Answer
$\dfrac{27-5\sqrt 5}{6}$
Work Step by Step
Equation of line joining the two points $(0,0)$ and $(1,2)$:
$\dfrac{y-0}{x-0}=\dfrac{2-0}{1-0}$
or, $y=2x$
The limits in the region $D$: $0 \leq x \leq 1; 0 \leq y \leq 2x$
Surface area, $S=\iint_S dS=\iint_D\sqrt {1+f_x^2 +f_y^2} dA$
or, $=\int_0^1\int_0^{2x} \sqrt {1+(2x)^2 +(2)^2} dy dx$
or, $=\int_0^1[y\sqrt {5+4x^2}]_0^{2x}dx$
or, $=\int_0^1(2x)\sqrt {5+4x^2}dx$
Plug $4x^2+5 =u $ then we have $du=8x dx$
$S=(\dfrac{1}{4})\int_5^9 u^{1/2} du=(\dfrac{1}{4}) \cdot (\dfrac{2}{3})[u^{3/2}]_5^9$
Hence, $S=(\dfrac{1}{6})[9^{3/2}-5^{3/2}]=\dfrac{27-5\sqrt 5}{6}$
