Calculate the flow out of $(x,y,z)$ through the surface: $$z=\sqrt{x^2+y^2-15} ,\qquad 16 \leq x^2+y^2 \leq 40$$
To solve this, I tried to apply Gauss Divergence Theorem, I did the following:
First we note that the space is not compact, we solve this by adding a "roof" at z=5 and "flor" at z=1 thus we get the following: $$\iint_{Y_1}+\iint_{Y_2}+\iint_{Y_3} = \iiint_{D}div(\vec{F})dxdydz$$ Where $Y_2$: $z=1, x^2+y^2 = 16$ and $Y_3: z=5, x^2+y^2=40$.
If I solve these integrals one by one I get: $$\iint_{Y_2}\vec{F}\cdot \vec{N}dS=\iint_{Y_2}(x,y,1)\cdot(0,0,-1)dS = -\iint_{x^2+y^2\leq16}1dxdy = -16\pi$$ $$\iint_{Y_3}= (x,y,5)\cdot(0,0,1)dS =...=200\pi$$ Using that the normal vector is pointing outwards.
This is where I think I'm messing up, solving the triple integral. $$\iiint_{D}div(\vec{F})dxdydz = 3\iint_{16\leq x^2 +y^2 \leq 40}(\int_1^{5} dz)dxdy = ...= 288\pi$$
Which would give me the final answear: $\iint_{Y_1}=288\pi+16\pi-200\pi=104\pi$
Which is wrong.
Solving the integrals is fairly simple, where I think I'm wrong is in setting upp the integrals correctly, if someone could point me in the right direction that would be great :)
Thanks in advance