Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 3A (Vector Calculus)

Table of contents
- Solution 3A
- i 3A
  - Solution i
- ii 3A
  - Solution ii

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})

.

i

Solution

Here the area factor is

1 + 2^{2} + 3^{2} = 14

and the triangular base has area

3

. The surface area is therefore

314

.

ii

Solution

Write the upper cap as

z = 1 - x^{2} - y^{2}

. Its area factor is

(1 - r^{2})^{- 1/2}

, and the projected disc has radius

a

. Therefore

area (S) = 2 π \int_{0}^{a} \frac{r}{1 - r ^{2}} d r = 2 π (1 - 1 - a) .

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