Divergence Theorem Clarification. Given that $\textbf{F} =  \langle 3x,y^3,-2z^2 \rangle$, 
and the region bounded by $ x^2 + y^2 = 9$ and $z=0, \: z=5$
I'm trying to use the divergence theorem to find the line integral.
$$\iint_{D} \textbf{F} \cdot \textbf{N} \: dS = \iiint_E \nabla \cdot \textbf{F}\:d\textbf{V}$$

Attempt:
Top disk:
Since the normal vector of a disk on the xy-plane is $\langle 0, 0, 1 \rangle$ and z = 5 on the top disk,
Now,
$\iint_T \langle 3x , y^3, -2(5)^2 \rangle \cdot \langle 0 , 0, 1\rangle \: d\textbf{S} = \int_0^{2\pi} \int_0^3 -50 r\: dr\: d\theta  = -45\pi $. But this is not right!
Similarly, for the bottom disk, 
$\iint_B \langle 3x , y^3, -2(0)^2 \rangle \cdot \langle 0 , 0, -1\rangle \: d\textbf{S} = 0$
However, the answer given for both of the disk is $\frac{-45}{4}\pi$
Any help would be appreciated! Thank you
 A: Your method of calculating the surface integrals on the top and bottom surfaces is correct! (Although, as Doug pointed out, the top contribution should be $-450\pi$.)
But you also need to work out the surface integral on the curved surface of the cylinder!
Let's parametrise the curved surface by
$$ (x,y,z) = (3 \cos \phi, 3 \sin \phi, z), \ \ \ \ \ 0 \leq \phi < 2\pi , \ \ 0 \leq z \leq 5. $$
Then
$$ \vec F = (9\cos \phi, 27 \sin^3 \phi, -z^2), \ \ \ \vec n = (\cos \phi, \sin \phi, 0), \ \ \ \ \  dS = 3d\phi dz. $$
So the contribution from the curved surface is
$$ \iint \vec F. \vec n dS = \int_{z = 0}^{z = 5} \int_{\phi = 0}^{\phi = 2\pi} (9\cos^2 \phi + 27 \sin^4 \phi) 3d\phi dz = \frac{1755\pi}{4}.$$
I plugged this into mathematica, but you can work it out explicitly. Anyway, the sum of all three contributions is $$-450\pi + 0 - \frac{1755\pi}{4} = - \frac {45\pi}{4}.$$

Of course, you can also work out the volume integral of the divergence. Here, $$\nabla . \vec F = 3 + 3y^2 - 4z = 3 + 3 \rho^2 \sin^2 \phi - 4z$$ in cylindrical polars, and a quick calculation on mathematica gives
$$ \iiint \nabla . \vec F dV = \int_{\rho = 0}^{\rho = 3} \int_{z = 0}^{z = 5}  \int_{\phi = 0}^{\phi = 2\pi} (3 + 3 \rho^2 \sin^2 \phi - 4z) \rho d\phi dz d \rho= - \frac{45\pi}{4}$$
