Answer
$\dfrac{\pi}{6} (13 \sqrt {13}-1)$
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.
Our aim is to calculate the area of the given surface.
$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 $D$.
Apply the polar co-ordinates because of the part $x^2+y^2$ .
$A(S)=\int_{0}^{2 \pi} \int_{0}^{\sqrt 3} \sqrt {1+4r^2} r dr d \theta $
Set $1+4r^2 = u$ and $du= 8 r dr$
$A(S)=\int_0^{2 \pi} \int_{1}^{13} \dfrac{\sqrt u}{8} da d \theta \\=(\dfrac{1}{8}) (\dfrac{2}{3} ) \int_0^{2 \pi} u^{3/2} d \theta \\= \dfrac{1}{12} [ \theta (\sqrt {(13)^3}-1)] \\=\dfrac{2 \pi}{12} (13 \sqrt {13}-1) \\= \dfrac{\pi}{6} (13 \sqrt {13}-1)$
