Answer
$$4 \pi a^2 $$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy$
and, $\iint_{D} dA$ represents the projection of the surface on the $xy$-plane.
given: $z=\sqrt {a^2-(x^2+y^2}$ is the equation of the upper hemi-sphere.
The area of the given surface is: $A(S)=2 \iint_{D} \sqrt {1+(\dfrac{-x}{\sqrt {a^2-(x^2+y^2}})^2+(\dfrac{-y}{\sqrt {a^2-(x^2+y^2}})^2} dx \ dy \\$
or, $= 2 \times \int_{a}^a \int_{-\sqrt {a^2-y^2}}^{\sqrt {a^2-y^2}} \dfrac{a}{\sqrt {a^2-(x^2+y^2)}}dx \ dy$
Now, apply the polar co-ordinates for the part $x^2+y^2$ .
$$A(S)=2 \lim\limits_{t \to a^{-}} \int_{0}^{2 \pi} \int_{0}^{t} \dfrac{a}{\sqrt {a^2-r^2}} r \ dr \ d \theta \\ = 2 \times \int_{0}^{2 \pi} d \theta \times \lim\limits_{t \to a^{-}} \int_{0}^{t} \dfrac{ar}{\sqrt {a^2-r^2}} dr \\= 2 (2 \pi) \times \lim\limits_{t \to a^{-}} [-a \sqrt {a^2-r^2}]_0^t \\= 4 \pi \times (-a \sqrt {a^2-t^2} +a \sqrt {a^2}] \\= 4 \pi\times (a) \times (a) \\= 4 \pi a^2 $$
