Using line integrals and Green formula to calculate force? A force field $F$ = -y$^2$I$+x$j acts on a particle which moves on a closed loop formed by the sides of a triangle with vertices at $(0,0)$, $(1,0)$ and $(0,2)$ in the anticlockwise direction. Find the work done by the force field in two ways.
a)By taking the appropriate line integral 
b)By using Greens formula

I'm assuming the parametrisation is a quarter circle of radius 2, as that's the only shape which would cover the triangle 
  $$r(t) = (2\cos(t), 2\sin(t)), t \in [0,{\pi/2}]$$
  I have no idea how to do the Greens formula method

 A: Given two points $P$ and $Q$ you can parametrize the line between them by $r(t) = P + t(Q-P)$ for $t\in[0,1]$, this has the orientation going from $P$ to $Q$.
In your example you have three points, $P=(0,0),Q=(1,0)$ and $R= (0,2)$. You should draw the triangle that these points define and observe that the anti-clockwise direction starting at $P$ goes to $R$ and then to $Q$ and then back to $P$.
Given the vector field $F = (-y^2,x)$ the circulation of $F$ round the triangle $$\oint_{\text{triangle}} F\cdot ds$$
Can be computed as the sum of the three integrals
$$ \int_{\vec{PR}} F\cdot ds + \int_{\vec{RQ}} F\cdot ds + \int_{\vec{QP}} F\cdot ds$$
I'll do the first one for you. As I noted above $\vec{PR}$ can be parametrised by $r(t) = P + t(R-P) = (0,0) + t((0,2)-(0,0)) = (0,2t)$ for $t\in[0,1]$. So then
$$
\int_{\vec{PR}} F\cdot ds = \int_0^1 F(r(t)) \cdot r'(t) \, dt = \int_0^1 (-(2t)^2,0)\cdot(0,2) \, dt = \int_0^1 0 \, dt =0
$$
Note that this makes sense, because the path $\vec{PR}$ moves along the $y$-axis and as you move along the $y$ axis your direction of motion is perpendicular to the direction of $F$, so $F$ can do no work on you as you move along the $y$-axis.
Can you do the other two from here?
