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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014/10/b/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2014 / 10 / b / Solution by

Codex  0 2026-10-25

The original integration region is the triangle with vertices

(0, 0)

,

(0, 1)

and

(1/2, 1/2)

. Under

r = x + y

,

s = y - x

, its inverse map is

x = (r - s) /2

,

y = (r + s) /2

, with absolute inverse Jacobian determinant

1/2

. The boundaries

x = 0

,

y = x

,

x + y = 1

become

s = r

,

s = 0

,

r = 1

. Thus the transformed triangle is

0 \leq s \leq r \leq 1

, and

I = \frac{1}{2} \int_{0}^{1} \int_{0}^{r} \frac{s}{r ^{2} 1 - r ^{2}} d s d r = \frac{1}{4} \int_{0}^{1} \frac{d r}{1 - r ^{2}} = \frac{π}{8} .

The endpoint singularities are integrable after the inner integral:

r = 0

is only a collapsed vertex, while

r = 1

has the usual inverse-square-root behavior. The diagrams indicate the complete triangles, rather than rectangular bounds that would include points absent from the original region.

Figure 1.
Original and sum-difference-coordinate triangles for the transformed double integral
. Details.

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