I have a question about the work done by a vector field to a particle:
The red arrows represent the motion of the particle. The equation of the vector field is $F = [-2y, 3x]$.
The particle moves from $(0,0)$ to $(1,2)$, then to $(1,0)$, and then back to the origin.
Why my textbook integrates from $0$ to $1$ and not from $2$ to 0? Since for the 2nd curve (the one going from $(1,2)$ to $(1,0)$ the $y$ starting value is the $2$ and the end point is $0$ and $x$ is constant: $x=1$).
Note that for the integral from $(1,0)$ to $(0,0)$ we let $x=t$, $y=0$, $dx=dt$, and $dy=0$ so that
$$\begin{align} \int_{(1,0)}^{(0,0)}\left(-2\hat xy+3\hat yx\right)\cdot \left(\hat x\,dx+\hat y\,dy\right)&=\int_1^0 \left(-2\hat x (0) +3\hat y t\right)\cdot \left(\hat x\,dt+\hat y\,(0)\right)\\\\ &=0 \end{align}$$
For the integral from $(1,2)$ to $(1,0)$ we let $x=1$, $y=t$, $dx=0$ and $dy=dt$ so that
$$\begin{align} \int_{(1,2)}^{(1,0)}\left(-2\hat xy+3\hat yx\right)\cdot \left(\hat x\,dx+\hat y\,dy\right)&=\int_2^0 \left(-2\hat x t +3\hat y (1)\right)\cdot \left(\hat x\,(0)+\hat y\,dt\right)\\\\ &=\int_2^0 3\,dt\\\\ &=-6 \end{align}$$
