 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2013/6/solution

x = (u + 3 v) /7

y = (v - 2 u) /7

, with absolute inverse Jacobian determinant

1/7

. The two lines

y = - 2 x + 1, - 2 x + 4

become

v = 1, 4

, and

y = x /3, (x - 7) /3

become

u = 0, 7

\iint_{D} \frac{x - 3 y}{2 x + y} d x d y = \frac{1}{7} \int_{0}^{7} \int_{1}^{4} \frac{u}{v} d v d u = \frac{7}{2} ln 4 = 7 ln 2.

v

is bounded away from zero throughout the rectangle, so this transformation introduces no improper-integral issue.

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