Answer
$2a^2(\pi-2)$
Work Step by Step
In order to compute the surface area of the upper part of the hemi-sphere, we will have to multiply our answer by $2$.
We have $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dA$; $\iint_{D} dA$ is the area of the region $D$ and $D$ is the projection of the surface on the xy-plane.
And the equation of the upper part of the hemi-sphere is $z=\sqrt{a^2-y^2-x^2}$ and the points inside the circle are: $(r, \theta)$
Therefore, $A(S)=\int_{-\pi/2}^{\pi/2} \int_{0}^{\cos \theta} \dfrac{a}{\sqrt {a^2-r^2}} r dr d \theta =\int_{-\pi/2}^{\pi/2} [-a \sqrt {a^2-r^2}]_{0}^{\cos \theta} d \theta $
or, $=2 \int_{0}^{\pi/2} a^2-a^2 \sin \theta d\theta $
or, $=2a^2 [\theta+\cos \theta]_0^{\pi/2}$
Thus, the total area is:
$S_{A}=2 \times a^2(\pi-2)=2a^2(\pi-2)$
