Area of the surface on a sphere surrounded by three externally tangent circles?

Question

Let us draw three circles of radius $\dfrac23$ on a sphere of radius $1$, all of which are mutually tangent (externally).

How do I calculate the area of the surface surrounded by the three circles?

Answer 1

Let $r_1, r_2, r_3$ be the center of the three circles. Let $\triangle$ be the spherical triangle formed by $r_1, r_2, r_3$. It's known that it's a equilateral triangle with side length $\frac 43$. Using the cosine formula on the sphere, we have

$$(*)\ \ \ \ \ \cos \left( 4/3\right) = \cos \left( 4/3\right)\cos \left( 4/3\right) + \sin\left( 4/3\right)\sin \left( 4/3\right) \cos A,$$

where $A$ is the angle (all three angles are the same) of $\triangle$. Now the area of the shaded $C$ is found by

$$\text{Area}(C) = \text{Area}(\triangle) - 3\text{Area}(B),$$

where $B$ is the sector with radius $2/3$ and angle $A$. Both of them can be calculated: we have

$$B = \int_0^{2/3} \int_0^A \sin r d\theta dr= A(1-\cos 2/3),$$

and

$$\text{Area}(\triangle)= (A + A+ A - \pi).$$

Thus

$$\begin{split} \text{Area}(C) &= 3A - \pi - 3 A(1-\cos 2/3) \\ &= 3A \cos 2/3 - \pi \end{split}$$

Numnerically, its around 0.11.