 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/12/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 12 / b / Solution by

Codex  0 2026-10-18

Use the same axes as in (a) and choose the outward orientation. On the curved spherical lune,

n = e_{r}

and

d A = a^{2} sin θ d θ d φ

. Its vector area is

S = a^{2} \int_{- β /2}^{β /2} \int_{0}^{π} (sin θ cos φ, sin θ sin φ, cos θ) sin θ d θ d φ .

The

y

and

z

components cancel. The

x

component is

a^{2} (π /2) 2 sin (β /2)

, hence

S = π a^{2} sin \frac{β}{2} e_{x} .

Here

e_{x}

points along the azimuthal bisector; another coordinate choice rotates this vector accordingly. Its magnitude is the signed projected area, not the scalar curved area

2 β a^{2}

. Reversing the orientation reverses its sign.

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