Flux of $\mathbf{F}=(xz,yz,x^2+y^2)$ across paraboloid $z=4-x^2-y^2,\ z\geq 0$ I have calculated the flux of the vector field $\mathbf{F}=\begin{bmatrix}xz\\ yz\\ x^2+y^2\end{bmatrix}$ outward across the surface of the paraboloid $S$ given by $z=4-x^2-y^2,\ z\geq 0$ (with outward-pointing normal having positive $\mathbf{e}_3$ component but my answer doesn't agree with the one given by the book but I can't see what I am doing wrong. I would thus appreciate some feedback about what I am missing in my solution, thanks.

EDIT: there was an error in $D\mathbf{g}$ that's why the result was wrong. Now the calculation is correct.

What I have done:
We parametrize the surface by $\mathbf{g}:(0,2)\times (0,2\pi),\ \mathbf{g}\begin{pmatrix}r\\ \theta\end{pmatrix}=\begin{bmatrix}r\cos(\theta)\\ r\sin(\theta)\\ 4-r^2\end{bmatrix}$ so $D\mathbf{g}=\begin{bmatrix}\cos(\theta) & -r\sin(\theta)\\ \sin(\theta) & r\cos(\theta)\\ -2r & 0\end{bmatrix}$ and
\begin{align}
\text{flux}_{S}(\mathbf{F})&=\int_{S}\eta=\int_{(0,2)\times (0,2\pi)}\mathbf{g}^*\left(xz\ dy\wedge dz+yz\ dz\wedge dx+(x^2+y^2)\ dx\wedge dy\right)\\&=\int_{\theta=0}^{\theta=2\pi}\left(\int_{r=0}^{r=2}\left( r\cos(\theta)(4-r^2)(2r^2\cos(\theta))+r\sin(\theta)(4-r^2)(2r^2\sin(\theta))+r^2\cdot r \right)dr\right)d\theta\\&=\int_{\theta=0}^{\theta=2\pi} \left(\int_{r=0}^{r=2}(2r^3(4-r^2)+r^3dr\right)d\theta=\int_{\theta=0}^{\theta=2\pi} \left(\int_{r=0}^{r=2}(9r^3-2r^5)dr\right)d\theta\\ &=2\pi\cdot \left[\frac{9}{4}r^4-\frac{1}{3}r^6\right]_{r=0}^{r=2}=2\pi\left[36-\frac{64}{3}\right]=\frac{88\pi}{3}.
\end{align}

I got a different result,
$$\begin{align}
\iint_S \mathbf{F}\cdot d \mathbf{S}&=\iint_{x^2+y^2\leq 4}
\langle (xz,yz,x^2+y^2),(2x,2y,1)\rangle dx dy\\
&=\int_{0}^{2\pi}\int_{r=0}^2(2r^2(4-r^2)+r^2)r drd\theta\\
&=2\pi\int_{r=0}^2(9r^3-2r^5)\,dr=2\pi\left[9r^4/4-r^6/3\right]_0^2=\frac{88\pi}{3}.
\end{align}$$
The result is confirmed by applying the divergence theorem:
$$\iint_S \mathbf{F}\cdot d \mathbf{S}=
\iiint_D \text{div}(\mathbf{F})\,dxdydz-\iint_{S_2} \mathbf{F}\cdot d \mathbf{S}=\frac{64\pi}{3}-(-8\pi)=\frac{88\pi}{3}$$
where $D=\{(x,y,z):0\leq z \leq 4-x^2-y^2\}$ and $S_2=\{(x,y,0):x^2+y^2\leq 4\}$ is oriented downward.
