 Solution

For the surface area of a tricylinder, permutation symmetry shows that each of the three cylindrical surfaces contributes the same surface area to the boundary. On the face $x^2+y^2=1$, put $x=\cos \theta$, $y=\sin \theta$. The remaining two inequalities requireso $|z|\le \min (|\sin \theta |,|\cos \theta |)$. The surface area element on this unit circular cylinder is $d\theta dz$. There are eight congruent angular sectors, with the minimum equal to $\sin \theta$ for $0\le \theta \le \pi /4$. Thus the area of this face isEvery boundary point lies on at least one of these three cylindrical surfaces; their pairwise intersections are curves of zero surface area. Adding the three faces therefore gives