Answer
$$2\sqrt 2-1$$
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 $y=2t \implies \dfrac{dy}{dt}= 2$
So, $P(x,y) =ds=\sqrt { (2t)^2+ (2)^2 }dt= 2 \sqrt {t^2+1} \ dt$
The mass the wire can be computed as:
$$\int_C \rho \ ds= \dfrac{3}{2} \int_{0}^{1} 2 \sqrt {t^2+1} \ dt$$
Substitute $a=t^2+1 \implies da=2t$
So, $$\int_C \rho \ ds=\dfrac{3}{2} \int_{1}^{2} a^{1/2} \ da\\=\dfrac{3}{2}[\dfrac{2a^{3/2}}{2}]_1^2\\=2\sqrt 2-1$$
