Let $$c: \left[\frac{1}{2},1\right]\times[0,1]\to\mathbb R^2 \\ c(r,t)=\big(r\cos(2\pi t),r\sin(2\pi t)\big)$$$$\omega=\frac{1}{\sqrt{x^2+y^2}}dx\wedge dy$$ Calculate $\int_c\omega$.
Now we have
$c^*(dx)=d(c^*x)=d(r\cos(2\pi t))=\cos(2\pi t)\,\text{d}r-2\pi r\sin(2\pi t) \, \text{d}t$
$c^*(dy)=d(c^*y)=d(r\sin(2\pi t))=\sin(2\pi t)\,\text{d}r+2\pi r\cos(2\pi t)\,\text{d}t$
$\implies \ \ c^*(\text{d}x\wedge\text{d}y)=2\pi r\ \ \text{d}r\wedge\text{d}t$
$\implies c^*\omega=c^*(\frac{1}{\sqrt{x^2+y^2}} \, \text{d}x\wedge\text{d}y)=\frac{1}{r}2\pi r \ \ \text{d}r\wedge\text{d}t= 2\pi\ \ \text{d}r\wedge\text{d}t$
$$\implies\ \int_c \omega=\int_{[\frac{1}{2},1]\times[0,1]}2\pi\ \ \text{d}r \wedge \text{d}r = 2\pi\int_\frac{1}{2}^1 \, \text{d}r \int_0^1\text{d}t=\pi$$
Could anyone look at this calculation and tell me if it's correct? I feel quite uncertain if I applied all computation rules correctly. Thank you!