I have the following formula:
$$f = \oint\frac{ds}{C}$$
This integral happens over a (closed) circle with radius $r$, so normally the solution would be:
$$\oint\frac{ds}{C} = \frac{2\pi r}{C}$$ However now I have a circle that has 2 different constants: between $0$ and $\theta$ radians $C$ is changed.
My questions are:
- What is the exact solution
- How to write the boundaries/integral correctly.
For 1. I think: $$ \oint\frac{ds}{C_s} = \frac{r \theta}{C_0} + \frac{r (2\pi - \theta)}{C_1} $$
For the second I got as far as: $$\oint\frac{ds}{C_s} = \int_0^{\theta}\frac{ds}{C_0} + \int_\theta^{2\pi}\frac{ds}{C_1}$$ However this is obviously wrong, as the radius just dropped from the equation. So how do I solve this line integral - properly?
A small image of the problem:
The problem is with the integral notation, as you observe.
Your answer is fine.
For this case, it's probably better to say $$ \gamma_0: [0, \theta] \to \mathbb R^2 : t \mapsto (\cos(t), \sin(t)) \\ \gamma_1: [\theta, 2\pi] \to \mathbb R^2 : t \mapsto (\cos(t), \sin(t)) $$ and then say that your left hand integral is
$$ \int_{\gamma_0} \frac{ds}{C_0} + \int_{\gamma_1} \frac{ds}{C_1}. $$