# Testing my knowledge of differential forms

• I
• JonnyG

#### 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?

Last edited:

##\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##?

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: $$\int_C dx = \int_0^\pi -r \sin t dt$$
$$= \int_0^\pi r(-\sin t ) dt$$
$$= r(\cos \pi - cos 0)$$
$$= r(-r - r)$$
$$= -2r^2$$

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

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

Wow I see my mistake now. Thanks