Evaluate the work integral where $F(x,y)=\langle-y,x\rangle$ over a triangle with vertices $A(-2,-2)$, $B(2,-2)$, $C(0,1)$.
I am not sure how to approach this problem. I tried setting $AB(4,0)$, $BC(-2,3)$ and $CA(-2,-3)$ but I am not sure how to proceed.
Without using Green's theorem
On
The three segments $AB,BC,CA$ can each be parameterized on the interval $[0,1]$ as follows:
\begin{eqnarray} (1-t)A+tB=(1-t)\langle-2,-2\rangle+t\langle2,-2\rangle&=&\langle-2+4t,-2\rangle\\ (1-t)B+tC=(1-t)\langle2,-2\rangle+t\langle0,1\rangle&=&\langle2-2t,-2+3t\rangle\\ (1-t)C+tA=(1-t)\langle0,1\rangle+t\langle-2,-2\rangle&=&\langle-2t,1-3t\rangle \end{eqnarray}
The differential vectors $\langle dx,dy\rangle$ for the three respective segments are $\langle4,0\rangle dt,\langle-2,3\rangle dt,\langle-2,-3\rangle dt$
Thus
\begin{equation} \int_{AB}-y\,dx+x\,dy=\int_0^1(2)(4)+(2-4t)(0)dt=\int_0^18\,dt=8 \end{equation} \begin{equation} \int_{BC}-y\,dx+x\,dy=\int_0^1(2-3t)(-2)+(2-2t)(3)dt=\int_0^12\,dt=2 \end{equation} \begin{equation} \int_{CA}-y\,dx+x\,dy=\int_0^1(-1+3t)(-2)+(-2t)(-3)dt=\int_0^12\,dt=2 \end{equation}
Therefore,
\begin{equation} \int_{ABC}-y\,dx+x\,dy=8+2+2=12 \end{equation}
You will want to split this line integral into three. The first being the line integral from A to B, second from B to C, and third from C to A. The first line integral is along the line $y=-2$ and it is from $ x =-2$ to $x=2$. Thus, it can be parametrized as $x=t$ and $y=-2$ where $-2 \le t \le 2$. From this, we get the integral: $$\int_{-2 }^2 <2,t> . <1,0> dt$$ (The dot means dot product) Using this, I think you can do the rest! Comment if you need more information/help :)