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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 $D$.

  1. 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!

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\begin{eqnarray} \int_c\alpha&=&\int_1^2c^*\alpha=\int_1^2\frac{c_1(t)c_1'(t)+c_2(t)c_2'(t)}{c_1^2(t)+c_2^2(t)}\,dt=\int_1^2\frac{t\cos^2\phi+t\sin^2\phi}{t^2\cos^2\phi+t^2\sin^2\phi}\,dt\\ &=&\int_1^2\frac{t(\cos^2\phi+\sin^2\phi)}{t^2(\cos^2\phi+\sin^2\phi)}\,dt=\int_1^2\frac1t\,dt=\ln2. \end{eqnarray}

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

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