Outward Flux of Vector Field F across Surface S

[Differentiate1]

Question: Find the outward flux of the vector field F = i-2j-2k across the surface S defined by z = 4-x²-y² 0≤z≤4

At first, I used the Divergence Theorem to solve this problem. I took the divF and got the answer of 0. By definition, integrating 0 three times will still equal 0. Thus, the answer I wrote down currently is 0.

Now there's also another way to find the outward flux, and that is by taking the double integral of F⋅ndσ

n = [r_r X r_θ]/[|r_r X r_θ|]
dσ = |r_r Xr_θ|

I found r_r X r_θ here from another problem: http://i.imgur.com/UMj72Ub.png

Checking with WolframAlpha, I got this:
1st integration: http://www.wolframalpha.com/input/?i=integrate+(-2r^2cos(theta)+4r^2sin(theta)-2r)+dr+from+0+to+2

2nd integration: http://www.wolframalpha.com/input/?i=integrate+(-4/3)(-8sin(theta)+4cos(theta)+3)+from+0+to+2pi

I would greatly appreciate it if someone can explain why the answer using the Divergence theorem is not equal to the Outward Flux when taking the double integral.

Thanks,

Differentiate1

 

[Differentiate1]

Never mind, figured out my mistake. I should've not used the parametrized form to take the double integral. I fixed this by making n = [-2x-2y]/|-2x-2y| and dσ = |-2x-2y| (I acquired these values from finding the gradient of z).