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Shaft

radosr

Baruch MFE Faculty
Joined
7/18/07
Messages
660
Points
73
First, an interview classic to warm up: Two infinite cylindrical shafts of radius 1 intersect in 3-space so that their axes of symmetry intersect (at a point) at a right angle. What is the volume of the intersection of shafts?

Now, a question that was recently asked is the same question as above, except that we have three cylindrical shafts instead (you can visualize this by imagining 3 shafts, whose axes of symmetry are the x-, y-, and z-axis, respectively).
 
For the two-shaft case, the region of intersection is comprised of 4 congruent pieces that look somewhat like slices of an orange. Their total volume is (4\int_{-1}^1\frac{\sqrt{2(1-x^2)}^2}{2}dx=\frac{16}{3}\approx 5.333). (The flat surfaces of the orange slices are each half the ellipse (x^2+\frac{y^2}{2}=1).)

For the three-shaft case, the third cylinder shaves off a volume equal to (4\int_{\frac{\sqrt{2}}{2}}^1 4\sqrt{1-x^2}(x-\sqrt{1-x^2})dx=8\sqrt{2}-\frac{32}{3}) from the 4 orange slices in the previous case, so the answer is (16-8\sqrt{2}\approx 4.686). (This is tougher to visualize, so I won't explain, but if someone insists I might try.)
 
Warm up: it's the volume of a cylindar of radius 1 and height 1 = π * 1^2 *2 = 2π

For the question with three shafts, the three shafts limit the intersection to a sphere or radius 1. Volume is 4/3 * π * 1 = 4π/3


 
Warm up: it's the volume of a cylindar of radius 1 and height 1 = π * 1^2 *2 = 2π

For the question with three shafts, the three shafts limit the intersection to a sphere or radius 1. Volume is 4/3 * π * 1 = 4π/3
Actually, I think my first answer is wrong (visually it is). But I am pretty sure the second one is a sphere.
 
Actually, I think my first answer is wrong (visually it is). But I am pretty sure the second one is a sphere.

There's no way to get a sphere because cylinders have straight geodesics in a certain direction whereas spheres have circular ones in all directions.
 
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