Solution
= Solution
Write the upper cap as $z=\sqrt{1-x^2-y^2}$. Its area factor is $(1-r^2)^{-1/2}$, and the projected disc has radius $\sqrt a$. Therefore
$$
\boxed{\operatorname{area}(S)=2\pi\int_0^{\sqrt a}\frac r{\sqrt{1-r^2}}\,dr=2\pi(1-\sqrt{1-a}).}
$$