TITLE: Why do parametrizations to the normal of a sphere sometimes fail?
QUESTION [7 upvotes]: If I take the upper hemisphere of a sphere, $x^2 + y^2 + z^2 = 1$, to be $\sqrt{1 - x^2 - y^2}$, then the normal is given by $\langle -f_x, - f_y, 1\rangle$ at any point.
This leads to an odd result: on the plane $z = 0$, while one might expect all normals of the sphere to not have any $z$ component (i.e, to only point radially outwards), the $\vec{k}$ component is still 1.
A similar parametrization in cylindrical coordinates is:
$\langle \sin(\phi)^2 \cos(\theta), \sin(\phi)^2 \sin(\theta), \sin(\phi) \cos(\phi)\rangle$. 
At $\phi = 0$ and $\theta = 0$, which corresponds to the point $(0,0,1)$ in cartesian coordinates, the normal is $\langle 0,0,0 \rangle$ while one would expect $(0,0,1)$.

REPLY [3 votes]: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$Your intuition is probably based on unit normal vectors. The (outward) unit normal vector $\Vec{n}$ to the unit sphere at the point $(x, y, z)$ is $x\Vec{i} + y\Vec{j} + z\Vec{k}$, and indeed, the unit normals along the equator $\{z = 0\}$ have vanishing $z$ component while the unit normal at the north pole $(0, 0, 1)$ is $\Vec{k}$.
What happens in your examples is this: Let $\Vec{X}$ be a smooth parametrization of a surface $S$ in $\Reals^{3}$, and let $\Vec{X}_{u}$ and $\Vec{X}_{v}$ denote the partial derivatives of $\Vec{X}$. If the partials are linearly independent for parameter values $(u, v)$, their cross product
$$
\Vec{X}_{u} \times \Vec{X}_{v}
  = \|\Vec{X}_{u} \times \Vec{X}_{v}\|\, \underbrace{\frac{\Vec{X}_{u} \times \Vec{X}_{v}}{\|\Vec{X}_{u} \times \Vec{X}_{v}\|}}_{\Vec{n}}
$$
is normal to $S$ at $\Vec{X}(u, v)$, but has length $\|\Vec{X}_{u} \times \Vec{X}_{v}\|$.
Write $f(u, v) = \sqrt{1 - u^{2} - v^{2}}$. For the graph parametrization $\Vec{X}(u, v) = \bigl(u, v, f(u, v)\bigr)$, you have
$$
\Vec{X}_{u} \times \Vec{X}_{v} = -f_{u}\Vec{i} - f_{v} \Vec{j} + \Vec{k} = \frac{1}{f(u, v)} \underbrace{\bigl(-u \Vec{i} -v \Vec{j} + f(u, v)\Vec{k}\bigr)}_{\Vec{n}};
$$
the $z$ component of the unit normal approaches $0$ along the equator, as expected, while the magnitude of the cross product itself grows without bound in such a way that the $z$ component is identically equal to $1$.
For the spherical coordinates parametrization
$$
\Vec{X}(\theta, \phi) = a\cos\theta \sin\phi \Vec{i} + a\sin\theta \sin\phi \Vec{j} + a\cos\phi \Vec{k},
$$
the partial $\Vec{X}_{\theta} = a\sin\phi(-\sin\theta \Vec{i} + \cos\theta \Vec{j})$ vanishes at the poles $\phi = 0$ and $\phi = \pi$ (geometrically, longitudes all meet at the poles, and so latitude circles shrink to points), so the cross product vanishes at the poles, too:
$$
\Vec{X}_{\phi} \times \Vec{X}_{\theta} = a^{2} \sin\phi \underbrace{(\cos\theta \sin\phi \Vec{i} + \sin\theta \sin\phi \Vec{j} + \cos\phi \Vec{k})}_{\Vec{n}}.
$$
The unit normal, however, has the "expected" behavior even at the poles.