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Consider a continuous curve $\gamma$ in $\mathbb{R}^3$ parameterized by $\mathbf{r}(t)$ as $t:a \rightarrow b$. Now, it is my understanding that line integrals $$ \int_\gamma \phi \ ds $$ for a scalar field $\phi$ does not depend on the orientation of $\gamma$.

My confusion arises as follows. It is my understanding that $\mathbf{r}(t)$ still corresponds to the curve in question but with reversed orientation if we let $t:b \rightarrow a$. But if this is the case we get $$ \int_\gamma \phi \ ds = \int_a^b \phi \| \dot{\mathbf{r}} \| \ dt \neq \int_b^a \phi \| \dot{\mathbf{r}} \| \ dt = \int_{\gamma_{\text{rev}}} \phi \ ds. $$

Anyone see what I'm missing? Any help is appreciated!

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  • $\begingroup$ There should be a negative sign introduced when you switch from a forward parameterization to a backward parameterization. $\endgroup$
    – random0620
    Commented Apr 19 at 15:50
  • $\begingroup$ If you reverse orientation you are actually doing $\mathbf{r}(-t)$. The chain rule gives your missing minus sign. $\endgroup$
    – Randall
    Commented Apr 19 at 15:51
  • $\begingroup$ Hmm, okay. I guess I don't understand why $\mathbf{r}(t)$ when $t:b \rightarrow a$ would not suffice? Why am I using $\mathbf{r}(-t)$? If so, am I going from $a$ to $b$ or vice versa? $\endgroup$
    – Incubu121
    Commented Apr 20 at 8:31
  • $\begingroup$ Suppose you have $\mathbf{r}(t) = \langle f(t), g(t) \rangle$ for $a \leq t \leq b$, that is, domain $[a,b]$. Neither $b \leq t \leq a$ nor $[b,a]$ make sense, so when you go backwards you are actually changing to the new path $\mathbf{s}(t)=\mathbf{r}(-t) = \langle f(-t), g(-t) \rangle$ for $-b \leq t \leq -a$, that is, domain $[-b,-a]$. Now if you carefully write down the integrals your minus sign issue goes away. Your mistake is thinking that just saying $t: b \to a$ makes the path run backwards, when it actually doesn't make any sense when you look at the **definition ** of a path. $\endgroup$
    – Randall
    Commented Apr 21 at 0:26
  • $\begingroup$ @SamKirkiles How come? Is this just a definition? In other words, are we just defining $\int_\gamma \phi \, d s= \operatorname{sgn}(b-a) \int_a^b \phi(\mathbf{r}(t))\|\dot{\mathbf{r}}(t)\| \, d t$? $\endgroup$
    – Incubu121
    Commented Apr 21 at 6:50

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