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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.

AlephZero
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    You have a scalar (${\rm d}s$) line integral, not a vector ($\cdot{\rm d}\vec{r}$) line integral. Scalar ones are insensitive to orientation, so there's no contradiction here. In general, $\int_C \vec{F}\cdot{\rm d}\vec{r}$ and $\int_C \vec{F}\cdot\vec{T},{\rm d}s$ are not exactly the same thing. The reason being that ${\rm d}s = |\vec{r}'(t)|,{\rm d}t$ doesn't pick a negative sign from a change of orientation as ${\rm d}\vec{r} = \vec{r}'(t),{\rm d}t$ does, due to $|\cdot|$. – Ivo Terek Mar 27 '23 at 05:28
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    ok, I was confused and thought that scalar line integrals also depended on the orientation, thanks a lot. – AlephZero Mar 27 '23 at 05:41

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