 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/9b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

n = (a, b, c)

and

r = (x, y, z)

, the integrand is

\frac{1}{2} (n \times r) \cdot d r

. Since

\nabla \times [\frac{1}{2} (n \times r)] = n,

the Stokes theorem gives the signed planar area, and its absolute value gives the ordinary area.

The

x y

projection of the given curve is the astroid

x = 2 cos^{3} t

y = sin^{3} t

, whose area is

\frac{1}{2} \oint (x d y - y d x) = \frac{3 π}{4} .

z = 3 y

stretches area from its

x y

projection by

1 + 3 = 2

. Hence

Area = \frac{3 π}{2} .

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