Two cylinders of equal radius intersect at right angles. Find the volume of the intersection between the two cylinders, without using calculus!

Say the  first cylinder has height 2r, volume 2*pi*r^3, and falls exactly within an enclosing cube, say C, of volume (2r)^3, the difference between the two being 4 triangular wedges (with a curve on the hypotenuse) each of volume 1/2(4-pi)r^3. Let the second  cylinder be identical, except for its orientation, which is at right angles to the first. Both cylinders and C are centred on O.

We take a slice of the desired solid, S, perpendicular to both sets of wedges, at height P from O.
(1) Any such slice must be a square, say s^2, since the circumferences of the two cylinders that supply its vertices are of identical size, are at right angles to each other, and are centred on the same point, O.
(2) Since a horizontal slice of C is also square, the difference between C and S at height P is a difference of squares,  (2r)^2-s^2
(3) Now consider that each vertex of s at height P is also of constant distance, namely r, from O. So for any P chosen, r^2-s^2=P^2
(4)  As it happens, there is a equivalent solid that has the same property that at every height P its area is P^2, namely an inverted square pyramid, or, to be precise, two in this case, each with a base of (2r) and a height of r, giving a combined volume of (2r)^3/3
So the difference between (2r)^3 and the volume of S is (2r)^3/3, and the volume of S is therefore  (2r)^3-(2r)^3/3 = 2(2r)^3/3.

It follows that 3 such shapes would have exactly the same volume as two of the enclosing cubes. Whodathunkit.

Nice problem.

 

Edited on May 27, 2013, 5:39 am

Posted by broll on 2013-05-26 10:16:03