$B$ and $B'$ lie on a circle of latitude (black arc) of a unit sphere. The great arc between $B$ and $B'$ is also shown (the orange arc). Given:

  • the azimuthal difference $\alpha$ between $B$ and $B'$,
  • the zenith angle for the circle $\theta$ of latitude,

what is the area on the unit sphere between the two arcs between $B$ and $B'$?

I tried to approach the problem by first parametrizing the great circle and then integrating the length of the longitudinal arcs between $B$ and $B'$, but I don't know how to express the angle between the great circle and the $xOy$ plane.

I'm only aware of two special cases:

  • when $\alpha=\pi$, the area is half of a spherical cap $$ \displaystyle\int_0^{\theta}\pi\sin\theta\ \mathrm d\theta = \pi (1-\cos\theta),$$
  • when $\theta=\pi/2$, the area is $0$ because the two arcs overlap.

Answers that would help my understanding of the question could

  • express the angle between the great circle and $xOy$ plane and then look for the area, or

  • use other methods that are perhaps more suitable or elegant.

Illustration of the region whose area I am looking for


1 Answer 1



$$\cot \theta=a \cos(\phi-\beta)$$

Let the end points be $(\theta_{0},0)$ and $(\theta_{0},\phi_{0})$, then

$$\cot \theta_{0}=a \cos \beta=a \cos (\phi_{0}-\beta)$$


\begin{align*} \beta &= \frac{\phi_{0}}{2} \\ a &= \frac{\cot \theta_{0}}{\cos \frac{\phi_{0}}{2}} \\ dA &= \sin \theta \, d\theta \, d\phi \\ A &= 2\int_{0}^{\beta} (\cos \theta_{0}-\cos \theta) \, d\phi \\ &= 2\int_{0}^{\beta} \left( \cos \theta_{0}- \frac{\cot \theta}{\sqrt{1+\cot^2 \theta}} \right) d\phi \\ &= 2\int_{0}^{\beta} \left[ \cos \theta_{0}- \frac{a\cos (\phi-\beta)}{\sqrt{1+a^2 \cos^2 (\phi-\beta)}} \right] d\phi \\ &= 2\int_{0}^{\beta} \left( \cos \theta_{0}- \frac{a\cos \phi}{\sqrt{1+a^2 \cos^2 \phi}} \right) d\phi \\ &= 2\left[ \beta \cos \theta_{0}- \tan^{-1} \left( \frac{a\sin \beta}{\sqrt{1+a^2\cos^2 \beta}} \right) \right] \\ &= \phi_{0} \cos \theta_{0}- 2\tan^{-1} \left( \cos \theta_{0} \tan \frac{\phi_{0}}{2} \right) \end{align*}

Alternative Method

Considering a spherical triangle with $a=b=\theta_{0}$ and $C=\phi_{0}$.


\begin{align*} \tan \frac{A+B}{2} &= \frac{\cos \frac{a-b}{2}}{\cos \frac{a+b}{2}} \cot \frac{C}{2} \\[3pt] &= \frac{\cot \frac{\phi_{0}}{2}}{\cos \theta_{0}} \\[5pt] \Delta &= A+B+C-\pi \\[3pt] &= 2\tan^{-1} \frac{\cot \frac{\phi_{0}}{2}}{\cos \theta_{0}}+ \phi_{0}-\pi \\[3pt] &= \phi_{0}-2\tan^{-1} \left( \cos \theta_{0} \tan \frac{\phi_{0}}{2} \right) \\[3pt] \text{the required area} &= \Delta-\int_{0}^{\theta_{0}} \int_{0}^{\phi_{0}} \sin \theta \, d\theta \, d\phi \\[3pt] &= \Delta-(1-\cos \theta_{0}) \phi_{0} \\[3pt] &=\phi_{0} \cos \theta_{0}- 2\tan^{-1} \left( \cos \theta_{0} \tan \frac{\phi_{0}}{2} \right) \end{align*}

  • $\begingroup$ I'm new to spherical trigonometry, so I would like to ask why is $\cot\theta=a\cos(\phi-\beta)$? Furthermore, what do $a$ and $\beta$ represent? $\endgroup$
    – Frenzy Li
    Aug 11, 2016 at 10:00
  • $\begingroup$ They are integration constants for geodesic equations. $\endgroup$ Aug 11, 2016 at 10:06
  • 1
    $\begingroup$ Have you learnt calculus of variation? May refer to this $\endgroup$ Aug 11, 2016 at 10:08
  • $\begingroup$ Answer updated with corrections. $\endgroup$ Aug 11, 2016 at 11:19
  • $\begingroup$ Actually, an absolute value would take care of the sign. Thank you very much for showing these two different approaches. $\endgroup$
    – Frenzy Li
    Aug 11, 2016 at 11:27

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