Answer
$\pi a^{2} $
Work Step by Step
The area is:
$$S=\int_{0}^{\frac{a}{2}}2\pi x\sqrt{1+(\frac{dx}{dy})^{2}}~dy$$
The first derivative of $x$ with respect to $y$ is:
$$x'=-\frac{y}{\sqrt{a^{2}-y^{2}}}$$
$$S=\int_{0}^{\frac{a}{2}}2\pi x\sqrt{1+(-\frac{y}{\sqrt{a^{2}-y^{2}}})^{2}}~dy$$
$$S=\int_{0}^{\frac{a}{2}}2\pi x\sqrt{1+\frac{y^2}{a^{2}-y^{2}}}~dy$$
$$S=\int_{0}^{\frac{a}{2}}2\pi x\sqrt{\frac{a^2}{a^{2}-y^{2}}}~dy$$
$$S=\int_{0}^{\frac{a}{2}}2\pi x\frac{a}{\sqrt{a^{2}-y^{2}}}~dy$$
$$S=\int_{0}^{\frac{a}{2}}2\pi \sqrt{a^{2}-y^{2}}\frac{a}{\sqrt{a^{2}-y^{2}}}~dy$$
$$S=\int_{0}^{\frac{a}{2}}2\pi a~dy$$
$$S=[2\pi ay]_{0}^{\frac{a}{2}}=2\pi a\cdot\frac{a}{2}=\pi a^{2} $$
