 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/16/a/solution

The cube is

1 \leq x \leq 2

0 \leq y, z \leq 1

, so

∣ x ∣ = x

throughout it. Evaluate the six outward surface integrals directly.

x = 2

2

2

; on

x = 1

it is

- 1

- 1

. On

y = 1

2 x

\int_{1}^{2} \int_{0}^{1} 2 x d z d x = 3

; on

y = 0

it is zero. On

z = 1

e^{- 1}

and on

z = 0

it is

- 1

, so those unit-area faces give

e^{- 1}

and

- 1

Adding all six contributions yields

Φ = 2 - 1 + 3 + 0 + e^{- 1} - 1 = 3 + e^{- 1} .

Only direct surface integral was used, as requested.

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