Answer
$$A(S)=\sqrt{1+a^2+b^2} A(D) $$
Thus, the area has been verified.
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.
The area of the given surface is: $A(S)=\iint_{D} \sqrt{1+(a)^{2}+(b)^{2}} d A \\ =\iint_{D} \sqrt{1+a^2+b^2} dA $
and $A(S)=\sqrt{1+a^2+b^2} \iint_{D} dA $
Now, we are given that $A(D)$ is the area of that projection.
This implies that $A(D)=\iint_{D} dA$
So, we have:
$A(S)=\sqrt{1+a^2+b^2} A(D) $
Therefore, the result has been verified.
