Evaluate the integral by using Gauss divergence theorem. Evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$
by using the Gauss divergence theorem.
Where $S$ is the closed surface obtained from the surfaces $x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0$ on the top and the bottom and $z=4-\sqrt{x^2+y^2}$ on the side.
My calculation shows that answer is $\frac{40\pi}{3}$ but the answer should be $\frac{88\pi}{3}$.What is I am doing wrong?
I set $$\iint_S \bf{F}\cdot dS=\iiint_{V}\text{div}FdV=2\iiint_VzdV$$
In cylindrical coordinates the limits are $$0\leq z\leq 4-r,2\leq r\leq 4,0\leq \theta\leq 2\pi$$ 
What is wrong with these limits?

Anyway, this is what I get:
$$2 \int\int\int z dV = 2\int_0^2 \int_0^{4-z} \int_0^{2\pi} z r\ d\theta dr dz = 4\pi \int_0^2 \int_0^{4-z} zr\ dr dz$$
$$ = 2\pi \int_0^2  z(4-z)^2 dz = 2\pi \int_0^2 \bigg(z^3 - 8z^2 + 16z\bigg)dz$$
$$ = 2\pi \bigg( \frac{2^4}{4} - \frac{16(2^3)}{3} + \frac{16(2^2)}{2}\bigg) = \frac{88\pi}{3}$$
