$\def\vec{\overrightarrow}$I am stuck on the following problem that says:
If $$\vec F=x \hat i+y \hat j+(z^2-1) \hat k,$$ then prove that $$ \iint \vec{F} \cdot \vec{n} \,\mathrm{d} S=4 \pi,$$ where $S$ is the surface bounded by the planes $z=0,z=1$ and the cylinder $x^2+y^2=4$.
My Try: By Gauss divergence theorem we get $$ \iiint \nabla \cdot \vec{F} \,\mathrm{d}V = \iint \vec{F} \cdot \vec{n} \,\mathrm{d} S.$$
Here, $$\mathrm{div}\,\vec F=\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z^2-1)=2+2z.$$
Limits of volume $V$: $z=0 \,\,\,\text{to}\,\,\,z=1$. In $xy$-plane, $x=-2$ to $x=2$ and $y=-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$.
So, $$ \iiint\nabla \cdot \vec{F} \,\mathrm{d}V =\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{0}^{1}2(1+z) \,\mathrm dx\mathrm dy\mathrm dz= \cdots=12 \pi.$$
I skipped some steps but can someone please explain in detail where I went wrong in the solution as the answer should be $4 \pi$.
Thanks in advance for your time.
Check also by cylindrical coordinates
$$ \int \int \int \nabla \cdot \vec{F} \mathrm{d}V =\\=\int_0^1 dz \int_{0}^{2\pi} d\theta \int_{0}^{2}2r(1+z)\, dr=2\pi\int_0^1 [r^2]_0^2\cdot(1+z)\,dz =8\pi\left[z+\frac{z^2}2\right]_0^1=12\pi$$