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Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 3 / b / Solution 2026-10-25

The original double integral covers the triangle

0 \leq y \leq x \leq 1

. By Fubini's theorem, after reversing the order,

y

runs from

0

to

1

and

x

from

y

to

1

. Since a continuous function is integrable on this bounded region,

\int_{0}^{1} \int_{0}^{x} f (y) d y d x = \int_{0}^{1} \int_{y}^{1} f (y) d x d y = \int_{0}^{1} (1 - y) f (y) d y .

For the square pyramid, a horizontal section at height

z

is a square of side

2 (1 - z)

: each vertex moves linearly from the base towards

(1, 1, 1)

. Hence its cross-sectional area is

4 (1 - z)^{2}

, and its volume is

4 \int_{0}^{1} (1 - z)^{2} d z

. To use the identity just proved, choose

f (y) = 1 - y

:

Vol = 4 \int_{0}^{1} \int_{0}^{x} (1 - y) d y d x = 4 \int_{0}^{1} (x - \frac{x ^{2}}{2}) d x = 4 (\frac{1}{2} - \frac{1}{6}) = \frac{4}{3} .

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 6 / b / Solution 2026-10-18

Interpret the stated side lengths as all eight edges of the square pyramid. The equal lateral edges put its apex above the base centre. The base-centre-to-vertex distance is half the diagonal,

1/ 2 m

. The Pythagorean theorem therefore gives

h = 1^{2} - (1/ 2)^{2} = \frac{1}{2} m .

Using the pyramid volume formula with a

1 m^{2}

base,

V = \frac{1}{3} (1) \frac{1}{2} = \frac{1}{3 2} m^{3} \approx 0.2357 m^{3} .

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