1 Form Integral along the curve Let $\alpha$ be the $1$-form on $D=\mathbb{R}^2-\{(0,0)\}$ defined by,
$$
\alpha=\frac{xdx+ydy}{x^2+y^2},
$$
where $(x,y)$ are cartesian coordinates on $D$.


*

*Evaluate the integral of the $1$-form $\alpha$ along the curve $c$ defined $c(t)=(t\cos\phi,t\sin\phi)$, where $1 \le t \le 2$ and $\phi$ is a constant.


So far:
\begin{eqnarray}
c(t)&=&(t\cos\phi, t\sin\phi)\\
c'(t)&=& (\cos\phi, \sin\phi)\\
x^2+y^2 &=&t^2\cos^2\phi+t^2\sin^2\phi=t^2\\
\int a(c'(t))&=& \int^2_1{t^{-1}}dt = \ln{2}
\end{eqnarray}
I am not even slightly convinced about this, any input appreciated!

In polar coordinates $\alpha=\frac{1}{r}dr=d\ln r$ and the integral along the straight radial line is simply $\ln2-\ln1$.
