Answer
$ 9-\dfrac{5\sqrt{5}}{3}$
Work Step by Step
The surface Area of the part $z=f(x,y)$ can be wriiten 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+(x/2)^2+(-1/2)^2} dA $
and, $\iint_{D} dA$ is the area of the region inside $D$.
We have: $
D=\left\{ (x, y) | 0 \leq y \leq 2x, \ 0 \leq x \leq 2 \right\}
$
Therefore, $A(S)=\int_{0}^{2} \int_{0}^{2x} \sqrt {\dfrac{x^2}{2}+\dfrac{5}{4}} dy \ dx \\= \int_{0}^{2} x\sqrt {x^2+5} dx $
Let us suppose that $x^2+5=a \implies da = 2x dx$
Thus, $A(S)= \int_{5}^{0} \dfrac{\sqrt a da }{2} \\= [\dfrac{a\sqrt a}{2}]_{5}^{9} \\=\dfrac{1}{3} [9 \sqrt 9-5 \sqrt 5]\\=\dfrac{10 \sqrt{10}}{24} \\= 9-\dfrac{5\sqrt{5}}{3}$
