 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v/3/a/iii/solution

Parametrize the finite cylindrical surface patch by

R (t, z) = (cos t, sin t, z), - π \leq t \leq π, 0 \leq z \leq cosh t .

∣ R_{t} \times R_{z} ∣ d t d z = d t d z

. The seam

t = \pm π

has zero surface area, so it causes no overcounting. Therefore

A = \int_{- π}^{π} \int_{0}^{c o s h t} d z d t = 2 sinh π .

Unrolling 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.

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