Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 3F / d

Solution

ρ

then has Euclidean radius

R = tanh (ρ /2)

4 r d r d θ / (1 - r^{2})^{2}

, so

Area = \int_{0}^{2 π} \int_{0}^{R} \frac{4 r}{( 1 - r ^{2} ) ^{2}} d r d θ = 4 π (\frac{1}{1 - R ^{2}} - 1) .

Therefore the area is independent of the centre, and

Area = 4 π sinh^{2} (ρ /2) = 2 π (cosh ρ - 1) .

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