Calculate the flow of the vector field $F(x, y, z) = (x, y, z)$ out (away from the z-axis) through the cylinder surface $x^2 + y^2 = 4$, when $0 ≤ z ≤ 1$, by stop at the surface and use the divergence theorem.
The divergence is 3. And then I changed the coordinates to polar where $x=r\cos(\theta), y=r\sin(\theta), z=z$. The bounds are $0 \leq r \leq 2$ and $ 0 \leq \theta \leq 2\pi$ and $0 \leq z \leq 1$ And the area element is r so I get $\iiint_{V}3r \ d\theta dr dz = 12\pi$. But the answer is $8\pi$.
The surface of the cylinder consists of three parts: \begin{align} S_{\text{top}} = \{ (x, y, z) \in \mathbb R^3 : x^2 + y^2 < 4, z = 1 \}, \\ S_{\text{bottom}} = \{ (x, y, z) \in \mathbb R^3 : x^2 + y^2 < 4, z = 0 \}, \\ S_{\text{curved}} = \{ (x, y, z) \in \mathbb R^3 : x^2 + y^2 = 4, 0 \leq z \leq 1 \} . \end{align}
The question is asking you to compute $$ \iint_{S_{\text{curved}}} \mathbf F . d\mathbf A.$$
However, you have computed $$ \iiint_{V} \mathbf \nabla .\mathbf F \ dV,$$ which is equal to $$ \iint_{S_{\text{top}}} \mathbf F . d\mathbf A + \iint_{S_{\text{bottom}}} \mathbf F . d\mathbf A + \iint_{S_{\text{curved}}} \mathbf F . d\mathbf A.$$
So you probably want to compute $\iint_{S_{\text{top}}} \mathbf F . d\mathbf A$ and $\iint_{S_{\text{bottom}}} \mathbf F . d\mathbf A $, and subtract these from your answer for $ \iiint_{V} \mathbf \nabla .\mathbf F \ dV$.
You should find that $$ \iint_{S_{\text{top}}} \mathbf F . d\mathbf A = \iint_{x^2 + y^2 < 4 } 1 \ dx dy = 4\pi$$ and $$ \iint_{S_{\text{top}}} \mathbf F . d\mathbf A = \iint_{x^2 + y^2 < 4} 0 \ dxdy = 0.$$
I hope you can fill in the details.