Answer
a) We need to write the parametrization function. $x(t)=0; y(t)=t^2$ where $1 \leq t \leq 3$
Since, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
We have: $x=0 \implies \dfrac{dx}{dt}=0$
and $y=t^2 \implies \dfrac{dy}{dt}=2 t$
So, $P(x,y) =ds=\sqrt {(0)^2+(2 t)^2} dt=2t \ dt$
b) $\dfrac{208}{3}$
Work Step by Step
a) We need to write the parametrization function. $x(t)=0; y(t)=t^2$ where $1 \leq t \leq 3$
Since, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
We have: $x=0 \implies \dfrac{dx}{dt}=0$
and $y=t^2 \implies \dfrac{dy}{dt}=2 t$
So, $P(x,y) =ds=\sqrt {(0)^2+(2 t)^2} dt=2t \ dt$
b) We need to re-write the line integral in terms of $t$ to simplify the integral by using parametric equations.
Now, the line integral is:
$\int_C (x+4\sqrt y) ds= \int_1^{3} 4\sqrt {t^2} dt(2t \ dt) \\=[8\dfrac{t^3}{3}]_{1}^{3}\\=72-\dfrac{8}{3}\\=\dfrac{208}{3}$
