From a solid unit cube, remove an eighth of a sphere of radius 1, centered at one corner of the cube.
From the opposite corner, do the same.
These scoops will intersect. The remaining shape will have a circular hole.
(1) What is the diameter of this hole?
(2) What is the remaining volume?
Geogebra doesn't like solids, only surfaces, but here's a rendering.
TinkerCad is for 3D printing, so it can also be used for another rendering.
The two spheres overlap so as to "cap" each other. Since the long diagonal of the cube (with endpoints at the sphere centers) is sqrt(3) and each sphere envelops it from its own center out to a length of 1, the overlapped segment of the diagonal is of length:
overlap = 1-(sqrt(3)-1)=2-sqrt(3).
The sphere surfaces intersect to form a circle or "hole" perpendicular to the diagonal in the middle of overlap segment, so each cap has a height h of half the overlap segment length:
h = 1-sqrt(3)/2.
From Pythagorus, the radius of the cap base (or hole) is
a = sqrt(r^2 - (r-h)^2), where here r=1. so:
a = 0.5
The volume of the pristine cube is C=1
The volume of each cap is Vc = (1/3) pi h^2 (3r - h), with r=1
The area of each one-eighth sphere is Vs =(1/8) (4/3) pi r^3, with r=1
Subtracting-off the volume of the two spheres doubly counts the two caps, so this volume must be added back into the total volume T.
The remaining volume is:
T = 1 - 2 Vs + 2 Vc
= 1 - 2 (1/8) (4/3) pi + 2 (1/3) pi (1-sqrt(3)/2)^2 (3-h)
Edited on August 25, 2021, 9:11 am
soln
