 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv-resit/3/b/solution

r_{s} = (t cos t, t sin t, 2), r_{t} = s (cos t - t sin t, sin t + t cos t, 0) .

r_{s} \times r_{t} = s (- 2 (sin t + t cos t), 2 (cos t - t sin t), t^{2}) .

The mixed terms cancel in its squared Euclidean norm:

∣ r_{s} \times r_{t} ∣^{2} = s^{2} (4 (1 + t^{2}) + t^{4}) = s^{2} (t^{2} + 2)^{2} .

Since

s > 0

s (t^{2} + 2) d s d t

. There is no repeated coverage:

z = 2 s

recovers

s

, and

x^{2} + y^{2} = s t

then recovers

t

. Therefore the surface area is

area (S) = \int_{0}^{2 π} \int_{0}^{1} s (t^{2} + 2) d s d t = \frac{1}{2} [\frac{t ^{3}}{3} + 2 t]_{0}^{2 π} = \frac{4 π ^{3}}{3} + 2 π .

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