Answer
$\dfrac{1}{6} (17^{3/2}-1)$
Work Step by Step
Since, we have $ds=\sqrt{(dx)^2+(dy)^2} dt$
or, $ds=\sqrt{1+4x^2} dx$
Now, the line integral is:
surface area $\int_C (x+\sqrt y) ds=\int_{0}^{2} x+\sqrt {x^2}(\sqrt{1+4x^2} dx)$
or, $= 2 \int_{0}^{2} (x\sqrt{1+4x^2}) dx$
Let $1+4x^2 a \implies x dx =\dfrac{da}{8}$
Then, $=\dfrac{1}{4} [\dfrac{2}{3} (u)^{3/2}]_{1}^{17}$
Thus, surface area $=\dfrac{1}{6} (17^{3/2}-1)$
