A spherical area bounded by three curves

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 characterised by y = z, and the blue arc as the great arc (center at the origin) characterised by x = y.

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

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

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

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

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

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

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

Hence, the integral becomes,

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

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

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

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

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

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