= Solution
Orient the triangle by the path $A\to B\to C\to A$. The <cross product> $(B-A)\times(C-A)=(0,1,0)$ gives the positive <normal vector>. In the plane $y=1$, the triangular region is $x,z\geq0$, $x+z\leq1$. The surface integral in <Stokes theorem> is therefore
$$
\int_0^1\int_0^{1-z}4xz^2\,dx\,dz=2\int_0^1z^2(1-z)^2\,dz=\frac1{15}.
$$
For the boundary <line integral>, $AB$ has $z=0$, so contributes zero. On $BC$, $x=0$ and only $dz$ is nonzero, so its contribution is also zero. Parametrize $CA$ by $(x,y,z)=(t,1,1-t)$, $0\leq t\leq1$. Its contribution is
$$
\int_0^1[2t(1-t)^3-t^2(1-t)^2]dt=\frac1{15}.
$$
Thus \b[both oriented integrals equal $1/15$]. Reversing the orientation would reverse both signs.
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