 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2013-resit/5/solution

z = 6 - 3 x - 2 y

. Its projection on the

x y

0 \leq x \leq 2

0 \leq y \leq 3 - 3 x /2

d Σ = 1 + z_{x}^{2} + z_{y}^{2} d x d y = 14 d x d y .

Thus the weighted surface integral is

\iint_{Σ} y d Σ = 14 \int_{0}^{2} \int_{0}^{3 - 3 x /2} y d y d x = 314 .

A geometric check uses the triangular face's area

314

y

coordinate

(0 + 3 + 0) /3 = 1

. Multiplying them gives the same result, because

y

is an affine function on this planar triangle. Its edges have zero surface measure, so strict or non-strict octant inequalities do not change the integral.

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