Calculate the integral with divergence theorem 
Given $F=(xz,y,z^2)$ and $E=\{(x,y,z)\;|\;y\geq0,x^2+y^2+z^2\leq4\}$ calculate the integral $\int_{\partial E} \langle F,N\rangle$ in two ways using the divergence theorem.

First $\text{div}(F)=3z+1$ .
Second $\int_E \text{div}(F)=\int_{\partial E} \langle F,N\rangle$ , the left part I calculated and got $16\pi$ ,for right part.
$$\int_{\partial E} \langle F,N\rangle=\int_{y=0}\langle F,N_1\rangle+\int_{x^2+y^2+z^2=4}\langle F,N_2\rangle$$
with $N_1=(0,-y,0),N_2=\frac{1}{2}(x,y,z)$. For the integral with $N_1$ wasn't sure how to proceed from here.
 A: As already pointed out by Math Lover, the left part is
$$\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}(3z+1)\, dV=\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}1\, dV=\frac{1}{2}\left(\frac{4 \pi}{3}\cdot 2^3\right)=\frac{16 \pi}{3}$$
where the integral of $3z$ is zero by symmetry.
For the right part, the integral over the disc is
$$\iint_{y=0,x^2+z^2\leq 4}(xz,y,z^2)\cdot(0,-1,0)\,dxdz=-\iint_{y=0,x^2+z^2\leq 4}y\,dxdz=0.$$
As regards the integral over the half surface of the sphere, by the using spherical coordinates
$$x=2\sin(\phi)\cos(\theta),\quad y=2\sin(\phi) \sin(\theta),\quad z= 2\cos(\phi),\quad dS=2\sin(\phi)d\theta d\phi,$$ we get
$$\begin{align}\iint_{y\geq 0,x^2+y^2+z^2= 4}(xz,y,z^2)\cdot\frac{(x,y,z)}{2}\,dS
&=\iint_{y\geq 0,x^2+y^2+z^2= 4}\frac{x^2z+y^2+z^3}{2}\,dS\\
&=\iint_{y\geq 0,x^2+y^2+z^2= 4}\frac{y^2}{2}\,dS\\
&=\int_{\phi = 0}^{\pi} \int_{\theta = 0}^{\pi}4\sin^2(\phi) \sin^2(\theta)\,\sin(\phi)d\theta d\phi\\
&=4\int_{0}^{\pi} \sin^3(\phi)\, d\phi \cdot\int_{0}^{\pi}\sin^2(\theta)\,d\theta \\
&=4\,\frac{4}{3}\,\frac{\pi}{2}=\frac{16\pi}{3}.
\end{align}$$
Note that the integral of $\frac{x^2z+z^3}{2}$ is zero by symmetry.
