Answer
$\dfrac{27-5\sqrt 5}{6}$
Work Step by Step
The surface area can be found as: $S=\iint_S dS=\iint_D\sqrt {1+s_x^2 +s_y^2} dA$
The equation of the line joining the two points $(0,0)$ and $(1,2)$:
$\dfrac{y-0}{x-0}=\dfrac{2-0}{1-0}$
or, $y=2x$
$S=\iint_S dS=\int_0^1\int_0^{2x} \sqrt {1+(2x)^2 +(2)^2} dy dx$
or, $=\int_0^1[y\sqrt {5+4x^2}]_0^{2x}dx$
or, $=\int_0^1(2x)\sqrt {5+4x^2}dx$
Suppose $4x^2+5 =k \implies dk=8x dx$
Then $S=(\dfrac{1}{4})\int_5^9 [k^{1/2}] dk$
or, $=(\dfrac{1}{4})(\dfrac{2}{3})[k^{3/2}]_5^9$
Hence, $S=\dfrac{27-5\sqrt 5}{6}$
