Answer
$ 9-\dfrac{5\sqrt{5}}{3}$
Work Step by Step
We have: $
D=\left\{ (x, y) | 0 \leq y \leq 2x , \ 0 \leq x \leq 2 \right\}
$
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+(x/2)^2+(-1/2)^2} dA $
Now, $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 $
Consider $x^2+5=t$ and $dt = 2x dx$
So, $A(S)= \int_{5}^{0} \dfrac{\sqrt t dt }{2} \\= [\dfrac{t\sqrt t}{2}]_{5}^{9} \\=\dfrac{1}{3} [9 \sqrt 9-5 \sqrt 5]\\=\dfrac{1}{24} \times 10 \sqrt{10}\\= 9-\dfrac{5\sqrt{5}}{3}$
