Four cylinders with radius 1, with axes in the same plane intersect at 90-degree and 45-degree angles. What is the volume of their intersection, to the nearest percent?
Note : Try this problem first.
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See the solution at the related problem . The only difference is that in this case, for each z , the four long rectangles intersect in an octagon of height 2 1 − z 2 . The area of an octagon of height h is 2 ( 2 − 1 ) h 2 and so our volume is: ∫ − 1 1 2 ( 2 − 1 ) ⋅ 4 ( 1 − z 2 ) d z = 8 ( 2 − 1 ) [ − 1 1 z − 3 z 3 ] = 3 1 6 ⋅ 2 ( 2 − 1 ) ≈ 4 . 4 2
Naturally, this is the volume of two unit cylinders intersecting at right angles (16/3) times the ratio of the area of an octagon to that of a square of the same height.