Answer
$\dfrac{\pi}{3}$
Work Step by Step
The surface area is given by:
$S=\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA$
This implies that
$S=(2) \int_{-1/2}^{1/2} \int_{0}^{1/2}\dfrac{1}{\sqrt {1-x^2} }dy dx$
Thus, $S=[\sin^{-1}x]_{-1/2}^{1/2}=\dfrac{\pi}{3}$
