How do you derive the parametric equations for a sphere? \begin{align} x & = r \cos(\theta)\sin(\varphi), \\ y & = r \sin(\theta)\sin(\varphi), \\ z & = r \cos(\varphi), \end{align} where $\theta$ is from $0$ to $2\pi$ and $\varphi$ is from $0$ to $\pi$. There are no good explanations online.
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$\begingroup$If you're familiar with surfaces of revolution, the derivation is easy. A circle that is rotated around a diameter generates a sphere.
The parametric equations for a surface of revolution are:
$$ \left(f(u)\cos v, f(u)\sin v, g(u)\right) $$
Where $\left(f(u), g(u)\right)$ are the parametric equations of the rotated curve. For a circle, they are $\left(r \cos u, r \sin u\right)$. Therefore, the parametric equations of a sphere are:
$$ \left(r \cos u \cos v, r \cos u \sin v, r \sin u\right) $$
Change the variables from $(u, v)$ to $(\frac{\pi}{2} - \varphi, \theta)$ to get the equations in your question.
$\endgroup$ 2 $\begingroup$$\phi$ is the angle which polar radius makes with the $z$ axis. Hence, multiplying by $\sin \phi$ you find the projection of the polar radius on the $xy$ plane. Then multiplying by $\cos \theta$ and $\sin \theta$ respectively you find projections on axes $x$ and $y$
$\endgroup$ $\begingroup$I thought that it would be interesting to parameterise a sphere by using only 2 parameters $r$ and $\theta$ as follows\begin{align} x&=r\sin(\theta)\cos(n\theta),\\ y&=r\sin(\theta)\sin(n\theta),\\ z&=r\cos(\theta), \end{align}where $\theta$ is from $0$ to $\pi$ and $n$ is a reasonably large integer e.g. 64. This results in a spiraling curve which describes a sphere with radius r. Check out the visualisation here.
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