I had to evaluate the surface intregral $\int\int_s \operatorname{curl}\vec{F} \cdot \hat{n}\ dS $ where S is that part of surface of paraboloid   $z= 1 - x^2 -y^2$
for which $z \geqslant 0 $ and $ \vec{F} = y\hat{i} + z\hat{j} + x\hat{k}  $
I found the curl of $\vec{F}$ which was correct. But for finding $\hat{n}$ I assumed $\phi = z + x^2 + y^2 -1$ and used
$$ \hat{n} = \frac{\operatorname{grad}\phi}{|\operatorname{grad} \phi|} = \frac{2x \hat{i} +2y\hat{j}+\hat{k}}{3}$$
but in the book it is written that $\hat{n}$ is "obviously" $\hat{k}$ 
Is it correct or wrong and why?
 A: The normal vector of that surface should be
$$\frac{2x \hat{i} +2y\hat{j}+\hat{k}}{\sqrt{(2x)^2+(2y)^2+1}}$$
However I think the book is using Stoke's theorem. Instead of considering the surface $S$, you can consider the surface on the $xy$ plane that has the same boundary as $S$, which is $x^2+y^2=1, z=0$. The normal vector of that surface is $\hat{k}$.
A: To expand on the answer by @KittyL , recall that 
$$\nabla \cdot \nabla \times \vec A=0 \tag 1$$
for any sufficiently smooth $\vec A$.  
Using the Divergence Theorem along with $(1)$ reveals that 
$$\int_V \nabla \cdot \nabla \times \vec F dV=\oint_S \nabla \times \vec F \cdot \hat n\,dS=0 \tag 2$$
This implies that the closed surface integral of the Curl of a vector (here, $\vec F$) is zero.  
And in turn this means that we can split the closed surface integral on the right-hand side of $(2)$ into components (open surface integrals) that sum to zero.  Therefore, we have
$$\oint_S \nabla \times \vec F \cdot \hat n\,dS=\int_{S_1} \nabla \times \vec F \cdot \hat n\,dS+\int_{S_2} \nabla \times \vec F \cdot \hat n\,dS=0 \tag 3$$
where $S_1+S_2=S$.  Finally, we have from $(3)$
$$\int_{S_1} \nabla \times \vec F \cdot \hat n\,dS=-\int_{S_2} \nabla \times \vec F \cdot \hat n\,dS \tag 4$$
For the problem at hand, we view the original surface as $S_1$ and the surface $S_2$ as the surface defined by $z=0$, $x^2+y^2\le 1$.  Note that for $S_2$ the unit normal would be $-\hat z$.  Upon absorbing the minus sign on the right-hand side of $(4)$ yields the desired result
$$\int_{S_1=S} \nabla \times \vec F \cdot \hat n\,dS=\int_{S_2=\{z=0, x^2+y^2\le 1\}} \nabla \times \vec F \cdot \hat z\,dS$$
and we are done.
