Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

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 $\theta \in (0,2\pi)$, $\rho \in (0,R)$, $\phi \in (0,\arcsin(\frac h R))$

share|cite|improve this question
$\rho$ 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
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 $h_1 \le z \le h_2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.