Arc length of a Polar curve as a Riemann sum 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. 

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. 
 A: The problem lies in that which kind of approximation is really sound and right from the point of mathematical view, not just from tuitition, the latter is often right but sometimes goes wrong.
When some (geometrical or physical) quantity, say U, is evaluated using the definite integral on [,], the routine procedure is
a) partition: [,] is divided into  small intervals [−1,], for each =1,2,⋯,, denote by Δ the true value of the quantity $U$ correpsonding to [−1,].
b) approximation: this step is of essence and extremly important, suppose that we can find some continous function, say $()\in C[,]$, satisfying 
$$ 
   \min_{x\in [x_{i-1},x_i]}f(x)\Delta x_i 
    \leqslant
 \Delta U_i
\leqslant 
\max_{x\in [x_{i-1},x_i]}f(x)\Delta x_i .
$$
c) summation: 
$$\sum_{i=1}^n
\min_{x\in [x_{i-1},x_i]}f(x)\Delta x_i \leqslant U=\sum_{i=1}^n \Delta U_i \leqslant \sum_{i=1}^n \max_{x\in [x_{i-1},x_i]}f(x)\Delta x_i 
$$
d) take the limit(note that ()∈[,]): 
$$
U = \int_a^b f(x)dx. 
$$
Step b) leads to Step d) , and it is mathematically rigorous, as ()∈[,].
Now we look at what happens to the area  and arc length  in your reasoning :
(I) The area A: step b) can be made rigorously by
$$
   \min_{x\in [x_{i-1},x_i]}\frac12 r^2(\theta)\Delta\theta_i 
\leqslant 
   \Delta A_i
\leqslant
 \max_{x\in [x_{i-1},x_i]}\frac12 r^2(\theta)\Delta \theta_i
$$
The above inequality is geometrically right.
(II) The arc length s: step b) cannot be made rigorously by 
$$
  \min_{x\in [x_{i-1},x_i]} r(\theta_i)\Delta \theta_i
\leqslant
 \Delta s_i
\leqslant 
  \max_{x\in [x_{i-1},x_i]} r(\theta_i)\Delta \theta_i
$$
The above inequality does not hold geometrically. 
I hope this might be helpful!
