How to Calculate the flux of the Vector Field on the surface $z = 1-x^2-y^2$ ( getting normal vector $(0,0,0)$ at the point $(0,0,1)$ ?!!! )

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Let $S$ be the surface $z = 1-x^2-y^2 , 0\leq z$.

Find $\int_{S} x^2z~dydz + y^2z~dzdx + (x^2+y^2)~dxdy$. Choose the direction of the normal upwards.

so i calculated the flux and i got that it is $\boxed{\frac{\pi}{2}}$. i am not sure if that is right

Parametrization $R(\theta,r) = (r\cos(\theta),r\sin(\theta),1-r^2)$

and the normal is $(2r^2\cos(\theta),2r^2\sin(\theta),r)$ i feel like its not right but i calculated it many times at the point $(0,0,1) ~ r = 0$ but the normal is $(0,0,0)$

$\int_{S}\vec{F} _\dot{} \vec{n}$ = $\int_{\theta = 0}^{\theta = 2\pi}\int_{r=0}^{r=1}~~(2r^4-2r^6)(\sin^3{\theta}+\cos^3{\theta}) + r^3~~drd\theta$