Parametric Equation

The parametric specification for a sphere consists of the three equations

$x = r \cos\theta \sin\phi$ $y = r cos\phi$, $z = r \sin\theta sin\phi$

where $(\theta, \phi) \in W$, $r$ is radius of the sphere

The parameter space $W$ being a subset of the plane $\mathbb{R}^{2}$

Normals

We shall now proceed with deriving the normals for points $p(x, y, z)$ of the above parametric equation.

Normals for parametric surfaces are computed by the equation below:

$\Large{\hat{n} = \frac{\frac{\partial{P}}{\partial{\theta}} \times \frac{\partial{P}}{\partial{\phi}}}{|\frac{\partial{P}}{\partial{\theta}} \times \frac{\partial{P}}{\partial{\phi}}|} }$

so

$\frac{\partial{x}}{\partial{\phi}} = -r\sin\theta\sin\phi$, $\frac{\partial{z}}{\partial{\phi}} = 0$ $\frac{\partial{y}}{\partial{\phi}} = r\cos\theta\sin\phi$

and

$\frac{\partial{x}}{\partial{\theta}} = r\cos\theta\cos\phi$, $\frac{\partial{z}}{\partial{\theta}} = -r\sin\phi$ $\frac{\partial{y}}{\partial{\theta}} = r\sin\theta\cos\phi$

haven found the partial deriviatives above we will now apply cross product on them to find the normal we get

$nx = r^2\cos\theta\sin^2\phi$ $ny = r^2\sin^2\theta\cos\phi\sin\phi + r^2\cos^2\theta\cos\phi\sin\phi$ $nz = r^2\sin\theta\sin^2\phi$

Lets try to simplify $ny$ by extracting common terms:

$ny = r^2\cos\phi\sin\phi(\sin^2\theta + 2\cos^2\theta)$

We know from trig identites that $(\sin^2\theta + 2\cos^2\theta) = 1$ therefore:

$ny = r^2\cos\phi\sin\phi$

We have found $n$ we now need to find $\hat{n}$ which can be found by dividing $n$ by its magnitude $|n|$

$|n| = \sqrt{(r^2\cos\theta\sin^2\phi)^2 + (r^2\cos\phi\sin\phi)^2 + (r^2\sin\theta\sin^2\phi)^2}$

$\sqrt{r^4\cos^2\theta\sin^4\phi + r^4\cos^2\phi\sin^2\phi + r^4\sin^2\theta\sin^4\phi}$

$\sqrt{r^4(\cos^2\theta\sin^4\phi + \cos^2\phi\sin^2\phi + \sin^2\theta\sin^4\phi)}$

$\sqrt{r^4((\cos^2\theta\ + \sin^2\theta)sin^4\phi + \cos^2\phi\sin^2\phi)}$

$\sqrt{r^4(sin^4\phi + \cos^2\phi\sin^2\phi)}$

$\sqrt{r^4\sin^2\phi(sin^2\phi + \cos^2\phi)}$

$\sqrt{r^4\sin^2\phi}$

$r^2\sin\phi$

we have found the magnitude of $n$ to be $r^2\sin\phi$, now we need to divide $n$ by it to get $\hat{n}$

$\hat{nx} = \frac{r^2\cos\theta\sin^2\phi}{r^2\sin\phi}$ $\hat{ny} = \frac{r^2\cos\phi\sin\phi}{r^2\sin\phi}$ $\hat{nz} = \frac{r^2\sin\theta\sin^2\phi}{r^2\sin\phi}$

$\hat{nx} = \cos\theta\sin\phi$ $\hat{ny} = \cos\phi$ $\hat{nz} = \sin\theta\sin\phi$

Turns out that the normal of a point on a sphere of any radius is the same point on a unit sphere 😉