 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016/10/b/ii/solution

The plane cuts the positive coordinate axes at

(3, 0, 0)

(0, 3, 0)

and

(0, 0, 3/2)

(1, 1, 2) / 6

, pointing away from the origin. In the parametrization

(x, 3 - x - 2 z, z)

, the vector surface element for this orientation is

(1, 1, 2) d x d z

. Thus

A \cdot d S = (x, - x, z) \cdot (1, 1, 2) d x d z = 2 z d x d z,

and the triangular projection gives

\iint_{S} A \cdot d S = \int_{0}^{3} \int_{0}^{(3 - x) /2} 2 z d z d x = \frac{9}{4} .

For the opposite orientation the answer is

- 9/4

. The normal direction must be stated because this open surface has no prescribed orientation in the question.

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