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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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What about $x^2+y^2+z^2 = 1$? – Pedro Tamaroff May 28 '12 at 21:41

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.

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$\phi$ 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 $y$

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