Calculate the perimeter of a circle with a continuously increasing radius I have a circle (if that would even be the correct name for this shape), with a radius function equal to $R=\frac 2\theta + 1$, where $\theta$ is the angle in radians. The domain is between $\theta = 0.25$ and $\theta = 2\pi$.
How do I calculate the total outside perimeter of this shape? I tried to divide it into individual sectors, and find the arc length, but couldn't find a way due to the two radii (in each sector) being different.
The main question is what is a generalised method that could be used to calculate the perimeter of the shape, with an increasing radius (which increases with the angle relative to a starting direction)? The exact function isn't too important.
 A: Remember how one derives the formula for the arc length for $y=f(x)$:
$$L=\int_{x_1,f(x_1)}^{x_2, f(x_2)}dl=\int_{x_1,f(x_1)}^{x_2, f(x_2)}\sqrt{(dx)^2+(dy)^2}=\int_{x_1}^{x_2}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$
We can start with the same formula, but we switch to polar coordinates:
$$x=r\cos\theta\\y=r\sin\theta$$
Since our integration variable is $\theta$, you get $$\frac{dx}{d\theta}=\frac{dr}{d\theta}\cos\theta-r\sin\theta\\\frac{dy}{d\theta}=\frac{dr}{d\theta}\sin\theta+r\cos\theta$$
Square the above equations and add the together. Then you get $$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=r^2+\left(\frac{dr}{d\theta}\right)^2$$
Then $$L=\int_{\theta_1}^{\theta_2}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} d\theta$$
A: There are three main ways to describe a curve


*

*by a cartesian equation $y=f(x)$,

*by parametric equation(s) $x=x(t), \ y=y(t)$,

*by a polar equation $r=r(\theta)$. We are here in this third case with
$$r(\theta)=\frac{2}{\theta}+1$$
Each "way" has its own formulas for the computation of areas, lengths, etc. which IMHO have not to be established again each time one uses them :
In particular, the length formula (6) for polar equations is 
$$L=\int_{\theta_1}^{\theta_2}\sqrt{r^2+\left(\dfrac{dr}{dθ}\right)^2} d\theta$$
Which here becomes :
$$\int_{0.25}^{2 \pi}\frac{\sqrt{(\theta^2+2\theta)^2+4}}{\theta^2}d\theta$$
which has no simple expression... You must use numerical methods for getting an approximate result. I found $16.325196$ with Matlab, which looks good when seeing the curve below (the equivalent of a circle with perimeter $2 \pi R$,  $R\approx 1.5$ + a "line" from $(1.5, 2.3)$ to $(8.5,2)$). 

A: Since $x=r\cos\theta$ and $y=r\sin\theta$, using $\theta$ as the parameter, one have $x=(\frac 2 {\theta} +1)\cos \theta$ and $y=(\frac 2 {\theta} +1)\sin \theta$. Therefore $$ds=\sqrt{(\frac {dx} {d\theta})^2+(\frac {dy} {d\theta})^2}d\theta=\sqrt{(-2\theta^{-2}\cos\theta-(\frac 2 \theta+1)\sin\theta)^2+(-2\theta^{-2}\sin\theta+(\frac 2 {\theta}+1)\cos\theta)^2}d\theta$$.
Now evaluate the integral $$\int _{0.25} ^{2\pi} ds$$.
