Answer
$\dfrac{2 \pi}{8} [\dfrac{2}{3} \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$
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.
$A(S)=\iint_{D} \sqrt {1+(-2x)^2+(2y)^2} dA= \iint_{D} \sqrt {1+4(x^2+y^2)} dA$
Apply the polar co-ordinates because of the part $x^2+y^2$
$A(S)=\int_{0}^{2 \pi} \int_{1}^2 \sqrt {1+4r^2} r dr d\theta $
Consider $1+4r^2 =u$ and $8 r dr = du$
Thus, $A(S)=(1/8) \int_{0}^{2 \pi} [\dfrac{2}{3} u^{3/2} du]_1^2 d\theta \\ =\int_{0}^{2 \pi} d\theta \dfrac{1}{8} [\dfrac{2}{3} (1+4(2)^2)^{3/2} -\dfrac{2}{3} (1+4(1)^2)^{3/2}] \\ =\dfrac{2 \pi}{8} [\dfrac{2}{3} \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$
