On
I’m going to suggest an intuitive, non-rigorous way of thinking about the problem.
For each $y\in[0,a/2]$ you’re revolving a little bit of arc, $ds$, around the $y$-axis. The distance of that bit of arc from the $y$-axis is $\sqrt{a^2-y^2}$, its $x$-coordinate, so it’s going travel in a circle around the $y$-axis, covering a distance of $2\pi\sqrt{a^2-y^2}$, and generate a bit of surface with an area of $$dA=2\pi\sqrt{a^2-y^2}ds\;.\tag{1}$$
Now what is $ds$? You get that from the arc length formula: $ds=\sqrt{(dx)^2+(dy)^2}$. In this case you have $x$ given as a function of $y$, so $y$ is your independent variable, and you should ‘divide’ and ‘multiply’ the arc length formula by $dy$ to get
$$ds=\sqrt{\left(\frac{dx}{dy}\right)^2+\left(\frac{dy}{dy}\right)^2}\,dy=\sqrt{\left(\frac{dx}{dy}\right)^2+1}\,dy\;.\tag{2}$$
Now substitute $(2)$ into $(1)$ to get $$dA=2\pi\sqrt{a^2-y^2}\sqrt{\left(\frac{dx}{dy}\right)^2+1}\,dy$$ and integrate: $$\begin{align*}A&=\int_0^{a/2}2\pi\sqrt{a^2-y^2}\sqrt{\left(\frac{dx}{dy}\right)^2+1}\,dy\\ &=2\pi\int_0^{a/2}\sqrt{(a^2-y^2)\left(\left(\frac{dx}{dy}\right)^2+1\right)}\,dy\;. \end{align*}$$
We have formula $$\text{Surface area } S = \int 2\pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy$$ we have $x= \sqrt{a^2-y^2}$ and $\frac{dx}{dy}= \frac{-y}{\sqrt{a^2-y^2}}$. Hence we have, $$S=\int_0^{\frac{a}{2}}2\pi \sqrt{a^2-y^2}\sqrt{1+ \frac{y^2}{{a^2-y^2}}} dy$$ $$=\int_0^{\frac{a}{2}}2\pi \sqrt{a^2-y^2} \frac{a}{\sqrt{a^2-y^2}}dy $$ $$=\int_0^{\frac{a}{2}}2\pi .a dy$$ $$=\pi .a^2$$