Solution
= Solution
For $a/|b|\ll1$, apply the <binomial series> to the result in (c):
$$
\left(1+\frac{4a^2}{b^2}\right)^{3/2}
=1+\frac{6a^2}{b^2}+\frac{6a^4}{b^4}+O(a^6/b^6).
$$
Therefore
$$
\boxed{A=\frac{\alpha b^2}{2}+\frac{\alpha a^2}{2}+O(a^4/b^2)
\sim\frac{\alpha b^2}{2}}.
$$
For $\alpha=2\pi$, the leading term is $\pi b^2$, the <area> of the base disk. A very shallow cap is nearly planar; its first <area> excess is $\pi a^2$.