This might be a stupid question, but I think I am misunderstanding something. I was working on a calculus question which requires the calculation of a vector normal to a paraboloid. $z=a^2-x^2-y^2$ describes the surface of the paraboloid, with $z > 0$. I am required to calculate $$\iint_S \text{curl}(\mathbf{G})\cdot \mathbf{n} \;dS$$ Where $\mathbf{G}(x,y,z)=y\mathbf{i}+z\mathbf{j}+x\mathbf{k}$. For the sake of practice, I want to do this without using Stokes' theorem.
The first thing I did was parametrize the surface; $$\mathbf{r}(r,\theta) = r\cos\theta\mathbf{i} + r\sin\theta\mathbf{j} + (a^2 - r^2)\mathbf{k},\qquad 0\leq r\leq a, \quad0\leq\theta\leq 2\pi$$ Then, as far as I've been taught, I can find a vector normal to this surface with $$\mathbf{n} = \mathbf{r}_r\times\mathbf{r}_\theta$$ Where $\mathbf{r}_r$ is the partial derivative of $\mathbf{r}$ with respect to $r$, and $\mathbf{r}_\theta$ is the partial derivative with respect to $\theta$. This gives $$\mathbf{n} = (2r^2\cos\theta, 2r^2\sin\theta, r)$$ But this doesn't seem right to me. If $r=0$, we are clearly talking about the top of this paraboloid - but this gives $\mathbf{n} = (0,0,0)$, where I might have expected $(0,0,1)$. Where is my misunderstanding?
The problem is that the normal vector you found by taking a cross product is not a unit normal vector. The vector field it generates on the surface of the paraboloid vanishes at $r=0$.
If you normalize $\vec{n}$ to get $\vec{\nu} = \frac{1}{||\vec{n}||}\vec{n}$, you'll find \begin{align} \vec{\nu} &= (r^2(1+4r^2))^{-1/2}(2r^2\cos\theta,2r^2\sin\theta,r)\\ &= \frac{1}{r\sqrt{1+4r^2}}(2r^2\cos\theta,2r^2\sin\theta,r) \\ &= \frac{1}{\sqrt{1+4r^2}}(2r\cos\theta,2r\sin\theta,1), \end{align} for all $r\neq0$. Now this unit normal vector field does not vanish at $r=0$, and in fact varies continuously. The limit at $r=0$ is exactly the unit normal you're looking for, i.e. $(0,0,1)$.