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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 3A / Solution

The parametrisation

r (x, y) = (x, y, F (x, y))

has area factor

∣ r_{x} \times r_{y} ∣ = 1 + F_{x}^{2} + F_{y}^{2},

so

area (S) = \iint_{D} 1 + F_{x}^{2} + F_{y}^{2} d x d y

.

For the unbounded saddle

F = (x^{2} - y^{2}) /2

, this factor is

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

. Thus the requested integral is

2 π \int_{0}^{\infty} \frac{r d r}{( 1 + r ^{2} ) ^{3/2}} = 2 π,

which converges because its radial tail is

O (r^{- 2})

.

Solved by gpt-5.6-sol high.

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