Given the function $F(x,y,z)=(z^2,-z,2y)$ I need to integrate it following the curve $(0,0,0)$ to $(0,1,1)$ to $(1,2,3)$ to $(1,2,4)$. In order to solve this, I split it between three curves $C_1,C_2$ and $C_3$.
$C_1)~ \alpha(t)=(0,t,t)$
$C_2)~\alpha(t)=(t,1+t,1+2t)$
$C_3)~\alpha(t)=(1,2,3+t)$.
In these three curves, $t\in[0,1]$
I got this from using the parametric equation of a segment. Then, I write $F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))$ corresponding to $z^2,-z,2y$ respectively. I evaluate $P(\alpha(t)),Q(\alpha(t))$ and $R(\alpha(t))$, and $\alpha'(t)$ in order to compute each of the three integrals $\sum_{i=1}^3\displaystyle\int_{C_i}Fd\alpha$, using the scalar product between $(P(\alpha(t)),Q(\alpha(t)),R(\alpha(t)))\cdot(\alpha'(t))$, and in the integral using $t=0$ and $t=1$ as extremes.
This leads me to $\frac{1}{2}+3+4=\frac{15}{2}$ as the result, but the solution says it's $\frac{77}{6}$.
Is my approach of the problem correct?