= Solution
The <divergence> of the position <vector field> is $\nabla\cdot\mathbf r=3$. At height $z$, the horizontal cross-section is a <disk> of <area> $\pi(1-z^2)$. By the <divergence theorem>, the outward <flux integral> is three times the enclosed <volume>:
$$
\Phi(z_0)=3\pi\int_{-1}^{z_0}(1-z^2)\,dz
=3\pi\left[z-\frac{z^3}{3}\right]_{-1}^{z_0}.
$$
Therefore
$$
\boxed{\Phi(z_0)=\pi(2+3z_0-z_0^3)=\pi(1+z_0)^2(2-z_0)}.
$$
A direct check separates the two pieces of the <surface integral>. On the unit <sphere>, the outward <unit normal> is $\mathbf r$, so $\mathbf r\cdot\mathbf n=1$ and the curved contribution is $2\pi(z_0+1)$. On the top <disk>, $\mathbf n=\mathbf e_z$, giving $\pi z_0(1-z_0^2)$. Their sum is the same <flux>.
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