 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/16/b/solution

\nabla \cdot F = 2 (y + 3) + 2 z = 2 y + 2 z + 6.

For the tetrahedral domain, convenient integration bounds are

0 \leq x \leq 1, 0 \leq y \leq 1 - x, 0 \leq z \leq 2 (1 - x - y) .

Consequently its volume integral is

I = \int_{0}^{1} \int_{0}^{1 - x} \int_{0}^{2 (1 - x - y)} (2 y + 2 z + 6) d z d y d x .

Putting

h = 2 (1 - x - y)

, the inner integral is

(2 y + 6) h + h^{2} = (16 - 4 x) (1 - x - y)

. Therefore

I = \int_{0}^{1} (16 - 4 x) \frac{( 1 - x ) ^{2}}{2} d x = \int_{0}^{1} (8 - 18 x + 12 x^{2} - 2 x^{3}) d x = 8 - 9 + 4 - \frac{1}{2} .

Thus

I = \frac{5}{2} .

There is an independent centroid check. The vertices are

(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 2)

1/3

(1/4, 1/4, 1/2)

. An affine function averages to its value at the centroid, so

I = (1/3) (6 + 2/4 + 2/2) = 5/2

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