Using divergence theorem, it's well known that $\int_V \nabla \phi \space dV = \int_S \phi $ dS, for a differentiable function $\phi$, where $V$ is a volume and $S$ is it's boundary.
Consider the cone x$=(\rho \cos(\theta), \rho \sin(\theta), 2 \rho)$; $0\le \rho \le 1$, $0 \le \theta \le 2\pi$ and $\phi(x, y, z)=z^2$.
Now obviously
dS$= \frac{\partial x}{\partial \theta} \times \frac{\partial x}{\partial \rho} = (-\rho \sin(\theta), \rho \cos(\theta), 0) \times (\cos(\theta), \sin(\theta), 2 )= (2\rho \cos(\theta), 2\rho \sin(\theta), -\rho)$
and
$\nabla \phi=(0, 0, 2z)$
So
$ \int_V \nabla \phi \space dV = \int_{\theta=0}^{2\pi}\int_{\rho=0}^{1}\int_{z=0}^{2} (0, 0, 2z) \space \rho \space dz d{\rho}d\theta=(0, 0, 4\pi)$
on the other hand
$\int_{S{_1}} \phi $ dS $= \int_{\theta=0}^{2\pi}\int_{\rho=0}^{1} 4\rho^2 (2\rho \cos(\theta), 2\rho \sin(\theta), -\rho) \space d{\rho}d\theta= (0, 0, -2\pi)$ where $S_1$ is the surface of the cone.
$\int_{S{_2}} \phi $ dS $= \int_{\theta=0}^{2\pi}\int_{\rho=0}^{1} 4(0, 0, \rho) \space d{\rho}d\theta= (0, 0, 4\pi)$ where $S_1$ is the surface of the disc $z=2$.
The answears don't agree and i really donțt understand where is the mistake. Can someone please help me out. Thank you.
You're missing the integral over the flat surface of the cone, i.e. the disk with $z=2$, $\rho<1$. Also, you seem to have written $\nabla \phi = (0,0,4\rho)$ in the volume integral, which doesn't agree with the rest of your solution (remember inside the volume, you don't have $z=2\rho$).