= Solution
Integrate the <surface charge density> using the <surface integral> in <spherical coordinates>:
$$
Q=2\pi a^2\frac{q(a-r_0^2/a)}{4\pi}
\int_{-1}^1\frac{du}{(a^2+r_0^2-2ar_0u)^{3/2}}.
$$
Since $r_0>a$, the elementary integral is
$$
\frac1{ar_0}\left[\frac1{\sqrt{a^2+r_0^2-2ar_0u}}\right]_{-1}^1
=\frac1{ar_0}\left(\frac1{r_0-a}-\frac1{r_0+a}\right)
=\frac2{r_0(r_0^2-a^2)}.
$$
Substitution yields the total induced <electric charge>
$$
\boxed{Q=-\frac{qa}{r_0}=q^\star.}
$$
The equality with the <image charge> also follows by applying <Gauss's law> to the exterior <electric field>: the physical and fictitious descriptions give the same <electric flux> through a sphere enclosing the conductor but excluding the physical <point charge>.
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