 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/16/b/ii/solution

O A C

y = 0

and the outward unit normal is

- e_{y}

. Its flux is

Φ_{O A C} = - \int_{0}^{1} \int_{0}^{1 - x} (x + z)^{2} d z d x = - \frac{1}{3} \int_{0}^{1} (1 - x^{3}) d x = - \frac{1}{4} .

OBC

x = 0

and the outward unit normal is

- e_{x}

. Hence

Φ_{OBC} = - \int_{0}^{1} \int_{0}^{1 - y} (y + z) d z d y = - \frac{1}{2} \int_{0}^{1} (1 - y^{2}) d y = - \frac{1}{3} .

A BC

, write

z = 1 - x - y

over

x \geq 0

y \geq 0

x + y \leq 1

(1, 1, 1)^{T} d x d y

, pointing away from the origin. Here

x + y + z = 1

, so

F = (1, 1, 0)^{T}

. Thus

Φ_{A BC} = \int_{0}^{1} \int_{0}^{1 - x} 2 d y d x = 1 .

The unnormalized normal vector in the oriented surface element already incorporates the factor

3

from the surface area; no additional factor is needed.

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