## Thomas' Calculus 13th Edition

$\dfrac{1}{6} (17^{3/2}-1)$
As we know that the surface area is: $ds=\sqrt{(dx)^2+(dy)^2} dt$ or, $ds=\sqrt{1+4x^2} dx$ Now, $S=\int_C (x+\sqrt y) ds=\int_{0}^{2} x+\sqrt {x^2}(\sqrt{1+4x^2} dx)$ or, $S= 2 \int_{0}^{2} (x\sqrt{1+4x^2}) dx$ Let $1+4x^2 =p \implies x dx =\dfrac{1}{8} dp$ or, $S=\dfrac{1}{6} [(p)^{3/2}]_{1}^{17}$ Thus, surface area $S=\dfrac{1}{6} (17^{3/2}-1)$