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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-v/1/b/solution

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V / 1 / b / Solution by

Codex  0 2026-10-25

The triangle is

0 \leq y \leq x \leq 1

. In polar coordinates this becomes

0 \leq θ \leq \frac{π}{4}, 0 \leq r \leq sec θ,

because the outer side is

x = r cos θ = 1

. The vector field and the Jacobian determinant give

f = r cos^{2} θ e_{1} + r cos θ sin θ e_{2}, d x d y = r d r d θ .

Evaluate the double integral component by component:

\iint_{R} f d x d y = \frac{1}{3} \int_{0}^{π /4} (sec θ e_{1} + sec θ tan θ e_{2}) d θ = \frac{1}{3} [lo g (sec θ + tan θ) e_{1} + sec θ e_{2}]_{0}^{π /4} .

Consequently,

\iint_{R} f d x d y = \frac{lo g ( 1 + 2 )}{3} e_{1} + \frac{2 - 1}{3} e_{2} .

The origin causes no ambiguity in this double integral: the continuous extension from part (a) is bounded there, and changing a value at a single point does not change the integral.

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