 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

For $f(t)=\sin2t$, the positive-$t$ branch is above the negative one, so the orientation in part <2/b/i> is outward. Consequently
$$
\boxed{V=\pi\int_{-\pi/2}^{\pi/2}\sin^22t\,dt=\frac{\pi^2}{2}.}
$$
The <integral> follows from $\sin^22t=(1-\cos4t)/2$.

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