Answer
The area of the surface obtained by rotating the curve about the y-axis is:
$$
\begin{aligned}
S&=\int 2 \pi x d s \\
&=\int^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^{2}-y^{2}} ) \sqrt {\left(\frac{a^{2}}{{a^{2}-y^{2}}}\right)} dy \\
&= \pi a^{2}
\end{aligned}
$$
Work Step by Step
$$
x= \sqrt{a^{2}-y^{2}}, \quad \quad 0 \leq y \leq \frac{a}{2}
$$
$\Rightarrow$
$$
\frac{dx}{dy}= \frac{-y}{\sqrt{a^{2}-y^{2}}}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(\frac{dx}{dy}\right)^{2}} dy \\
&=\sqrt {1+\left(\frac{-y}{\sqrt{a^{2}-y^{2}}}\right)^{2}} dy\\
&=\sqrt {1+\left(\frac{y^{2}}{{a^{2}-y^{2}}}\right)} dy \\
&=\sqrt {\left(\frac{a^{2}}{{a^{2}-y^{2}}}\right)} dy
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the y-axis is:
$$
\begin{aligned}
S&=\int 2 \pi x d s \\
&=\int^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^{2}-y^{2}} ) \sqrt {\left(\frac{a^{2}}{{a^{2}-y^{2}}}\right)} dy \\
&=\int^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^{2}-y^{2}} )\frac{a}{ \sqrt {a^{2}-y^{2}}} dy \\
&=2 \pi \int^{\frac{a}{2}}_{0} a dy\\
&=2 \pi a\left[y\right]_{0}^{\frac{a}{2}}\\
&=2 \pi a\left[ \frac{a}{2}\right]\\
&= \pi a^{2}
\end{aligned}
$$
