# A Line integrals of differential forms

1. Oct 29, 2016

### spaghetti3451

Consider a curve $C:{\bf{x}}={\bf{F}}(t)$, for $a\leq t \leq b$, in $\mathbb{R}^{3}$ (with $x$ any coordinates). oriented so that $\displaystyle{\frac{d}{dt}}$ defines the positive orientation in $U=\mathbb{R}^{1}$. If $\alpha^{1}=a_{1}dx^{1}+a_{2}dx^{2}+a_{3}dx^{3}$ is a $1$-form on $\mathbb{R}^{3}$, then its integral or line integral over $C$ becomes

$\displaystyle{\int_{C}\ \alpha^{1} = \int_{C}\ \sum\limits_{i}a_{i}(x)dx^{i}}$

$\displaystyle{=\int_{a}^{b} F^{*}\bigg[\sum\limits_{i}a_{i}(x)dx^{i}\bigg]}$

$\displaystyle{=\int_{a}^{b} F^{*}\bigg[\sum\limits_{i}a_{i}(x(t))\frac{dx^{i}}{dt}\bigg]dt}$

Thus $\displaystyle{\int_{(U,o;F)}\alpha^{p}:=\int_{(U,o)}F^{*}\alpha^{p}}$ is the usual rule for evaluating a line integral over an oriented parameterized curve! We may write this as

$\displaystyle{\int_{C}\alpha^{1}=\int_{a}^{b}\alpha^{1}\left(\frac{d{\bf{x}}}{dt}\right)dt}$

and so the integral of a $1$-form over an oriented parameterized curve $C$ is simply the ordinary integral of the function that assigns to the parameter $t$ the value of the $1$-form on the velocity vector at ${\bf{x}}(t)$.

What does it mean for $\displaystyle{\frac{d}{dt}}$ to define the positive orientation in $U=\mathbb{R}^{1}$

2. Oct 29, 2016

### Orodruin

Staff Emeritus
$d/dt$ is a vector on the real line.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted