I'm trying to evaluate the line integral for $F(x,y) = \langle \cos x, \sin x\rangle $ for C the triangle with vertices (0,0), $(\pi/3,0)$, $(0,\pi)$.
Using Green's Theorem I'm getting an answer of $3-3\cos(\pi/3)$.
However, when I try and do a more brute force way, I can't get the same answer. This is what I'm doing ..
Paramaterize the curve:
$C_1$ is $r_1 (t) = \langle t, 0\rangle, t\in [0,\pi/3]$
$C_2$ is $r_2 (t) = \langle t, \pi - 3t\rangle, t\in [0,\pi/3]$
$C_3$ is $r_3 (t) = \langle 0,t\rangle, t\in[0,\pi]$
Then
$\int_CF.dr = \int_{C_1}F.dr + \int_{C_2}F.dr + \int_{C_3}F.dr$
$= \int_{0}^{\pi/3} \langle \cos t, \sin t\rangle . \langle 1,0\rangle~dt + \int_{0}^{\pi/3}\langle \cos(t), \sin(t)\rangle.\langle 1,-3\rangle~dt + \int_{0}^{\pi} \langle 1,0\rangle.\langle 0,1\rangle~dt$
$=2\sin(\pi/3) + 3\cos(\pi/3)-3$
... I think I'm setting the problem up wrong.
$C_1$ is $r_1 (t) = \langle t, 0\rangle, t\in [0,\pi/3]$ is $\overline{\langle 0, 0\rangle\langle \pi/3,0\rangle}$
$C_2$ is $r_2 (t) = \langle t, \pi - 3t\rangle, t\in [0,\pi/3]$ is $\overline{\langle 0, \pi\rangle\langle \pi/3, 0\rangle}$
$C_3$ is $r_3 (t) = \langle 0,t\rangle, t\in[0,\pi]$ is $\overline{\langle 0,0\rangle\langle 0,\pi\rangle}$
The path $C_1C_2C_3$ does not trace the triangle, $\triangle\langle 0,0\rangle\langle \pi/3,0\rangle\langle 0,\pi\rangle$ in the right order.
You require line segments $\overline{\langle 0,0\rangle\langle \pi/3\rangle}$, $\overline{\langle \pi/3,0\rangle\langle 0,\pi\rangle}$, $\overline{\langle 0,\pi\rangle\langle 0,0\rangle}$