Answer
$2a^2(\pi-2)$
Work Step by Step
The surface area of the upper part of the hemi-sphere can be computed when we multiply our result by $2$ and the equation of the upper part of the hemi-sphere as follows: $z=\sqrt{a^2-y^2-x^2}$ and the points inside the circle are: $(r, \theta)$.
Therefore, $Surface \ area ; A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dA$; and $\iint_{D} dA$ is the area of the region $D$ and $D$ is the projection of the surface on the xy-plane.
Now, $Surface \ area ; A(S)=\int_{-\pi/2}^{\pi/2} \int_{0}^{\cos \theta} \dfrac{a}{\sqrt {a^2-r^2}} r dr d \theta \\=\int_{-\pi/2}^{\pi/2} [-a \sqrt {a^2-r^2}]_{0}^{\cos \theta} d \theta \\=2 \int_{0}^{\pi/2} a^2-a^2 \sin \theta d\theta \\=2a^2 [\theta+\cos \theta]_0^{\pi/2}$
So, the total area becomes: $$A(S)=2 \times a^2(\pi-2)=2a^2(\pi-2)$$
