 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v-resit/question-3/b/solution

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

1

and

x

from

y

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)

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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