Answer
$\dfrac{\pi}{6} [17 \sqrt {17}-1] $
Work Step by Step
The surface area of the surface can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$ and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt{1+(-2x)^2+(-2y)^2} dA \\= \iint_{D} \sqrt{1+4(x^2+y^2)} dA $
and, $\iint_{D} dA$ is the area of the region inside $D$.
We can use the polar co-ordinates because of the part $x^2+z^2$
Therefore, we have: $A(S)=\int_{0}^{2 \pi} \int_{0}^{2} \sqrt {1+4r^2} r dr d \theta \\= \int_{0}^{2 \pi} d \theta \times \int_{0}^{2} \sqrt {1+4r^2} r dr $
Substitute $1+4r^2 = a \implies da= 8r dr$
Now, $A(S)=2\pi \times \dfrac{1}{8} \int_{1}^{17} \sqrt a da \\=\dfrac{\pi}{4} [\dfrac{2 a^{3/2}}{3}]_1^{17} \\= \dfrac{\pi}{6} [17^{3/2} -1^{3/2}] \\= \dfrac{\pi}{6} [17 \sqrt {17}-1] $
