How to determine the outward normal for spehrical and cylindrical coordinates?

Question

Suppose I have a spherical surface $x^2+y^2+z^2=1$, and I correspondingly parametrize it as $r(\phi, \theta)=<\sin\phi\cos\theta,\ \sin\phi\sin\theta,\ \cos\phi>$. To find out the direction of normal, we compute $r_\phi\times r_\theta$ or $r_\theta\times r_\phi$, and they are in opposite direction. How do we know which is outward normal and how do we prove it?

Same situation in cylindrical coordinates. Suppose I have $x^2+y^2=1$ and I correspondingly parametrize it as $r(z,\theta)=<\cos\theta, \sin\theta, z>$. To find out the direction of normal, we compute $r_z\times r_\theta$ or $r_\theta\times r_z$. How do we know which is outward normal?

Answer 1

To find the orientation of a cross-product $a \times b$, point your right thumb in the direction of $a$ and point your right index finger in the direction of $b$. Your right middle finger will now be pointing approximately in the direction of $a \times b$.

Let us now imagine that the sphere is the surface of the Earth, so $r(\theta, \phi)$ is a city/village/whatever on the surface of the Earth. $r_\theta$ points in the direction of increasing $\theta$, and $r_\phi$ points in the direction of increasing $\phi$. In plain English, $r_\theta$ points south and $r_\phi$ points east.

If you now use the right-hand finger trick, you'll see that $r_\theta \times r_\phi$ points towards the sky. So $r_\theta \times r_\phi$ points radially outwards, and is the outward normal.

Answer 2

You should be able to determine which direction a vector has relative to the surface by checking the angle between the vector in question and the vector going from the origin to the surface point.

For a point on the spherical surface, the vector from the origin is

$$ \mathbf{r} = \begin{pmatrix} \sin \phi \cos \theta \\ \sin \phi \sin \theta \\ \cos \phi \end{pmatrix} $$

Calculating $\mathbf{r}_\phi$ and $\mathbf{r}_\theta$:

$$ \mathbf{r}_\phi = \begin{pmatrix} \cos \phi \cos \theta \\ \cos \phi \sin \theta \\ - \sin \phi \end{pmatrix} % \qquad % \mathbf{r}_\theta = \begin{pmatrix} - \sin \phi \sin \theta \\ \sin \phi \cos \theta \\ 0 \end{pmatrix} $$

Define the vector $\mathbf{v}$ as the cross product $\mathbf{r}_\phi \times \mathbf{r}_\theta$:

$$ \mathbf{v} = \mathbf{r}_\phi \times \mathbf{r}_\theta = \begin{pmatrix} \sin^2 \phi \cos \theta \\ \sin^2 \phi \sin \theta \\ \sin \phi \cos \phi \end{pmatrix} $$

Finding the magnitude, we obtain $|\mathbf{v}| = \sin \phi$ so $\mathbf{v}$ makes sense as a normal vector as long as $0 < \phi < \pi / 2$.

Now consider the angle $\alpha$ between $\mathbf{r}$ and $\mathbf{v}$, we can find its cosine using the scalar product:

$$ |\mathbf{v}| \cos \alpha = \mathbf{r} \cdot \mathbf{v} = \sin^3 \phi \cos^2 \theta + \sin^3 \phi \sin^2 \theta + \sin \phi \cos^2 \phi $$

which simplifies to $$ |\mathbf{v}| \cos \alpha = \sin \phi \quad \Rightarrow \cos \alpha = 1 \quad \Rightarrow \alpha = 0 $$

So, with the assumption that $0 < \phi < \pi$, we have shown that $\mathbf{v} = \mathbf{r}_\phi \times \mathbf{r}_\theta$ points in the same direction as $\mathbf{r}$. I.e. that $\mathbf{v}$ is an outward normal.

A similar method can be applied using cylindrical coordinates.