Answer
$\dfrac{2 \pi}{8} [\dfrac{2}{3} \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed 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.
Area of the surface will be: $A(S)=\iint_{D} \sqrt {1+(-2x)^2+(2y)^2} dA \\ = \iint_{D} \sqrt {1+4(x^2+y^2)} dA$
Next, we need to apply the polar co-ordinates for the part $x^2+y^2$
This implies that $A(S)=\int_{0}^{2 \pi} \int_{1}^2 \sqrt {1+4r^2} r dr d\theta $
Substitute $a =1+4r^2$ and $ 8 r dr = da$
Therefore, $A(S)=(\dfrac{1}{8})\times \int_{0}^{2 \pi} [\dfrac{2 a^{3/2} da}{3}]_1^2 d\theta \\ =\int_{0}^{2 \pi} d\theta \times \dfrac{}{8} [\dfrac{2}{3} (1+4(2)^2)^{3/2} -\dfrac{2}{3} (1+4(1)^2)^{3/2}] \\= \dfrac{2 \pi}{8} \times [\dfrac{2}{3} \times \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$
Now, $$A(S) =\dfrac{2 \pi}{8} [\dfrac{2}{3} \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$$
