# 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 $(0,0,1)$ ?!!! )

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 $$(0,0,0)$$

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

• anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0) – Mather Jan 15 at 22:47
• Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$. – Maxim Jan 16 at 4:30
• But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ? – Mather Jan 16 at 7:40
• The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r \cos \theta, 2 r \sin \theta, 1)/\sqrt{1 + 4 r^2}$. – Maxim Jan 16 at 8:04