Answer
$\dfrac{27-5\sqrt 5}{6}$
Work Step by Step
The surface area is calculated as: $S=\iint_S dS=\iint_D\sqrt {1+f_x^2 +f_y^2} dA$
The equation of line joining the two points $(0,0)$ and $(1,2)$ is given as:
$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}$
or, $\dfrac{y-0}{x-0}=\dfrac{2-0}{1-0} \implies y=2x$
The limits in the region $D$ can be defined as: $0 \leq x \leq 1; 0 \leq y \leq 2x$
Now, $S=\iint_S dS=\int_0^1\int_0^{2x} \sqrt {1+(2x)^2 +(2)^2} dy dx=\int_0^1 \sqrt {5+4x^2} dy dx$
This gives: $S=\int_0^1[y\sqrt {5+4x^2}]_0^{2x}dx=\int_0^1(2x)\sqrt {5+4x^2}dx$
Plug $4x^2+5 =p \implies dp=8x dx$
Then $S=\dfrac{1}{4}\int_5^9 p^{1/2} dP=\dfrac{1}{4}[\dfrac{2}{3}p^{3/2}]_5^9=(\dfrac{1}{6})[9^{3/2}-5^{3/2}]$
Hence, $S=\dfrac{27-5\sqrt 5}{6}$
