ \vec{F}=4x\bar{i}+3y\bar{j}-4z^2\vec{k} \\[2ex] \displaystyle \nabla{}\vec{F}=\frac{\partial{}}{\partial{}x}\left(4x\right)+\ \frac{\partial{}}{\partial{}y}\left(3y\right)-\frac{\partial{}}{\partial{}z}(4z^2) \\[2ex] \displaystyle =4+3-8z \\[2ex] \displaystyle =7-8z \\[2ex] $

$\displaystyle By\ Gauss\ Divergence\ theorem\ \\[2ex] \displaystyle \iint_S\vec{F}\bullet{}\hat{n}ds=\iiint_V\nabla{}\vec{F}dV \\[2ex] \displaystyle =\int_{z=0}^4\int_{y=0}^{\frac{\left(4-z\right)}{2}}\int_{z=0}^{\frac{\left(4-2y-z\right)}{2}}\left(7-8z\right)dxdydz\ \ \ \ \ \ \ \ \ \ \\[2ex] $

$ \displaystyle =\int_{z=0}^4\int_{y=0}^{\frac{\left(4-z\right)}{2}}\left(7-8z\right){\left[x\right]}_0^{\frac{\left(4-2y-z\right)}{2}}dydz \\[2ex] \displaystyle =\int_{z=0}^4\int_{y=0}^{\frac{\left(4-z\right)}{2}}\left(7-8z\right)\left[\frac{\left(4-2y-z\right)}{2}-0\right]dydz \\[2ex] \displaystyle =\int_{z=0}^4\int_{y=0}^{\frac{\left(4-z\right)}{2}}\left(7-8z\right)\times{}\frac{1}{2}\left[\left(4-z\right)-2y\right]dydz \\[2ex] \displaystyle =\frac{1}{2}\int_{z=0}^4\left(7-8z\right){\left[\left(4-z\right)y-\frac{{2y}^2}{2}\right]}_0^{\frac{\left(4-z\right)}{2}}dz \\[2ex] \displaystyle =\frac{1}{2}\int_{z=0}^4\left(7-8z\right)\left\{\left[\frac{{\left(4-z\right)}^2}{2}-\frac{{\left(4-z\right)}^2}{4}\right]-0\right\}dz \\[2ex] \displaystyle =\frac{1}{2}\int_{z=0}^4\left(7-8z\right)\times{}\frac{{\left(4-z\right)}^2}{2}dz \\[2ex] \displaystyle =\frac{1}{8}\int_{z=0}^4\left(7-8z\right)(16-8z+z^2)dz \\[2ex] \displaystyle =\frac{1}{8}\int_{z=0}^4(112-56z+7z^2-128z+64z^2-8z^3)dz \\[2ex] \displaystyle =\frac{1}{8}\int_{z=0}^4(112+71z^2-184z-8z^3)dz \\[2ex] \displaystyle =\frac{1}{8}{\left[\frac{-8z^4}{4}+\frac{71z^3}{3}-\frac{184z^2}{2}+112z\right]}_0^4 \\[2ex] \displaystyle =\frac{1}{8}\left\{\left[-2{(4)}^4+\frac{71{(4)}^3}{3}-92{(4)}^2+112(4)\right]-0\right\} \\[2ex] \displaystyle =\frac{-8}{3} \\[2ex] $