I want to evaluate the line integral $$\oint_{C} (3y)dx+(2x)dy$$ where $C$ is the boundary of $ 0 \le x \le \pi, \enspace 0 \le y \le \sin{x} $. I found with Green's theorem that the result is $-2:$ $$\oint_{C} (3y)dx+(2x)dy = \int\int_{R} (N_x - M_y) dA = \int\int_{R} (-1) dA = \int_{0}^{\pi}\int_{0}^{\sin{x}} (-1) dA = -2.$$ Then, I tried to parametrize the curve and took the integral as follows, but found $2$:$$x=t, y=\sin{t}, \enspace 0\le t \le \pi \implies r(t)=<t,\sin{t}> \text{ and } F=<3y,2x>=<3\sin{t},2t>$$ and $dr/dt = <1,\cos{t}>$ so that $$F \cdot dr/dt = 3\sin{t} + 2t \cos{t}.$$ Thus, $$\oint_{C} (3y)dx+(2x)dy= \oint_{C} F \cdot dr/dt = \int_{t=0}^{t=\pi} (3\sin{t} + 2t \cos{t}) dt = \int_{t=0}^{t=\pi} 3\sin{t} dt + \int_{t=0}^{t=\pi} 2t \cos{t} dt = 6 -4 = 2.$$
I think the problem is with the orientation of $C$ because in the second case, the orientation is clockwise. Do we always take line integrals with the counter-clockwise direction? Is writing $$\oint_{C} (3y)dx+(2x)dy= \oint_{C} F \cdot dr/dt = - \int_{t=0}^{t=\pi} (3\sin{t} + 2t \cos{t}) dt$$ fix all the problems?