Answer
$8 \sqrt 5 \pi^2$
Work Step by Step
Since, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
We have: $x=2 \cos t \implies \dfrac{dx}{dt}=-2 \sin t$
and $y=2 \sin t\implies \dfrac{dy}{dt}=2 \cos t$ and $z=t \implies \dfrac{dz}{dt}=1$
So, $P(x,y) =ds=\sqrt {4 \sin^2 t+4 \cos^2 t +1}dt=\sqrt 5 \ dt$
The mass of a spring can be computed as:
$\int_C \rho \ ds=\sqrt 5 \int_{0}^{4 \pi} t \ dt \\=\sqrt 5 [\dfrac{t^2}{2}]_{0}^{4 \pi}\\=8 \sqrt 5 \pi^2$
