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 Incoming links: Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 1 / a

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 1 / c / Solution 2026-10-25

The inequalities become the rectangle

1/2 \leq u \leq 1

,

1/2 \leq v \leq 1

in parabolic polar coordinates. The change of variables formula and the Jacobian determinant from a give the area

A = \int_{1/2}^{1} \int_{1/2}^{1} (u^{2} + v^{2}) d v d u = \frac{1}{2} \int_{1/2}^{1} u^{2} d u + \frac{1}{2} \int_{1/2}^{1} v^{2} d v = [\frac{s ^{3}}{3}]_{1/2}^{1} = \frac{7}{24} .

The middle panel in b shows the curved region in Cartesian coordinates; its straight edges in parabolic polar coordinates explain the simple integration limits.

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