Answer
$a^2 ( \pi-2)$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Our aim is to compute the surface, for which the area is the part of the hemi-sphere which lies above the xy-plane. Its equation can be represented as: $z=\sqrt {a^2-y^2-x^2}$ and the domain (D) is the collection of points which lie inside the circle $x^2+y^2=ax$.
Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt {1+(-y/z)^2+(-x/z)^2} dA= \iint_{D} \sqrt {\dfrac{x^2+y^2+z^2}{z^2}} dA$
We can use the polar co-ordinates because of "$x^2+y^2+z^2$". Thus:
$A(S)=\int_{- \pi/2}^{\pi/2} \int_{0}^{ a \cos \theta} \dfrac{a} {\sqrt {a^2-x^2-y^2}}\ r \ dr d\theta \\=\int_{- \pi/2}^{\pi/2} [ -a \sqrt {a^2-r^2}]_{0}^{ a \cos \theta} d\theta=\int_{- \pi/2}^{\pi/2} a^2-a^2 |\sin \theta| d\theta \\ = \int_{0}^{\pi/2} [a^2-a^2 |\sin \theta| ]+[a^2-a^2 |\sin (-\theta) | ] d \theta \\ = 2 [a^2 \theta+a^2 \cos \theta]_0^{ \pi/2} \\=a^2 ( \pi-2)$
