Undergrad Testing my knowledge of differential forms

[JonnyG]

I am test my knowledge of differential forms and obviously I am missing something because I can't figure out where I am going wrong here:

Let $C$ denote the positively oriented half-circle of radius $r$ parametrized by $(x,y) = (r \cos t, r \sin t)$ for $t \in (0, \pi)$. The value of $\int_C dx$ should give the net change in $x$ as we travel from the beginning to the end of the circle, right? This change should be $-2r$ since $C$ is of radius $r$. However, $x = r \cos t \implies dx = -r \sin t dt \implies \int_C dx = \int_0^\pi -rsint dt = -2r^2 \neq -2r$.

Where am I going wrong?

 

[Infrared]

$\int_0^\pi -rsint dt = -2r^2 \neq -2r$.
Where am I going wrong?

Check this again- where are you getting the extra factor of $r$?

 

[JonnyG]

Check this again- where are you getting the extra factor of $r$?

I made an edit - the half circle is parametrized for $0 \leq t \leq \pi$.

As for the extra factor of $r$, this is my calculation: $$\int_C dx = \int_0^\pi -r \sin t dt$$
$$= \int_0^\pi r(-\sin t ) dt $$
$$= r(\cos \pi - cos 0) $$
$$ = r(-r - r) $$
$$= -2r^2$$

 

[Office_Shredder]

Isn't $\cos(0)$ just 1 though?

 

[JonnyG]

Isn't $\cos(0)$ just 1 though?

Wow I see my mistake now. Thanks