Calculate the flux of $\vec F=\dfrac{1}{x^2+y^2}(x,y,z)$ through the cylinder $\{(x,y,z)\in\mathbb{R}^3∣x^2+y^2=2,−2\leq z\leq2\}$ by using Gauss Law (divergence theorem).
By calculation I obtain that $\nabla\cdot\vec F=\dfrac{1}{x^2+y^2}$
I call:
The flux through the closed cylinder $\{(x,y,z)\in\mathbb{R}^3∣x^2+y^2=2,−2\leq z\leq2\}$ for $C$.
The flux through the bottom plate $\{(x,y,z)\in\mathbb{R}^3∣x^2+y^2=2,z=2\}$ for $B$.
The flux trough the top plate $\{(x,y,z)\in\mathbb{R}^3∣x^2+y^2=2,z=-2\}$ for $T$.
What I want to calculate is $C-B-T$ which I call $K$.
$$K = \iiint_V\nabla\cdot\vec F \,dx\,dy\,dz - \iint_\text{top}\vec F\cdot\vec n\,dS - \iint_\text{bottom}\vec F\cdot\vec n\,dS$$ with the bounds being the entire cylinder, top face, and bottom face respectively.
$\vec n = (0,0,1)$ for $T$ and $\vec n= (0,0,-1)$ for $B$
$$C = \iiint\frac{1}{x^2+y^2}\,dx\,dy\,dz = \left(\iint_A\frac{1}{x^2+y^2}\,dxdy\right)\left(\int dz\right) = 4\iint_A\frac{1}{x^2+y^2}\,dxdy$$ where $A=\{(x,y)\in\mathbb{R}^2|x^2+y^2=2\}$ is a horizontal cross section of the cylinder.
$$B = \iint_A(0,0,-1)\cdot\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2},\frac{-2}{x^2+y^2}\right)dx\,dy = 2\iint_A \frac{1}{x^2+y^2}\,dx\,dy$$
$$T = \iint_A (0,0,1)\cdot\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2},\frac{2}{x^2+y^2}\right)\,dx\,dy = 2\iint_A\frac{1}{x^2+y^2}\,dx\,dy$$
$$K = 4\iint_A\frac{1}{x^2+y^2}\,dx\,dy - 2\iint_A\frac{1}{x^2+y^2}\,dxdy - 2\iint_A\frac{1}{x^2+y^2}\,dx\,dy = 0$$
But in the end I get $0$, while the answer should be $8\pi$. What went wrong?