I'm trying to evaluate the line integral of a circle. My first thought was I could use the fundamental theorem of line integrals to say the line integral is 0, since the starting and endpoint are the same. However, I don't think the vector field is conservative so I don't think I can use the fundamental theorem. I thought I could try solving it with direct parameterization, but I don't have an intuition for whether or not my answer makes sense.
Evaluate the line integral $\int_C (x-y) \,ds$ where C is the circle centered at (1, 2) with a radius of 1.
$r(t) = \langle \cos t + 1, \sin t+ 2 \rangle$ where $0 \leq t \leq 2 \pi$
$r'(t) = \langle-\sin t , \cos t \rangle$
$\lVert r'(t) \rVert = 1$
$\int^{2\pi}_0 f(r(t)) \lVert r'(t) \rVert \,dt = \int^{2\pi}_0 (\cos t +1 - \sin t -2)(1) \,dt = \int^{2\pi}_0 \cos t - \sin t - 1 \,dt$
$\sin t + \cos t - t \mid^{2\pi}_0 = -2\pi$