I'm tring to solve the following line integral: $$\int_C (2yx^2-4x) ds$$ where $C$ is the lower half of the circle centered at the origin of radius 3 with clockwise rotation.
However, if I use the parametrization $r_1(t)=(3\cos(t),-3\sin(t))$, $t\in[0,\pi]$, then the result I get is $$\int_0^\pi 3[2(-3\sin(t))(3\cos(t))^2-4(3\cos(t))]=\int_0^\pi 3[-54\sin(t)\cos^2(t)-12\cos(t)]\\ =3[18\cos^3(t)-12\sin(t)]\Big|_0^\pi=-108$$
But if I use the parametrization $r_2(t)=(3\cos(t),3\sin(t))$, $t\in[\pi,2\pi]$, which is the oposite orientation of $r_1$,then I also get $-108$: $$\int_\pi^{2\pi} 3[2(3\sin(t))(3\cos(t))^2-4(3\cos(t))]=\int_\pi^{2\pi} 3[54\sin(t)\cos^2(t)-12\cos(t)]\\ =3[-18\cos^3(t)-12\sin(t)]\Big|_\pi^{2\pi}=-108$$
Shouldn't the result be $108$? What is wrong? I already check the math several times and i'm pretty sure that is not it.