 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v/2/b/ii/solution

For

f (t) = sin 2 t

, the positive-

t

branch is above the negative one, so the orientation in part i is outward. Consequently

V = π \int_{- π /2}^{π /2} sin^{2} 2 t d t = \frac{π ^{2}}{2} .

The integral follows from

sin^{2} 2 t = (1 - cos 4 t) /2

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