Verify Divergence theorem by Surface integrals. Verify the divergenece theorem to $$\mathbf{F }=4xi-2y^2j+z^2k$$ for the region bounded by $x^2+y^2=4$ , $z=0$, $z=3$
I've already done the triple integral for the divergence $\iiint_R \operatorname{div}\bar F\;dV$ and the result I got is $84\pi$, but I'm having trouble solving it by surface integrals. I've defined $S_1$ as $x^2+y^2-4$ and the normal vector I got was $(2,2,0)$ but as I continue, the answer I keep getting is $4\pi$ 
Can you guys help me out with this?
 A: This region is a cylinder with radius $2$ around the $z$ axis, and so it can be parameterized as $r(\theta, z) = (2\cos \theta, 2 \sin \theta, z)$, where $\theta \in [0, 2\pi]$ and $z\in [0,3]$. Then
$$r_\theta = (-2\sin \theta, 2\cos \theta, 0)$$
$$r_z = (0,0,1)$$
$$\vec{N} = r_\theta \times r_z = (2\cos \theta, 2\sin \theta, 0)$$
(note that $\vec N$ points outwards). Now you need to evaluate 
$$I_1 = \int_{0}^{ 2\pi}\int_{0}^3 \vec{F}(r(\theta, z))\cdot \vec{N}\, dz\, d\theta$$
Now you need to integrate over the top and bottom on the integral. Since $z$ is constant, these integrals are equivalent to 
$$I_2=-\iint_S F(x,y,0)\, dA$$
$$I_3=\iint_S F(x,y,3)\, dA$$
where $S\subseteq \mathbb R^2$  is the disc centered at the origin with radius $2$. The negative sign in $I_2$ is due to the fact that the normal vector points outwards, which is downward on that part of the surface. Then your final answer is $I_1+I_2+I_3$, which should be equal to what you got using the divergence theorem. 
A: Parametrize the side of the cylinder of interest by
$$r(t,z)=\langle 2\cos (t), 2 \sin (t),z \rangle$$
With $t \in [0,2\pi]$ and $z \in [0,3]$. An outward pointing unit normal vector for the side is $\langle \cos(t), \sin(t),0 \rangle$ and also $dS=2dzdt$ hence we get that,
$$\iint_{S_1} \vec F \cdot \vec n dS$$
$$=2 \int_{0}^{2\pi} \int_{0}^{3} \langle 8\cos (t), -8\sin^2 (t), z^2 \rangle \cdot \langle \cos t, \sin t,0 \rangle dz dt$$
$$=16 \int_{0}^{2\pi} \int_{0}^{3} (\cos^2(t)-\sin^3(t)) dz dt$$
$$=48\pi$$

Next parametrize the top of the cylinder by $r(x,y)=\langle x,y,3 \rangle$ with $x^2+y^2 \leq 4$. An obvious outward pointing unit normal is $\langle 0,0,1 \rangle$. Furthermore it is easy to see that $dS=dA$ at the top of the cylinder, because the surface is flat parallel to the $xy$ plane.
From here,
$$\iint_{S_2} \vec F \cdot \vec n dS=\iint_{D} \langle ..,...,9 \rangle \cdot \langle 0,0,1 \rangle dA$$
$$=9\text{Area}({D})=9(\pi (2^2))=36 \pi$$

The bottom piece is much similar. It evaluates to zero. Really because $\vec F=\langle..,...,0 \rangle$ there and $\vec n=\langle 0,0,-1 \rangle$ there. Therefore $\vec F \cdot \vec n=0$ on the bottom surface.

Hence indeed we get,
$$\iint_{S} \vec F \cdot d\vec S=48\pi+36\pi=84\pi$$
As the divergence theorem claims.
