This is an exercise question from Spivak's calculus on manifolds chapter number 4 question 26.
Show that $\int_{C_{R,n}}d\theta=2\pi n$, and use stoke's theorem to conclude that $C_{R,n}\neq \partial c$ for any $2$ chain $c$ in $\mathbb{R}^2-0$ where $C_{R,n}$ is singular 1 cube $[0,1]\rightarrow \mathbb{R}^2-0$ defined as $t\mapsto (R\cos 2\pi n t,R\sin 2\pi n t)$
$d\theta$ is the $1$ form $$\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ I have tried the following.. I write $\tau$ for $C_{R,n}$.
We have $$\int_\tau d\theta=\int_{[0,1]}\tau^*(d\theta)$$
I am not very much sure how to compute $\tau^*(d\theta)$
$$\tau^*(d\theta)=\tau^*\left(\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right)$$
$$=\left(\frac{-y}{x^2+y^2}\circ \tau\right) \tau^*(dx)+\left(\frac{-x}{x^2+y^2}\circ \tau\right) \tau^*(dy)$$
Now, $$\tau^*(dx)=\frac{\partial(R\cos 2\pi nt)}{\partial(t)}dt=-2\pi nR\sin 2\pi nt$$ $$\tau^*(dy)=\frac{\partial(R\sin 2\pi nt)}{\partial(t)}dt=2\pi nR\cos 2\pi nt$$
$$\left(\frac{-y}{x^2+y^2}\circ \tau\right):[0,1]\rightarrow \mathbb{R}; t\rightarrow -\frac{1}{R}\sin 2\pi n t$$
$$\left(\frac{x}{x^2+y^2}\circ \tau\right):[0,1]\rightarrow \mathbb{R}; t\rightarrow \frac{1}{R}\cos 2\pi n t$$
So, $$\tau^*(d\theta)=2\pi n \sin^2(2\pi nt)+2\pi n \cos^2(2\pi nt)=2\pi n$$
We have $$\int_\tau d\theta=\int_{[0,1]}\tau^*(d\theta)=\int_{[0,1]}2\pi n=2\pi n$$
Suppose we have $C_{R,n}\neq \partial c$ for some $2$ chain $c$ in $\mathbb{R}^2-0$ then
$$2\pi n=\int_{C_{R,n}}=\int_{\partial c}d\theta=\int_{c}d(d\theta)=\int_c 0=0$$
So, we have $2\pi n=0$ a contradiction.. Thus, there can be no $2$ chain $c$ such that $\partial c=C_{R,n}$
Please let me know if this is correct way of solving ...