Answer
$\frac{4}{3}(10^{\frac{3}{2}} − 1)$
Work Step by Step
$x = t^{2},y=2t, 0 ≤ t≤ 3$
$\int_{C}y ds$ =
$\int_{0}^{3}2t\sqrt {{(\frac{dx}{dt})^{2}}+{(\frac{dy}{dt})^{2}}}ds$=
$\int_{0}^{3}2t\sqrt {{(2t)^{2}}+{(2)^{2}}}ds$=
$\int_{0}^{3}2t\sqrt {{4(t)^{2}}+{4}}ds$=
$\int_{0}^{3}4t\sqrt {{t^{2}}+{1}}ds$=
$\frac{4}{3}(t^{2}+1)^{\frac{3}{2}}|_{0}^{3}$=
$\frac{4}{3}(10^{\frac{3}{2}} − 1)$
