Suppose we have a curve in polar plane satisfying the equation $r=f(\theta)$ with $\theta\in[a,b].$ To find the area enclosed by this curve in this range of $\theta$ using Riemann integrals, we partition $[a,b]$ into sub-intervals such that $a=\theta_0<\theta_1<\cdots<\theta_{n-1}<\theta_n=b$ and, then the area is given by (defined by) limit of the Riemann summation $$\lim_{\max\Delta\theta_k\to 0}\sum_{k=0}^n\dfrac12(f(\theta_k^*))^2\,\Delta\theta_k$$ where $\theta_k^*$ is an arbitrary point in $[\theta_k, \theta_{k+1}]$ and $\Delta\theta_k=\theta_{k+1}-\theta_{k}.$ Therefore the problem reduce to evaluate the definite integral $$\dfrac12\int_{a}^{b} r^2\,d\theta.$$

The whole idea this process is that we can approximate area covered by the graph on $[\theta_k, \theta_{k+1}]$ by a (infinitesimal) circular sectors shown as in the below picture.

enter image description here

If we suppose that arc length of the curve $\delta S_k$ on $k$-th sub-interval is approximately equal to the arc length of (area approximating) circular sector, then $\delta S_k\approx f(\theta_k^*)\Delta\theta_k$ for any $\theta_k^*\in[\theta_k, \theta_{k+1}].$ This leads to the conclusion that arc length of the curve is given by the Riemann summation $$\lim_{\max\Delta\theta_k\to 0}\sum_{k=0}^nf(\theta_k^*)\,\Delta\theta_k$$ which is (as a notation) equivalent to the definite integral $$\int_{a}^{b} r\,d\theta.$$

However, this is completely different from the standard formula that we used to compute polar arc lengths given as $$\int_{a}^{b} \sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta.$$ I know this formula can be derived from the arc length formula of Cartesian coordinates, but I would like to know an explanation in terms of Riemann summations. Also I am curious why the "approximating by circular arcs" approach does not leads to the same formula. Thank you in advance for your consideration.

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    $\begingroup$ You are encountering the "staircase paradox" : math.stackexchange.com/q/12906 $\endgroup$ – Jean Marie Feb 5 at 0:53
  • $\begingroup$ Your reaction ? Of course, here, it is not in the case of a circle that you will have a paradoxical situation because your formula and the classical formula give the same result, $2\pi$. $\endgroup$ – Jean Marie Feb 5 at 8:47
  • $\begingroup$ @JeanMarie: Thank you very much for your link. I see your point, but still couldn't understand why second formula is correct over first one. $\endgroup$ – Bumblebee Feb 5 at 16:14

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