 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/13/b/ii/solution

The arc length factor is the norm of the tangent:

\frac{d r}{d t} = 36 sin^{2} 6 t + 36 cos^{2} 6 t + 64 = 10.

Hence

d s = 10 d t

and

Φ = x y = cos 6 t sin 6 t = \frac{1}{2} sin 12 t

\int_{Γ} Φ d s = 5 \int_{0}^{π /12} sin 12 t d t = \frac{5}{6} .

This uses scalar arc length, rather than the oriented displacement used in the vector line integrals of part (a).

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