Here is the problem: "Verify Gauss Theorem for $F:\mathbf{R}^3 \rightarrow\mathbf{R}^3$ with $F(x,y,z)=(xz,0,0)$ in the positively oriented solid $S$, bounded by the paraboloid $\{x^2+y^2\}$ and the plane $\{z=1\}$"
Also, here it is the definition of the theorem i have:
Gauss Theorem: Let $F=(P,Q,R):U \subset \mathbf{R}^3\rightarrow{\mathbf{R}^3}$ which is $C^1$ in $U$, and $\Omega \subset U$ a Jordan measurable set such that $S=Fr(\Omega)=\partial \Omega$ it is a surface by pieces and $\Omega \cup S \subset U$ then we have that: $$\int_{\Omega}^{} \nabla \cdot \, F=\int_{S=\partial \Omega}^{}F \cdot d\sigma $$ where $\sigma$ is a parametrization.
The thing is that i think it verifies Gauss Theorem but i don't get that when i do the integrals, here is my attempt:
Let's define: $$\Omega=\{(x,y,z): 0 \leq z \leq x^2+y^2 \}, z \in [0,1] $$ We want to calculate: $$\int_{\partial \Omega }^{}F \cdot d\sigma $$ We define the following parametrization: $\sigma_{1,2}:B\subset\mathbf{R}^2\xrightarrow{}\mathbf{R}^3$ $$\sigma_1 (u,v)=(u,v,u^2+v^2) $$ $$\sigma_2(u,v)=(u,v,0) $$ We then proceed to find the normal vectors $\frac{\partial \sigma_i}{\partial u} \times \frac{\partial \sigma_i}{\partial v} $
$$\frac{\partial \sigma_1}{\partial u} \times \frac{\partial \sigma_1}{\partial v}=(-2u,-2v,1) $$
Then i multiply this one for a $-1$ because i need it to point the other direction, so what i get is $(2u,2v,-1)$. Also $$\frac{\partial \sigma_2}{\partial u} \times \frac{\partial \sigma_2}{\partial v}=(0,0,1) $$ We do calculate $F\cdot d\sigma_1=2u^2(u^2+v^2)$ and $F\cdot d\sigma_2=0$. We proceed to calculate the integral (which i will do it in polar coordinates): $$ \int_{B}^{} F \cdot d\sigma_1 \cdot (\frac{\partial \sigma_1}{\partial u} \times \frac{\partial \sigma_1}{\partial v})=\int_{0}^{2 \pi}\int_{0}^{1} 2 r^5 cos^2 \theta dr d\theta= \frac{\pi}{3}$$ (i also did use symbolab to compute it just to make sure i did not make an error myself). By the other hand, we need to calculate the following integral: $$\int_{\Omega}^{} \nabla \cdot F $$ $\Omega$ is the same as before, we then proceed to calculate the integral in cylindrical coordinates, we also have that $\nabla \cdot F= z$ so $$\int_{\Omega}^{} \nabla \cdot F=\int_{0}^{2 \pi}\int_{0}^{1} \int_{0}^{r^2} hr \, dh dr d \theta=\frac{\pi}{6}$$
And as i said at the beginning, i do not know what i am doing wrong, did i messed up at the parametrization? or it does not satisfies the conditions of the theorem?... any hint will be useful, thank you.
Please note your second integral is set up incorrectly. Otherwise you working is fine.
So it should be $\, \displaystyle \int_{0}^{2 \pi}\int_{0}^{1} \int_{r^2}^{1} \, h \, r \, dh \, dr \, d\theta$
Note the limits of $h$.