What is the area of intersection between two adjacent spherical caps?

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SUMMARY

The discussion focuses on calculating the surface area of a sphere intersecting with a cube, specifically when the sphere's radius (r) varies. Three distinct scenarios are outlined based on the radius: when the sphere is entirely within the cube (0 < r < L/2), when parts of the sphere protrude from the cube (L/2 < r < L/√2), and when only the corners of the cube are visible (L/√2 < r < L√3 / 2). The equations for the total area of the sphere (A = 4πR²) and the area of the spherical caps (C = 2πR(R - L/2)) are established, while the challenge remains in determining the overlap area (E) between adjacent spherical caps.

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  • Familiarity with spherical cap geometry and surface area calculations.
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morjo
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Ok, so a cube and a sphere are both centered at origo. The cube has side lengths L. I need to know the surface area of the part of the sphere that is inside the cube, for all possible r. There will be three different equations, one for 0 &lt; r &lt; L/2, when the entire sphere is inside the cube, one for L/2 &lt; r &lt; L/\sqrt(2), when six spherical caps of the sphere stick out of the cube sides , and one for L/\sqrt(2) &lt; r &lt; L\sqrt(3) / 2, when only the corners of the cube is visible. I have figured out the two first, but not the last.

Any help would be greatly appreciated!
 
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You need to show some work on your own before we know where you are stuck and how to help.
 
OK. The reason I did not show my work is because i believe there must be a much simpler way. But her it is.

If I am not mistaking, the area X of the sphere INSIDE the cube is the total area A of the sphere, minus the area C of the six spherical caps, plus the area of the overlaps E between the caps near each of the 12 edges (because it has been subtracted twice). So X = A - 6C + 12E.

I have found A and C:
A = 4\pi R^2
C = 2\pi R h = 2 \pi R (R-L/2)

Now I only need E, the intersection / overlap between two adjacent spherical caps. In the attached drawing, the colored area is HALF of E. E has been divided in the middle by the arc of a great circle. Finding the colored area have proved more difficult than I expected, though.
 

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