Use Gauss Divergence Theorem to compute $$\int \int \limits_S F\cdot n \,dS$$ where $n$ is the outward normal for the following:
$S$ is the surface $x^2 +y^2=z^2$ and $z \in [0,1]$, and $F(x,y,z)=(x,xy,z)$
The $div(F)=2+x$. I let z be z and $x=r\cos\theta$ and $y=r\sin\theta$. If you sub these into the surface equation, it ends up as z=r so I made my limits be $r \in [0,z]$ and $\theta \in [0,2\pi]$. Then integrated $$\int \limits_{\theta=0}^{2\pi} \int \limits_{z=0}^1 \int \limits_{r=0}^z (2+r\cos\theta) r\,dr\,dz\,d\theta$$ and then got $2\pi/3$.
But in the solution, it was $2\pi/3-\pi$ which I don't understand why.
$S$ is not a closed surface, so you need to include the integral over the part of the boundary the volume that is not part of $S$: in particular, in your case this is the disc $x^2+y^2<1$, $z=1$.
$F \cdot n$ is equal to (choosing the outward normal!) $z$, so it is constant and equal to $1$. Thus the surface integral is just the area of the disc, which is $\pi$. You subtract it because it should start off on the same side as the integral over the rest of the surface.
(Don't worry, everyone makes this mistake in vector calculus...)