# In the area of sphere, which angle with $0\leq \alpha<2\pi$ and which angle with $0\leq \beta < \pi$?

I am lacking the skill of visualizing the problem, picture here, to decide the right intervals. The way I do it currently is to try things but the technique fails with anything more complicated to 2D. So how do you know which angle correspond to which angle range?

[update]

Let's break this into parts. Suppose there are two nodes, one $N$ in the north pole and one $S$ in the south pole. The distance $N-S$ is $\pi$. The distance $N-N$ is either $0$ or $2\pi$. Now, please, break this problem with nodes and graph-theoretically. See the paths are clearly different, the ending points vary: one is even and one is odd (trivially even if $N-N=0$ length accepted).

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Should it be $\beta \leq \pi$ or are you trying to determine the area of some particular subset of the sphere? – Myself May 17 '11 at 17:38
To use your notation, $\alpha$ is longitude, and $\beta$ is co-latitude. – J. M. May 17 '11 at 17:39
No what I mean is that by letting $\beta$ run to $\phi$ you are doing something odd... It's like when you're integrating the area of a unit square and integrating $\int_{0}^1 \int_{0}^x dx\,dy$. Now my hint to visualise this would be to draw a rectangle of size $\pi\times 2\pi$ and try to imagine how it folds around the sphere like a world map with coordinates $\phi,\theta$. – Myself May 17 '11 at 17:45
@Myself: sorry, typo, I meant $\pi$, not $\phi$. – hhh May 17 '11 at 18:30

## 2 Answers

Look at a globe: To fix a position of a point on earth we give its geographical latitude $\theta$ and its geographical longitude $\phi$. The latitude $\theta$ runs from $90^\circ$ south (equivalent to $\theta=-{\pi\over2}$) at the south pole to $90^\circ$ north (equivalent to $\theta={\pi\over2}$) at the north pole. Along the equator the geographical latitude is $0^\circ$ or simply $0$. The geographical longitude $\phi$ is constant along meridians, it runs from $180^\circ$ west (equivalent to $\phi=-\pi$) to $180^\circ$ east (equivalent to $\phi=\pi$). The "dateline" $\phi=-\pi$ resp. $\phi=\pi$ is actually the same meridian on $S^2$, and the meridian $\phi=0$ is the meridian of Greenwich. We see that the sphere $S^2$ is covered essentially one-one by a $(\phi,\theta)$-coordinate system where $\phi$ runs from $-\pi$ to $\pi$ and $\theta$ from $-{\pi\over2}$ to ${\pi\over2}$. Note that under this representation the $\phi$-values of the north and south poles are undefined.

In many physical or technical situations, e.g. when studying the fine movement of a spinning top, it is more practical to have $\theta=0$ at the north pole. In this setup the angle $\theta$ takes values between $0$ and $\pi$. The formulas converting $(x,y,z)$-coordinates into spherical coordinates $(r,\phi,\theta)$ look a bit differently then, and the obvious symmetry $\theta\mapsto -\theta$ is not visible anymore.

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If you extend it to a globe, $d\theta$ represents the interval between two meridians of longitude, so should go from $0$ to $2\pi$. $d\phi$ represents the interval between two parallels of latitude, so should go over a range of $\pi$. On the globe, it is from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$, but usually we measure from the pole and range over $0$ to $\pi$

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@hhh: Since we measure from the pole and not from the equator, in spherical coordinates, we speak of "co-latitude" and not latitude. – J. M. May 17 '11 at 17:42