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
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 $$ \iiint \nabla . \vec F dV = \int_{\rho = 0}^{\rho = 3} \int_{z = 0}^{z = 5} \int_{\phi = 0}^{\phi = 2\pi} (3 + 3 \rho^2 \sin^2 \phi - 4z) \rho d\phi dz d \rho= - \frac{45\pi}{4}$$