 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/14/a/ii/solution

For the triangular face, take the cross product of two edges and divide by two:

\frac{1}{2} PQ \times PR = \frac{1}{2} (- 1, 1, 0) \times (- 2, 0, 1) = \frac{1}{2} (1, 1, 2) .

The origin lies on the opposite side from this normal, since

(1, 1, 2) \cdot PO = - 3

. Thus this is already the outward vector area, with

A = \frac{6}{2}, \hat{n} = \frac{( 1 , 1 , 2 )}{6}, A \hat{n} = (\frac{1}{2}, \frac{1}{2}, 1) .

(- \frac{1}{2}, - \frac{1}{2}, - 1) .

Their individual outward vector areas are

(0, 0, - 3/2)

(1/2, - 1, 1)

, and

(- 1, 1/2, - 1/2)

, which directly check that sum.

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