A spherical area bounded by three arcs

We're going to consider the area shaded in light blue in the figure above. It is $\dfrac{1}{6}$ of the required

area (from symmetry). To compute its area we'll define the green arc as the great arc (center at the origin) and characterized by $y = z$ , and the blue arc as the great arc (center at the origin) characterized by $x = y$ .

$\text{Blue Area} = \large \displaystyle \int_{\phi= \phi_1}^{\frac{\pi}{4}} \int_{ \theta = \theta_1 }^{ \theta_0 } \sin \theta d\theta d\phi$

where $\phi_1$ is characterized by $\sin \theta_0 \sin \phi_1 = \cos \theta_0$

Therefore, $\phi_1 = \sin^{-1} ( \cot \theta_0 )$

And $\sin \theta_1 \sin \phi = \cos \theta_1$ (since $y = z$ ), thus $\theta_1 = \tan^{-1} (\dfrac{1}{\sin \phi} )$

After integration with respect to $\theta$ , the above integral becomes,

$\text{Blue Area} = \large \displaystyle \int_{\phi = \phi_1 }^{\frac{\pi}{4}} (\cos \theta_1 - \cos \theta_0) d\phi$

Since $\tan \theta_1 = \dfrac{1}{\sin \phi}$ then $\cos \theta_1 = \dfrac{\sin \phi}{\sqrt{1 + \sin^2 \phi} }$

Hence, the integral becomes,

$\text{Blue Area} = \large \displaystyle \cos \theta_0 (\phi_1 - \frac{\pi}{4}) + \int_{\phi =\phi_1}^{ \frac{\pi}{4}} \dfrac{\sin \phi}{\sqrt{1 + \sin^2 \phi} } d\phi$

Substituting $\sin^2 \phi = 1 - \cos^2 \phi$ , the integral becomes,

$\text{Blue Area} = \large \displaystyle \cos \theta_0 (\phi_1 - \frac{\pi}{4}) + \int_{\phi = \phi_1}^{\frac{\pi}{4}} \dfrac{\sin \phi}{\sqrt{2 - \cos^2 \phi }} d\phi$

$\text{Blue Area} = \large \displaystyle \cos \theta_0 (\phi_1 - \frac{\pi}{4}) + \int_{\phi = \phi_1}^{\frac{\pi}{4}} \dfrac{\sin \phi}{\sqrt{2}\sqrt{1 - (\cos \phi / \sqrt{2} )^2 }} d\phi$

Using the substituting $u = \dfrac{\cos \phi}{\sqrt{2}}$ , the integral becomes

$\text{Blue Area} = \cos \theta_0 (\phi_1 - \frac{\pi}{4}) + \sin^{-1} (\cos \phi_1 / \sqrt{2} ) - \dfrac{\pi}{6}$

Hence, finally, the required area is $6$ times the above expression, and is given by

$\text{Area} = 6 \left( \cos \theta_0 (\phi_1 - \frac{\pi}{4}) + \sin^{-1} (\cos \phi_1 / \sqrt{2} ) - \dfrac{\pi}{6} \right)$

Monte Carlo with Phyton.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

import  random
import  math
n=10000000
A=math.cos(math.pi*81/180)
f=0
for k in range(n):
  t=random.uniform(0,1)*2*math.pi
  z=random.uniform(0,1)*2-1
  x=math.sqrt(1-z*z)*math.cos(t)
  y=math.sqrt(1-z*z)*math.sin(t)
  if z>A and  x>A and y>A:
    f=f+1
print(f/n*4*math.pi)

0.907925202907