Line integral doesn't depend on parametrization
Let $g:[a,b] \to \mathbb{R}^n$ be a parametric curve
Let $ \phi : [c,d] \to [a,b] \in C^1$
And let $h = g \circ \phi$.
- If $\phi(c) = a$, $\phi(d) = b$, let say that $h$ doesn't change orientation of $g$
- If $\phi(c) = b$, $\phi(d) = a$, let say that $h$ changes the orientation of $g$
--- I want to prove that a line integral over $h$ and $g$ are equal if they don't change orientation, and changes sign if they change orientation.
Let say $f : A \subseteq \mathbb{R}^n \to \mathbb{R}^n$, with $g([a,b]) = h([c,d]) \subseteq A$
$ \int_h f = \int_c^d f(h(t)) h'(t) dt$
$ = \int_c^d f(g(\phi(t))) g'(\phi(t)) \phi'(t) dt$
By a change of variables
$ u = \phi(t) $
$ du = \phi'(t) dt$
If it doesn't change orientation
$ = \int_a^b f(g(u)) g'(u) du$
$ = \int_g f$
If it changes orientation
$ = \int_b^a f(g(u)) g'(u) du$
$ = - \int_g f$
is this proof ok?
Now my real question is when dealing with a scalar field $f : A \subseteq \mathbb{R}^n \to \mathbb{R}$. I don't see how having $|| g'(u) ||$ implies that a change of orientation doesn't change the sign of the integral. Any help?
$\endgroup$ 11 Answer
$\begingroup$- $\phi$ is monotone increasing, $\hat{g}$ is the curve with the new parametrization:
$\begin{align}\int_a^bf(g(t))||g'(t)||dt=\\ =\int_c^df(g(\phi(u))||g'(\phi(u))||\phi'(u)du=\\ =\int_c^df(g(\phi(u))||g'(\phi(u))\phi'(u)||du=\\ =\int_c^df(\hat{g}(u))||\hat{g}'(u)||du \end{align}$
(We used that $\hat{g}'=g'(\phi(u))\phi'(u)$ by the chain rule
- $\phi$ is monotone decreasing:
$\begin{align}\int_a^bf(g(t))||g'(t)||dt=\\ =\int_d^cf(g(\phi(u))||g'(\phi(u))||\phi'(u)du=\\ =-\int_d^cf(g(\phi(u))||g'(\phi(u))\phi'(u)||du=\\ =\int_c^df(\hat{g}(u))||\hat{g}'(u)||du \end{align}$
$\endgroup$ 1
