Specific example of integrating a 1-form over a curve

Question

I was given the following definition in my course but no corresponding examples:

Supppose $\gamma:[a,b]\rightarrow{M}$ is a smooth curve and $\omega$ a 1-form on $M$ (so $\omega:M\rightarrow{T^*M}$). Then we get a smooth function $t\mapsto\omega(\gamma(t))\gamma'(t)$ (which makes sense since $\omega(\gamma(t))\in{T_{\gamma(t)}^*}M$ and so must act on $\gamma'(t)\in{T_{\gamma(t)}}M$ to return a real number). Then define: \begin{equation} \int_{\gamma}\omega=\int_a^b\omega(\gamma(t))\gamma'(t)\mathrm{d}t \end{equation}

where the right hand integral is the integral over $\mathbb{R}$ in the usual sense.

Now I am trying to compute this in the case of $\gamma(t)=(\cos(t),\sin(t))$ being the unit circle and $\omega$ being the following 1-form: \begin{equation} \omega=\frac{x\,\mathrm{d}y-y\,\mathrm{d}x}{x^2+y^2} \end{equation}

My problem is that I have no idea how to 'plug in' the curve $\gamma(t)$ into the 1-forms $\mathrm{d}x$ and $\mathrm{d}y$ as in the definition and get something I can understand/compute. This isn't an assessed problem but if you're uneasy doing the whole question for me then please can you explain how to generally approach these problems, perhaps through a similar example. Thanks in advance!

Answer 1

This is actually just calc 3, substitute $x=\cos t$ $y=\sin t$ and use $dx=-\sin t dt $ etc.It all simplifys to $\int dt$.

Answer 2

We first note that $\omega(\gamma(t))$ (the $1$-form $\omega$ evaluated at $\gamma(t)$) is simply $\omega(\gamma(t))\gamma'(t)\,dt=\dfrac{x(t)dy-y(t)dx}{x(t)^2+y(t)^2}$. But $dx,dy$ map $\gamma'(t)=(x'(t),y'(t))$ to its components, so $\omega(\gamma(t))\gamma'(t)=\dfrac{x(t)y'(t)-y(t)x'(t)}{x(t)^2+y(t)^2}.$ Hence one regains the 'calc 3' notion of $(x,y)$ being replaced by $(x(t),y(t))$ and differentiated appropriately to get an integral over $t$.