Answer
$$\dfrac{k(41\sqrt {41}-27)}{12}$$
Work Step by Step
Since, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
We have: $x= t^2 \implies \dfrac{dx}{dt}=2t$
and $y=2t \implies \dfrac{dy}{dt}= 2$ and $z=t \implies \dfrac{dz}{dt}=1$
So, $P(x,y) =ds=\sqrt { (2t)^2+ (2)^2 +1^2}dt= \sqrt {4t^2+5} \ dt$
We have: $\rho=kz=kt$
So, the mass the wire can be computed as:
$$\int_C \rho \ ds= k \int_{1}^{3} t \sqrt {4t^2+5} \ dt$$
Substitute $a=4t^2+5 \implies da=8t \ dt$
So, $$\int_C \rho \ ds=k \int_{1}^{3} t \sqrt {4t^2+5} \ dt\\=\dfrac{k}{8} \int_9^{41} [\dfrac{2a^{3/2}}{3} ]_9^{41}\\=\dfrac{k(41\sqrt {41}-27)}{12}$$
