= Solution
For a continuously differentiable <vector field> on a volume $V$ with piecewise smooth boundary $S$, the <divergence theorem> states
$$
\oiint_S\mathbf A\cdot\hat{\mathbf n}\,dS=\iiint_V\nabla\cdot\mathbf A\,dV,
$$
where $\hat{\mathbf n}$ is the \b[outward] unit normal. For the polynomial field here,
$$
\nabla\cdot\mathbf A=\partial_x(4xz)+\partial_y(-y^2)+\partial_z(yz)=4z-y.
$$
Therefore the outward <flux> through the unit cube is
$$
I=\int_0^1\int_0^1\int_0^1(4z-y)\,dx\,dy\,dz=2-\frac12=\boxed{\frac32}.
$$
As a direct boundary check, the $x=1$ face contributes $2$, the $y=1$ face contributes $-1$, and the $z=1$ face contributes $1/2$. All three opposite faces contribute zero, reproducing the result and fixing the orientation sign.
Back to article page
