Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V / 3

Table of contents
- a 3
  - i a
    - Solution i
  - ii a
    - Solution ii
  - iii a
    - Solution iii
- b 3
  - Solution b
- c 3
  - Solution c

a

i

Solution

For the given wrapped hyperbolic cosine curve,

x^{2} + y^{2} = cos^{2} t + sin^{2} t = 1, z = cosh t \geq 1 > 0,

so

C \subset S

. The curve makes one full turn around the circular cylinder, descending from

(- 1, 0, cosh π)

to

(1, 0, 1)

and ascending to its initial point. Its two parameter endpoints coincide; their tangent vectors

(0, - 1, - sinh π)

and

(0, - 1, sinh π)

do not agree, so the resulting closed curve has a corner there.

Figure 1.
The hyperbolic cosine curve wrapped around a unit circular cylinder, and the same surface unrolled
. The shaded finite patch extends from $z = 0$ to $z = cosh t$ . On the unrolled cylinder, its horizontal coordinate is $t \in [- π, π]$ ; the two vertical edges are identified. Details.

ii

Solution

The velocity vector of the parametric curve is

r^{'} (t) = (- sin t, cos t, sinh t) .

Its speed is

∣ r^{'} (t) ∣ = 1 + sinh^{2} t = cosh t,

where

cosh t > 0

fixes the sign of the square root. Hence the arc length is

ℓ (C) = \int_{- π}^{π} cosh t d t = 2 sinh π .

The coincidence of the endpoints does not duplicate any positive arc length.

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 .

The surface area element is

∣ 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.

b

Solution

At a fixed

z \in [- 1, 1]

, both cylinder inequalities become

∣ x ∣ \leq 1 - z^{2}, ∣ y ∣ \leq 1 - z^{2} .

Thus the cross section is a square of side

2 1 - z^{2}

and area

4 (1 - z^{2})

. Integrating these cross-sectional areas gives the volume of this bicylinder:

Vol (R_{1}) = \int_{- 1}^{1} 4 (1 - z^{2}) d z = 4 [z - \frac{z ^{3}}{3}]_{- 1}^{1} = \frac{16}{3} .

c

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 θ

,

y = sin θ

. The remaining two inequalities require

z^{2} \leq 1 - y^{2} = cos^{2} θ, z^{2} \leq 1 - x^{2} = sin^{2} θ,

so

∣ z ∣ \leq min (∣ sin θ ∣, ∣ cos θ ∣)

. The surface area element on this unit circular cylinder is

d θ d z

. There are eight congruent angular sectors, with the minimum equal to

sin θ

for

0 \leq θ \leq π /4

. Thus the area of this face is

A_{1} = \int_{0}^{2 π} 2 min (∣ sin θ ∣, ∣ cos θ ∣) d θ = 16 \int_{0}^{π /4} sin θ d θ = 8 (2 - 2) .

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

Area (\partial R_{2}) = 24 (2 - 2) .

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