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D is a spherical segment, obtained by intersecting a sphere (with center the origin and radius R) with plans z=0, $z=h<R$. What is the range of variation of $\rho, \phi, \theta$ (spherical coordinates)?

PS. I believe that $\theta \in (0,2\pi)$, $\rho \in (0,R)$, $\phi \in (0,\arcsin(\frac h R))$

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$\rho$ will depend on $\phi, h, R$ in quite an ugly way. If it were an option, I'd use cylindrical coordinates. –  user20266 Jun 17 '12 at 17:42
    
how would you use cylindrical coordinates? –  Mark Jun 17 '12 at 20:35

1 Answer 1

up vote 3 down vote accepted

I do assume there are two parallel planes intersecting your ball, at height $h_1< h_2$, like in your picture, and a ball centered at the origin with radius $R$. Things are a bit simpler if $h_1=0$

In cylindrical coordinates $(r,\theta,z)$ (with $\theta$ describing a circle in the $(x,y)$ plane) the set you are looking at can be described as $$ \{ x = (r,\theta,z): \theta \in [-\pi,\pi), z\in [h_1,h_2], 0 \le r \le \sqrt{R^2-z^2 }\} $$

Depending on whether the planes are to be considered parts of your set you may want to replace the closed interval for the $z$ coordinate by an open interval.

In spherical coordinates, $\theta \in [-\pi,\pi) $ as before, and all these $\theta$ have to be taken into account. I'll assume that $\phi\in (-\pi,\pi)$ where $\pm \pi $ correspond to north and south pole respectively. With this convention you'll need to consider $\phi$ in the range $$\arcsin(h_1/R)\le \phi\le \pi$$ but have to distinguish between the cases $$\phi\le \arcsin(h_2/R)$$ and $$\phi\ge \arcsin(h_2/R)$$ In the first case you'll have, for given $\phi$, $$R\ge \rho \ge\rho_1 =\frac{ h_1}{\sin(\phi)} $$

(that is the part of a ray from the origin with angle $\phi$ relative to the $(x,y)$ plane from the plane $z=h_1$ to the boundary of the ball) whereas in the second case you'll have

$$ \frac{ h_2}{\sin(\phi)}= \rho_2\ge \rho \ge \rho_1 =\frac{ h_1}{\sin(\phi)}$$

which is the part of a ray with angle $\phi$ relative to the $(x,y)$ plane between $h_1 \le z \le h_2$

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