= Solution
{c}
Parametrize the <surface of revolution> by
$$
\mathbf X(t,\theta)=(\cos t\cos\theta,\cos t\sin\theta,f(t)),\quad
-\pi/2\leq t\leq\pi/2,\quad0\leq\theta<2\pi.
$$
An oriented <vector surface area element> is
$$
\mathbf X_\theta\times\mathbf X_t
=(\cos t f'(t)\cos\theta,\cos t f'(t)\sin\theta,\sin t\cos t).
$$
Use the <vector field> $\mathbf F=(x/2,y/2,0)$, whose <divergence> is one. If this <parametrization> has the outward orientation on an enclosing surface, the <divergence theorem> gives
$$
\boxed{V=\pi\int_{-\pi/2}^{\pi/2}\cos^2t\,f'(t)\,dt
=\pi\int_{-\pi/2}^{\pi/2}f(t)\sin2t\,dt.}
$$
The second equality is <integration by parts>, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi/2$.
The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the <flux> formula is an algebraic signed <volume>. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric <volume> is instead
$$
V=2\pi\int_0^{\pi/2}\sin t\cos t\,|f(t)-f(-t)|\,dt.
$$
This explicitly handles a reversal or change in the order of the two branches.
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