# Area of sphere inside cube

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 $0 < r < L/2$, when the entire sphere is inside the cube, one for $L/2 < r < L/\sqrt(2)$, when six spherical caps of the sphere stick out of the cube sides , and one for $L/\sqrt(2) < r < L\sqrt(3) / 2$, 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!

phinds
Gold Member
2019 Award
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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