Answer
$A(S)=\sqrt{1+a^2+b^2} A(D) $
The result has been proved.
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
We are given that $A(D)$ is the area of that projection.
This implies that $\iint_{D} dA=A(D) ...(1)$
Our aim is to calculate the area of the given surface.
Next we will use calculator to compute the result as follows:
$A(S)=\iint_{D} \sqrt{1+(a)^{2}+(b)^{2}} d A \\ =\iint_{D} \sqrt{1+a^2+b^2} dA \\=\sqrt{1+a^2+b^2} \iint_{D} dA $
Use equation(1), so $$A(S)=\sqrt{1+a^2+b^2} A(D) $$
So, the result has been proved.
