I have a curve defined by following parametric equation: \begin{equation} \gamma(\theta) = 1 + 0.5 \times \cos (N \theta) (\cos(\theta),\sin(\theta)), 0 \leq \theta \leq 2 \pi \ \end{equation} I need to calculate the length of the curve between say θ = 0 to θ = 1.0 Formula for calculating the arc length of a curve in polar form is well known (see http://mathworld.wolfram.com/ArcLength.html Eq. 17 therein). Using this formula \begin{equation} s = \int\limits_{\theta = 0}^{\theta = 1.0} \sqrt{r^2 +\left(\frac{dr}{d\theta}\right)^2} \end{equation} where for N = 10, r is given as \begin{equation} r = 1 + 0.5 \times \cos(10 \theta) \end{equation} and \begin{equation} \frac{dr}{d\theta} = -0.5 \times \sin(10\theta) \end{equation} The problem is I can't evaluate the integral in Eq. 17 from the Mathworks link for the above equation explicitly. Obviously one can use Matlab ( with trapz command ) to evaluate the integral for given limits but I want an explicit expression for the indefinite integral in terms of generic θ1 and θ2 ( I have hundreds of such integrals to evaluate and difference between θ1 and θ2 is not constant) How do I evaluate the integral? Is there a substitution possible? Many thanks for help.
I missed dθ in Eq. (2) in my post above. The second equation should read \begin{equation} s = \int\limits_{\theta = 0}^{\theta = 1.0} \sqrt{r^2 +\left(\frac{dr}{d\theta}\right)^2} d\theta \end{equation}
The desired integral can't be expressed in terms of elementary functions; you need special functions, specifically the incomplete elliptical integral of the second kind. See this WolframAlpha link and this Mathworld page. In order to evaluate this special function, you can do this in MATLAB or this in Mathematica.
There was an error in dr/dθ \begin{equation} \frac{dr}{d\theta} = -5*\sin(10\theta) \end{equation} I tried http://www.wolframalpha.com/input/?i=integrate+sqrt%28%281%2B.5*cos%28N*theta%29%29^2%2B%28-N%2F2*sin%28N*theta%29%29^2%29+d%28theta%29 but the page times out. Do I need a pro version? Can this be done in Maxima?