Verify Gauss’s Divergence Theorem I have this assignment which we have not tackled and am getting mixed up in the divergence theorem tutorials like this one http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/diverg/diverg.html
Verify Gauss’s Divergence Theorem for F= 4 xiˆ− 2y jˆ+ zkˆ  over the region bounded by ρ = 4, z = −2, and z = 2.
Help me get it solved. Thanks!
 A: I would say that on the contrary:
$$\iint \vec{F}\, \vec{dS} = \iiint \operatorname{div} \vec{F}\, dV$$
Edit -  reneval:
$$S = S_1 \cup  S_2 \cup S_3$$
$S_1=$ part of a cylindrical surface $\varrho = 4 \equiv x^2+y^2=16\,, z  \in \langle-2,2\rangle$
$$S_2 = \operatorname{cover} \varrho \le 4\,, z=2\,;   S_3 = \operatorname{bottom} \varrho \le 4\,, z=-2$$
$$\iint_S \vec{F}\, \vec{dS}=\iint_{S_1} \vec{F}\, \vec{dS}+\iint_{S_2} \vec{F}\, \vec{dS}+\iint_{S_3} \vec{F}\, \vec{dS}$$
$\underline{1.\,  S1:}\,  \vec{n}=$ unit outer normal vector
$$\iint_{S_1} \vec{F}\, \vec{dS}=\iint_{S_1} \vec{F}\cdot \vec{n}\, dS,\text{ substitution:} x = 4\cos t\,, y = 4\sin t,\, z = z$$
$$\Rightarrow n=(\cos t,\sin t,0)\,, F=(16\cos t,32\sin^2t,z^2)\,, \vec{F}\cdot \vec{n}=(16\cos^2t+32\sin^2t+ 0),\, dS = 4\,dt\,dz $$
$$\Rightarrow\, =32\int_{-2}^{2} \int_{0}^{2\pi}(2\cos^2t+4\sin^3t)\,dt\,dz= \cdots = 256\pi$$
$\underline{2. S2,S3:}\,  \vec{n_2}=(0,0,1)\, \vec{n_2}=(0,0,-1)$ unit outer normal vectors
A similar calculation is determined that $\iint_{S_1}+\iint_{S_2}=0$
so $\underline{\iint \vec{F}\, \vec{dS} = 256\pi}$
$\underline{3.}$
$\operatorname{div} \vec{F} = 4 + 4y + 2z$
$$\underline{\iiint \operatorname{div} \vec{F}\, dV}=\int_{-2}^{2}\int_{-4}^{4}\int_{_{-\sqrt{16-x^2}}}^{^{\sqrt{16-x^2}}}(4 + 4y + 2z)\,dy\,dx\,dz=\cdots =\underline{256\pi}$$
A: Your field is $F(x,y,z) = (4x,2y^2,z^2)$? And when you say $\rho = 2$, that means you have to do your integral in spherical coordenates??
