Let $B$ be the region of $\mathbb{R}^3$
$$B = \{ (x,y,z) |\ x+y+z \le 1,\ x\geq0,\ y\geq0,\ z\geq0 \}$$
and $\mathbf{F}$ be the field $$\mathbf{F}(x,y,z) = (x,y,z)$$
My guess was use spherical coordinates for the triple integral, I integrate $3*\rho^2* \sin(\phi)$, which is the result of the divergence of $\mathbf{F}$ with $0\leq \theta\leq1/2\pi$, $0\leq\phi\leq1/2\pi$ and $0\leq\rho\leq1$ but it didn't work.
Maybe it can be even easier and for the other part of the theorem the sum of the double integrals I've got 2 surfaces the $z\geq0$ and $z\le1-x-y$ but didn't gave me the same result as triple integral.
For the first one I integrate from $0\leq \theta\leq1/2\pi$, $0\leq\phi\leq1/2\pi$ $\cos(\phi)*\sin(\phi) d\phi d\theta$
and gave me $\pi/4$, for the $x+z+y \le 1 $ I integrate $0\leq \theta\leq1/2\pi$, $0\leq\phi\leq1/2\pi$ $\sin(\phi) d\phi d\theta$
and gave me $\pi/2$ the sum is $3 \pi/4$ while the triple is $\pi/2$