= Solution
Differentiate the <flux>:
$$
\boxed{\Phi'(z_0)=3\pi(1-z_0^2)},\qquad
\Phi''(z_0)=-6\pi z_0.
$$
On $-1<z_0<1$, $\Phi$ is strictly increasing. Its limiting endpoint values are $\Phi(-1)=0$ and $\Phi(1)=4\pi$, and $\Phi(0)=2\pi$. It has horizontal limiting <tangents> at the endpoints, curves upward for $z_0<0$, curves downward for $z_0>0$, and has an <inflection point> at zero.
The <derivative> graph is a downward-opening <parabola>, approaching zero at both endpoints and attaining its maximum $3\pi$ at $z_0=0$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2-q17-flux.png]
{title=Flux through the truncated unit sphere and its derivative}
{description=Both curves are scaled by $\pi$. Endpoint circles show limiting values because the prescribed truncation parameter lies in $(-1,1)$. The <flux> increases from zero to $4\pi$, while its <derivative> is largest at the equator.}
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