Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 10

Table of contents
- a 10
  - Solution a
- b 10
  - Solution b

a

Solution

For the outward orientation of a spherical cap, its vector area element is

d S = \hat{r} a^{2} sin θ d θ d ϕ, \hat{r} = (sin θ cos ϕ, sin θ sin ϕ, cos θ) .

In the surface integral defining the vector area, the

x

and

y

components vanish by integration over a complete azimuth. The remaining component is

S_{z} = 2 π a^{2} \int_{0}^{π /4} cos θ sin θ d θ = \frac{π a ^{2}}{2} .

Consequently

S = (0, 0, \frac{π a ^{2}}{2}) .

This also equals the area of the cap's projection onto the horizontal plane, directed upwards.

b

Solution

The direction must first be represented by a unit vector:

\hat{q} = (1, 1, 1) / 3

. The scalar projection of the vector area is therefore

S \cdot \hat{q} = \frac{π a ^{2}}{2 3} .

If a vector projection is intended, multiply this scalar by

\hat{q}

:

proj_{q} S = \frac{π a ^{2}}{6} (1, 1, 1) .

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