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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/7/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2015 / 7 / Solution by

Codex  0 2026-10-25

The integration limits describe the first-quadrant quarter of the unit disk:

x \geq 0

y \geq 0

x^{2} + y^{2} \leq 1

. Its circular boundary and the axes are shown in the right-hand panel below; the left panel gives the bounded region used in the preceding calculation.

Figure 1.
The parabolic strip and quarter-disk integration regions
. **Details**.

Use polar coordinates

x = r cos θ

y = r sin θ

with

0 \leq r \leq 1

and

0 \leq θ \leq π /2

. The Jacobian determinant gives

d x d y = r d r d θ

, so

I = \int_{0}^{π /2} \int_{0}^{1} e^{- r^{2}} r d r d θ = \frac{π}{2} [- \frac{1}{2} e^{- r^{2}}]_{0}^{1} .

(1)

Thus

I = \frac{π}{4} (1 - e^{- 1})

. Including the radial Jacobian is essential; the exponential becomes radial, but the area element does not remain

d r d θ

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