= Solution
For $h>0$, parametrize the curved surface by $(r\cos\theta,r\sin\theta,r^2/a)$, with $0\leq r\leq\sqrt{ah}$ and $0\leq\theta\leq2\pi$. The magnitude of the <cross product> of its coordinate tangents gives the <area element of a surface>:
$$
dS=r\sqrt{1+\frac{4r^2}{a^2}}\,dr\,d\theta.
$$
Integrating gives
$$
\begin{aligned}
S&=2\pi\int_0^{\sqrt{ah}}r\sqrt{1+4r^2/a^2}\,dr\\
&=\frac{\pi a^2}{6}\left[(1+4r^2/a^2)^{3/2}\right]_0^{\sqrt{ah}}\\
&=\boxed{\frac{\pi a^2}{6}\left[\left(1+\frac{4h}{a}\right)^{3/2}-1\right].}
\end{aligned}
$$
This is the curved <surface area>, not the area of a lid across the horizontal opening. The expression tends to zero at $h=0$ and has dimensions of area. The <volume and centroid of a parabolic cup> use the same geometry.
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