= Solution
{c}
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 require
$$
z^2\leq1-y^2=\cos^2\theta,\qquad
z^2\leq1-x^2=\sin^2\theta,
$$
so $|z|\leq\min(|\sin\theta|,|\cos\theta|)$. 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 $0\leq\theta\leq\pi/4$. Thus the <area> of this face is
$$
A_1=\int_0^{2\pi}2\min(|\sin\theta|,|\cos\theta|)\,d\theta
=16\int_0^{\pi/4}\sin\theta\,d\theta
=8(2-\sqrt2).
$$
Every <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
$$
\boxed{\operatorname{Area}(\partial R_2)=24(2-\sqrt2).}
$$
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