Answer
$\dfrac{7}{3}$
Work Step by Step
Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= t dt$
This implies that
$\int_C f(x,y) ds=\int_{0}^{\sqrt 3} (1+t^2)^{1/2} t dt$
Thus, we have
$ (\dfrac{1}{3})[(1+t^2)^{3/2}]_{0}^{\sqrt 3}=\dfrac{7}{3}$
