 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/i/solution

On the face

OB A

z = 0

and the outward unit normal is

- e_{z}

. The flux integral is therefore

Φ_{OB A} = - \int_{0}^{1} \int_{0}^{1 - x} (1 - x - y) d y d x = - \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x = - \frac{1}{6} .

The negative sign comes from the outward orientation, which points below the tetrahedron on this face.

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