 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014-resit/5/solution

u = x - 3 y

v = 2 x + y

, with inverse

x = (u + 3 v) /7

y = (v - 2 u) /7

\frac{\partial ( u , v )}{\partial ( x , y )} = 7, \frac{\partial ( x , y )}{\partial ( u , v )} = \frac{1}{7} .

The two lines of slope

- 2

become

v = 1, 4

, while the two lines of slope

1/3

become

u = 0, 7

. Thus the parallelogram becomes

0 \leq u \leq 7

1 \leq v \leq 4

. Since

v

stays positive, the transformed integrand has no singularity there. The double integral is

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

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