Let $C_1$ be the plane curve defined in polar coordinates as $r(\theta)=\theta, 0 \leq \theta \leq \pi$, and F$: \mathbb{R}^2 \to \mathbb{R}^2$ the vector field F$(x,y)=(2xy-y + \sin x, e^{y^2}+x^2)$. Evaluate $\int_{C_1}$F $d$s orienting the curve such that it starts at the origin and ends at $(-\pi,0)$.
I tried evaluating the integral directly, but the integrand was ghastly. So I checked whether F was conservative, but it wasn't. What I ended up doing was closing the curve to try to apply Green's Theorem
$$C=C_1 \cup C_2$$
With $C_2$ the image of the parametrization $\sigma(t)=(\pi t - \pi,0), 0 \leq t \leq 1$. Let $D=\partial C$. Since $D$ is compact with $C$ piecewise-smooth and F$(x,y)=M(x,y)$i $+ N(x,y)$j is a continuously differentiable vector field, we know that
$$\int_C M dx + N dy = \int_{C_1} M dx + N dy + \int_{C_2} M dx + N dy = \iint_D (N_x - M_y) dA$$
Given that
$$\iint_D (N_x - M_y) dA = \iint_D dA =\int_0^\pi \int_0^\theta r drd\theta=\frac{\pi^3}{6}$$
and that
$$\int_{C_2} M dx + N dy = \int_0^1 \pi \sin (\pi t- \pi)=0 $$
We conclude
$$\int_{C_1} M dx + N dy = \frac{\pi^3}{6}$$
This is probably all wrong. Ugh, where did I trip?