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 $(0,0,1)$.

enter image description here

  • 2
    $\begingroup$ The problem is worse than that: On the intersection of the plane with the sphere, the partial derivatives $f_x$ and $f_y$ aren't even defined (check this yourself), so the formula for the normal isn't even meaningful there. Of course, this is consequence of the fact that our parameterization isn't well-defined at such points. $\endgroup$ Commented Dec 9, 2015 at 19:48

3 Answers 3


I think a large part of the difficulty here is the variation in the magnitudes of your "normals" in general.

When you give a "normal" in the form $\langle -f_x, - f_y, 1\rangle$, basically what you have is a radial projection of the hemisphere parameterized by $\langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi\rangle$ (for $0 \leq \phi \leq \frac\pi2$) onto the plane $z = 1$. That is, you have a projection that takes a point $P$ on the surface of the hemisphere to a projected point $P'$ on the plane such that $P$ and $P'$ are collinear with the center of the sphere. (This is also called a gnomonic projection of the sphere.) The coordinates of your "normals" are the coordinates of the projected points.

As long as $0 \leq \phi < \frac\pi2$, each point of your sphere does in fact project onto that plane, although as $\phi$ approaches $\frac\pi2$ the magnitudes of your "normals" grow without bound. And obviously the boundary of your hemisphere, where $\phi = \frac\pi2$, does not project on to the plane $z = 1$ at all.

If you change your choice of $f$ as recommended in the answer by H.R., this problem goes away. All your normals then will have the same magnitude, and they will be defined at all points of the hemisphere.

When you give your "normals" in the form $\langle\sin^2\phi\cos\theta, \sin^2\phi\sin\theta, \sin\phi\cos\phi\rangle$, you again have non-uniform magnitudes, but this time the magnitudes go to zero as $\phi$ approaches zero. In effect, you are radially projecting the hemisphere onto a kind of degenerate torus given by $r = \sin\phi$.

But notice that the three components of your "normals" all have the common factor $\sin\phi$. You can normalize the magnitudes of all your "normals" (except for the case $\phi = 0$) by multiplying by the scalar $1/\sin\phi$. If you do this, you get the vectors $\langle\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi\rangle$, that is, the coordinates of each vector (for $0 < \phi \leq \frac\pi2$) are simply the coordinates of the point on the sphere. If you define these vectors as the normals for all points such that $0 < \phi \leq \frac\pi2$, you may see that you can use the same formula to define the normal for $\phi = 0$ as well, and it works very nicely.

The formula ${\bf r}_\phi \times {\bf r}_\theta$ works nicely for the measurement of area (which is what it is used for in the example you quoted) precisely because its magnitude does go to zero as $\phi$ goes to zero and does so in the same way that the area element $r \,d\phi\,d\theta$ goes to zero. But as you observed, this property is not so desirable when you are trying to construct a set of normal vectors rather than trying to integrate some scalar function over area.

  • $\begingroup$ This response is revealing! Thanks. When you project the hemisphere given by $\langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi\rangle$ onto the plane $z = 1$, why isn't the parametrization of this projection simply given by $\langle \sin\phi \cos\theta, \sin\phi \sin\theta, 1\rangle$. In other words, how are the coordinates of the projected points the coordinates of the normals? $\endgroup$
    – Muno
    Commented Dec 10, 2015 at 0:22
  • $\begingroup$ These are not orthogonal projections, as if by a light source infinitely far away; they are radial projections, as if by a light source at the center of the sphere. I have modified the answer to indicate more clearly what kind of projections I mean. $\endgroup$
    – David K
    Commented Dec 10, 2015 at 1:36
  • $\begingroup$ You wrote that "The formula ${\bf r}_\phi \times {\bf r}_\theta$ works nicely for the measurement of area (which is what it is used for in the example you quoted) precisely because its magnitude does go to zero as $\phi$ goes to zero." In spite of some parametrizations being more well suited than others for certain purposes, is it true that all parametrizations can work for all purposes. For example, if I try to use the magnitude of $\nabla f$ where $$\nabla f=\frac{x}{\sqrt{1-x^2-y^2}}{\bf{i}}+\frac{y}{\sqrt{1-x^2-y^2}}{\bf{i}}+{\bf{k}}$$ in calculating a sphere's surface area, would work? $\endgroup$
    – Muno
    Commented Dec 10, 2015 at 19:43
  • $\begingroup$ I raise this thought, because $$ |\nabla f| = \frac{1}{1-x^2-y^2}$$ and so the surface area element around the perimeter of the hemisphere's base would be undefined, since $x^2 + y^2 == 1$ there. $\endgroup$
    – Muno
    Commented Dec 10, 2015 at 19:45
  • $\begingroup$ For surface area, it's OK to integrate a function that is not defined at the boundary. The boundary has zero area. It's even OK for the function to be undefined at some points inside the boundary. What you need is that the function be defined on a part of the hemisphere over which you can integrate, and that the area of this part of the hemisphere equals the area of the whole hemisphere. $\endgroup$
    – David K
    Commented Dec 10, 2015 at 21:11

Cartesian Coordinates

I think you should look closer to understand why this happens. If we write the implicit equation of the upper semi-sphere as


and then compute the $\nabla f$ to get

$$\nabla f=\frac{x}{\sqrt{1-x^2-y^2}}{\bf{i}}+\frac{y}{\sqrt{1-x^2-y^2}}{\bf{i}}+{\bf{k}}$$

then it is evident that this is not defined on the circle $x^2+y^2=1$ which lie on the plane $z=0$. However, there is a better way to define $f(x,y,z)$


and hence

$$\nabla f=2x{\bf{i}}+2y{\bf{i}}+2z{\bf{k}}=2{\bf{x}}$$

Spherical Coordinates

The radial unit vector in spherical coordinate is normal to the surface of unit sphere and it is given by

$${\bf{e}}_r= \sin\phi \cos\theta {\bf{i}}+ \sin\phi \sin\theta {\bf{j}}+ \cos\phi {\bf{k}}$$

and when $\theta=0$ and $\phi=0$ we get

$${\bf{e}}_r= {\bf{k}}$$

so in this case you we have no problems and everything is well-defined.

enter image description here

About the Example in the Book

You should note that

$${\bf{r}}_{\phi} \times {\bf{r}}_{\theta}=a^2 \sin\phi {\bf{e}}_r $$

and when just $\phi=0$ you will get ${\bf{r}}_{\phi} \times {\bf{r}}_{\theta}={\bf{0}}$. This is because the intersection of the $\phi=0$ (the positive part of $z$ axis) with the sphere is just a point and evaluating tangent for a point is meaning-less. You see that ${\bf{r}}_{\theta}$ which is meant to be the tangent to the curves of $\theta=\text{Const}$ at $\phi=0$ is $\bf{0}$ because there is no curves there actually! You can observe the same thing for $\phi=\pi$!

  • $\begingroup$ Would you know why ⟨sin(ϕ)2cos(θ),sin(ϕ)2sin(θ),sin(θ)cos(θ)⟩ doesn't work? This can be shown to give the normal if we parametrize $r$ (the surface) to be the standard equations from spherical to cartesian, ⟨sin(ϕ)cos(θ),sin(ϕ)sin(θ),cos(θ)⟩, and then take the cross product: $r_{\theta} \times r_{\phi}$ $\endgroup$
    – Muno
    Commented Dec 9, 2015 at 20:28
  • $\begingroup$ I think you have the wrong third coordinate: it should be $\cos \phi$, not $\cos \theta$. $\endgroup$
    – David K
    Commented Dec 9, 2015 at 20:40
  • $\begingroup$ @Muno: your parameterization is wrong! :) $\endgroup$ Commented Dec 9, 2015 at 20:41
  • $\begingroup$ @H.R. I updated my question with the parametrization I had in mind. $\endgroup$
    – Muno
    Commented Dec 9, 2015 at 20:54
  • 1
    $\begingroup$ @H.R. I meant both of my earlier comments to apply to Muno's first comment on this answer. I should have tagged him; my apologies for forgetting to do that. $\endgroup$
    – David K
    Commented Dec 10, 2015 at 15:07

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$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.


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