= Solution
The <divergence> is
$$
\nabla\cdot\boldsymbol F=2(y+3)+2z=2y+2z+6.
$$
For the tetrahedral domain, convenient integration bounds are
$$
0\leq x\leq1,\qquad 0\leq y\leq1-x,\qquad
0\leq z\leq2(1-x-y).
$$
Consequently its <volume integral> is
$$
I=\int_0^1\int_0^{1-x}\int_0^{2(1-x-y)}(2y+2z+6)\,dz\,dy\,dx.
$$
Putting $h=2(1-x-y)$, the inner <integral> is $(2y+6)h+h^2=(16-4x)(1-x-y)$. Therefore
$$
\begin{aligned}
I&=\int_0^1(16-4x)\frac{(1-x)^2}{2}\,dx\\
&=\int_0^1(8-18x+12x^2-2x^3)\,dx
=8-9+4-\frac12.
\end{aligned}
$$
Thus
$$
\boxed{I=\frac52.}
$$
There is an independent <centroid> check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An <affine function> averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the <integral of an affine function over a simplex>.
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