Answer
$\dfrac{4}{3}(10^{3/2} − 1)$
Work Step by Step
Consider $I=\int_{C}y ds=\int_{0}^{3}(2t) \times \sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}ds=\int_{0}^{3}2t\sqrt {{(2t)^{2}}+{(2)^{2}}}ds=
\int_{0}^{3}2t\sqrt {{4t^2}+{4}}ds=\dfrac{4}{3}|(t^{2}+1)^{3/2}|_{0}^{3}=\dfrac{4}{3}(10^{3/2} − 1)$
