I need a help with the calculation of the following line integral:
$$\int_{\kappa} arctg\frac{y}{x}dy-dx,$$
where $\kappa$ is the part of a parabola $y=x^2$ from point (1,1) to (0,0).
So far, I have tried to parametrize the parabola:
$(x-m)^2=2p(y-n)$ for $V=[m,n] =[0,0]$ and $A=[x,y]=[1,1]$, from which I got that $p=1/2$.
I have now $(x-0)^2=1/2(y-0)$ and by setting $x=t$ I obtain $y=2t^2$. But I am stuck here, unable to calculate the start and end interval for the integration. Even if I knew the interval, I am still unsure how to finalize the calculation. Any help will be much appreciated.
For the case of the parabola $y=x^2$, parameterize $\kappa$ by $\vec r(t)=\langle t,t^2\rangle$ with $0\le t\le1$. Then
$$\begin{align} \int_\kappa\left(\arctan\frac yx\,\mathrm dy-\mathrm dx\right) &= \int_0^1 \left\langle-1,\arctan\frac{t^2}t\right\rangle\cdot\frac{\mathrm d}{\mathrm dt}\left[\left\langle t,t^2\right\rangle\right]\,\mathrm dt\\[1ex] &=\int_0^1(2t\arctan t-1)\,\mathrm dt \end{align}$$
In case $V$ is supposed to denote the vertex of the parabola, there's only one other possible curve that fits the description, namely $\vec r(t)=\langle t^2,t\rangle$, in which case the integral would be
$$\begin{align} \int_\kappa\left(\arctan\frac yx\,\mathrm dy-\mathrm dx\right)&=\int_0^1\left\langle-1,\arctan\frac t{t^2}\right\rangle\cdot\frac{\mathrm d}{\mathrm dt}\left[\left\langle t^2,t\right\rangle\right]\,\mathrm dt\\[1ex] &=\int_0^1\left(\arctan\frac1t-2t\right)\,\mathrm dt\\[1ex] &=\int_0^1(\operatorname{arccot}t-2t)\,\mathrm dt \end{align}$$