Answer
$\dfrac{\pi}{6} (13 \sqrt {13}-1)$
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} \\=\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 $x^2+y^2$
Therefore, we have:
$$A(S)=\int_{0}^{2 \pi} \int_{0}^{\sqrt 3} \sqrt {1+4r^2} r dr d \theta $$
Substitute $1+4r^2 = a \implies da= 8 r dr$
$$A(S)=\int_0^{2 \pi} \int_{1}^{13} \dfrac{a^{1/2}}{8} da d \theta \\=\dfrac{1}{8} \times \int_0^{2 \pi} \dfrac{2}{3} a^{3/2} d \theta \\= \dfrac{1}{12} [ \theta (\sqrt {(13)^3}-1)] \\=\dfrac{2 \pi}{12} (13 \sqrt {13}-1) \\=\dfrac{\pi}{6} (13 \sqrt {13}-1)$$
