Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 9A (Vector Calculus)

Table of contents
- a 9A
  - Solution a
- b 9A
  - Solution b
- c 9A
  - Solution c

a

Solution

The divergence theorem is

\iint_{\partial V} F \cdot n d S = ∭_{V} \nabla \cdot F d V

. Here

\nabla \cdot F = 3 z + 2 yz

. The

y

term integrates to zero by symmetry, while the ellipse has area

π / ab

. Hence the flux is

\frac{π}{ab} \int_{0}^{3} 3 z d z = \frac{27 π}{2 ab} .

b

Solution

The top and bottom fluxes cancel because

F_{z} = x^{2} + y^{2}

is independent of

z

. Parametrise the side by

(cos θ / a, sin θ / b, z)

; its outward vector area is

(\frac{c o s θ}{b}, \frac{s i n θ}{a}, 0) d θ d z .

The term involving

y^{2} z

integrates to zero, and the remaining integral is

\int_{0}^{3} \int_{0}^{2 π} \frac{3 z c o s ^{2} θ}{ab} d θ d z = \frac{27 π}{2 ab},

confirming part (a).

c

Solution

The curl is

(2 y - y^{2}, x, 0)

, so the field is not conservative. On

C

,

z = 1

and

d z = 0

\oint_{C} (3 x d x + y^{2} d y) = \oint_{C} d (\frac{3}{2} x^{2} + \frac{1}{3} y^{3}) = 0.

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