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The curved surface of a circular cylinder, excluding any flat end caps. A parametrization of a radius-$a$ cylindrical surface is $\mathbf{R}(\theta ,z)=(a\cos \theta ,a\sin \theta ,z)$, with surface area element $ad\theta dz$.

Parametrize the finite cylindrical surface patch byThe surface area element is $|{\mathbf{R}}_t\times {\mathbf{R}}_z|dtdz=dtdz$. The seam $t=\pm \pi$ has zero surface area, so it causes no overcounting. ThereforeUnrolling the unit circular cylinder preserves surface area, which explains the equality with the area under the hyperbolic cosine in the figure. The patch consists of the cylindrical surface alone; no flat base is included.

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

For three radius-$a$ circular cylinders with mutually perpendicular axes through one point, the boundary of their common intersection has three congruent cylindrical surface pieces. On $x^2+y^2=a^2$, the height restriction is $|z|\le a\min (|\sin \theta |,|\cos \theta |)$. Eight congruent sectors give surface area $8a^2(2-\sqrt{2})$ per piece. Pairwise intersections are curves of zero surface area, so the three pieces add.

For $-\pi \le t\le \pi$, this parametric curve closes with a corner on a unit circular cylinder. Its speed is $\cosh t$, so its arc length is $2\sinh \pi$. The cylindrical surface between the curve and $z=0$ has the same surface area, because unrolling gives the integral ${\int }_{-\pi }^{\pi }\cosh tdt$.