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Consider the plane curve $\gamma$ in polar coordinates: $ r=r_0+e^{\lambda\theta}, \quad \theta_1 \le \theta \le \theta_2, $ where $r_0,\lambda,\theta_1>0$. Is it possible to compute explicitly the length of $\gamma$?

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    @Mercy Actually, your question is "Is it possible to compute explicitly the length of $\gamma$?" The answer is "yes" and some posters have given some clues as how to go about the calculation.2012-07-17

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There is a general way of doing this, but I assume you are not familiar with it so let's look at your particular example. In Cartesian coordinates, your curve looks like

$ \gamma (\theta) = \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix} = \begin{pmatrix} \left(r_0 +e^{\lambda \theta}\right)r\cos\theta\\ \left(r_0 +e^{\lambda \theta}\right)r\sin\theta \end{pmatrix}$ Therefore, the tangent is $ \frac{d\gamma}{d\theta}= \left( \begin{array}{c} e^{\theta \lambda } \lambda \cos\theta-\sin\theta \left(e^{\theta \lambda }+r_0\right) \\ e^{\theta \lambda } \lambda \sin\theta+\cos\theta \left(e^{\theta \lambda }+r_0\right) \end{array} \right)\ ,$ and its norm is calculated easily using trigonometrical identities to be $\left|\frac{d\gamma}{d\theta}\right|=\sqrt{e^{2 \theta \lambda } \left(1+\lambda ^2\right)+2 e^{\theta \lambda } r_0+r_0^2}$

The length is given by integrating $\left|\frac{d\gamma}{d\theta}\right|$ from $\theta_1$ to $\theta_2$.

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    @Mercy I plugged it in Mathematica. It gives a nasty result which is much too long and cumersome to copy here, but it is a "closed form" as you hoped.2012-07-17
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This can be very slightly less excruciating. Every polar curve $r = f(\theta)$ can be written as the parametric equations $x(\theta) = f(\theta) \cos(\theta)$, $y(\theta) = f(\theta)\sin(\theta)$. You can do a chain rule calculation with a little trigonometric magic to crunch it down to $ds = \sqrt{x'(\theta)^2 + y'(\theta)^2}\,d\theta = \sqrt{r^2 + r'(\theta)^2}\,d\theta.$

You have $r(\theta) = r_0 + e^{\lambda\theta}.$ Then $r'(\theta) = \lambda e^{\lambda \theta},$ so $ds = \sqrt{(r_0 + e^{\lambda\theta})^2 + \lambda^2 e^{2\lambda\theta}}\,d\theta = \sqrt{(1 + \lambda^2)e^{2\lambda\theta} + 2r_0 e^{\lambda\theta} + r_0^2}\,d\theta$ This agreees with the form obtained by yohBS above.

You then need to embark down this path $\int ds = \int \sqrt{(1 + \lambda^2)e^{2\lambda\theta} + 2r_0 e^{\lambda\theta} + r_0^2}\,d\theta = \int {\sqrt{(1 + \lambda^2)e^{2\lambda\theta} + 2r_0 e^{\lambda\theta} + r_0^2}\over e^{-\lambda x}} e^{\lambda\theta}d\theta$ Now put $z = e^{\lambda \theta}$ and you get left with $\int ds = {\sqrt{(1 + \lambda^2)z^2 + 2r_0 z + r^2_0}\over z}\, dz $ Now try completing the square and doing a trigonometric substitution.