Define the vector field $F(x,y,z) = (2xy+z)\hat i + y^2\hat j - (x+3y) \hat k$ and let the surface S be the closed surface consisting of the boundry of the solid enclosed by the four planes: $2x+2y+z=6$ , $x=y=z=0$. Verify the divergence theorem.
So I was able to find the value for the divergence side of the equality and I got 216. I am have trouble solving for the side of: $$\iint_{S} \vec{F}\,\dot \,dS\,$$ I know that when $z=0$ then $0 \le x \le3$ and $0\le y\le 3-x$. Also when $y=0$ then $0 \le x \le3$ and $0\le z\le 6-2x$. And when $x=0$ then $0 \le y \le3$ and $0\le z\le 6-2y$. But I do not know how to do the left side of the equality still.
The boundary of the solid is given by four triangles. You have to evaluate the surface integrals over each of them with respect to the vector field $\vec{F}$:
1) Triangle in the $xy$-plane: $z=0$, $2x+2y\leq 6$, $x\geq 0$, $y\geq 0$ with exterior normal vector $\vec{n_1}=(0,0,-1)$ then $$\iint_{S_1} \vec{F}\cdot d\vec{S}=\iint_{S_1}\vec{F}\cdot \vec{n_1} \,dxdy =\int_{x=0}^3\int_{y=0}^{3-x}(x+3y) \,dydx=?$$
Can you take it from here?
P.S. Are you sure that the value for the divergence side of the equality is 216? Please check it again. $$\int_V\text{div}(\vec{F})dV=\int_{x=0}^3\int_{y=0}^{3-x}\int_{z=0}^{6-2x-2y}(2y+2y+0)dzdydx=?$$