= Solution
Take the <unit normal> to point outwards. The <divergence> of the field is
$$
\nabla\cdot\mathbf F=(9x^2-2x)y+(3y^2-4y+1)x+2z.
$$
The <divergence theorem> turns the closed <surface integral> into a volume integral over $[0,1]^3$:
$$
\begin{aligned}
I&=\int_0^1\int_0^1\int_0^1
\left[(9x^2-2x)y+(3y^2-4y+1)x+2z\right]dx\,dy\,dz\\
&=(3-1)\frac12+\frac12(1-2+1)+1
=\boxed{2}.
\end{aligned}
$$
For a direct check of the <surface integral>, on $x=1$ the outward component is $F_x=2y$, contributing $\int_0^1\int_0^1 2y\,dy\,dz=1$. On $x=0$ it vanishes. On both $y=0$ and $y=1$, $F_y=x y(y-1)^2=0$. On $z=1$, $F_z=0$, whereas on $z=0$ the field component is $F_z=-1$ and the outward normal is $-\widehat{\mathbf z}$, giving a contribution $1$. The six face contributions are therefore $1,0,0,0,0,1$, again totaling \b[2]. The positive bottom-face contribution is a consequence of its outward orientation.
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