Answer
$\dfrac{\pi}{3}$
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Since, $x^2+z^2 =4 $ and $ z=\sqrt {4-x^2}$
Our aim is to calculate the area of the given surface. $A(S)=\iint_{D} \sqrt{1+(\dfrac{-x}{\sqrt {4-x^2}})^2+(0)^2} \\=\iint{D} \sqrt {\dfrac{x^2}{4-x^2}+1} dA \\ =\int_{0}^{1} \int_{0}^{1} \sqrt {\dfrac{4}{4-x^2}}dy dx \\=2 \times \int_{0}^{1} \sqrt {\dfrac{1}{4-x^2}}\ dy \ dx$
Set $ 2 \sin \theta =x$ and $dx= 2 \cos \theta d \theta $
$A(S)=2\times \int_0^{ \pi/6} \sqrt {\dfrac{1}{4-4 \sin^2 \theta}} (2 \cos \theta d \theta) \\=2\int_0^{ \pi/6} \sqrt {\dfrac{1}{\cos^2 \theta}} \times (\cos \theta d \theta) \\=2 \int_0^{ \pi/6} d \theta \\= 2 \times (\dfrac{\pi}{6}-0) \\=\dfrac{\pi}{3}$
