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 [𝑎,𝑏], the routine procedure is
a) partition: [𝑎,𝑏] is divided into 𝑛 small intervals [𝑥𝑖−1,𝑥𝑖], for each 𝑖=1,2,⋯,𝑛, denote by Δ𝑈𝑖 the true value of the quantity $U$ correpsonding to [𝑥𝑖−1,𝑥𝑖].
b) approximation: this step is of essence and extremly important, suppose that we can find some continous function, say $𝑓(𝑥)\in C[𝑎,𝑏]$, 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 𝑓(𝑥)∈𝐶[𝑎,𝑏]):
$$
U = \int_a^b f(x)dx.
$$
Step b) leads to Step d) , and it is mathematically rigorous, as 𝑓(𝑥)∈𝐶[𝑎,𝑏].
Now we look at what happens to the area 𝐴 and arc length 𝑠 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!
I think this will work, but it will take some time to write out a full proof.
(See the update below.)
Let $\phi \colon [0, 1] \to [0, 1]$ be the Cantor function.
For real $a, b, k,$ and $h > 0,$ consider this function ($[b, b + k]$ means $[b + k, b]$ if $k < 0$):
$$
\phi^* = \Phi(a, h, b, k) \colon [a, a + h] \to [b, b + k], \ x \mapsto b + k\phi\left(\frac{x - a}h\right).
$$
I will take for granted, without proof for the moment, that $\phi^*$ is differentiable almost everywhere, with derivative zero where it is defined (this follows from well-known properties of $\phi,$ and so it needn't be proved here) and (I expect this to be not too hard to prove, perhaps by expressing $\phi^*$ as a uniform limit of "step-like" functions based on approximations to the Cantor set) the graph of $\phi^*$ is rectifiable, with arc length $h + |k|.$ (According to the Wikipedia article, this is known to be true when $h = k.$ It is intuitively quite clear why this is so, and the proof ought to generalise to the case of distinct $h, k.$)
For $n = 0, 1, 2, \ldots,$ let $s_n$ be the $n^\text{th}$ partial sum of the series:
$$
1 - \frac12 + \frac13 - \frac14 + \cdots = \log2.
$$
Construct a continuous function $f \colon [0, 1] \to [0, 1]$ by glueing together these functions, for $k = 0, 1, 2, \ldots$:
\begin{align*}
\Phi\left(1 - \frac1{2^{2k}}, \frac1{2^{2k+1}}, s_{2k}, \frac1{2k+1}\right) & \colon
\left[1 - \frac1{2^{2k}}, 1 - \frac1{2^{2k+1}}\right] \to [s_{2k}, s_{2k+1}], \\
\Phi\left(1 - \frac1{2^{2k+1}}, \frac1{2^{2k+2}}, s_{2k+1}, -\frac1{2k+2}\right) & \colon \left[1 - \frac1{2^{2k+1}}, 1 - \frac1{2^{2k+2}}\right] \to [s_{2k+2}, s_{2k+1}],
\end{align*}
where:
\begin{gather*}
f\left(1 - \frac1{2^{n}}\right) = s_n \quad (n = 0, 1, 2, \ldots), \\
f(1) = \log2.
\end{gather*}
Then $f$ is differentiable almost everywhere, with derivative zero wherever it is defined, therefore:
[As Paramanand Singh pointed out in the comments, and as I've only slowly come to understand, the expression on the left cannot be understood as a Riemann integral, therefore my answer does not strictly meet the terms of the question. (See the second update, below.)]
$$
\int_0^1\sqrt{1 + f'(x)^2}\,dx = 1.
$$
But for $n = 1, 2, 3, \ldots,$ the graph of the restriction of $f$ to the interval $[0, 1 - 2^{-n}]$ has arc length:
$$
1 - \frac1{2^n} + \left(1 + \frac12 + \frac13 + \cdots + \frac1n\right)
$$
and this is unbounded, therefore the graph of $f$ is not rectifiable.
Update
It turns out to be remarkably easy to prove that the arc length of the graph of $\phi^*$ is $h + |k|.$
@user856's beautifully simple answer to Arc length of the Cantor function says everything that is really needed, but it can be misunderstood, as can be seen from one of the comments on it. The same reservation applies to Dustan Levenstein's brief comment on Elementary ways to calculate the arc length of the Cantor function (and singular function in general), which I believe is a version of the same argument. In the hope of being easily understood, I will labour the proof. I'm sorry!
For $n = 1, 2, 3, \ldots,$ the $n^\text{th}$ stage of the traditional "middle third" construction of the Cantor set yields $m = 2^n - 1$ pairwise disjoint open intervals, the smallest of which have length $\left(\frac13\right)^n,$ and whose lengths sum to $1 - \left(\frac23\right)^n.$ Given $\epsilon > 0$ with $\epsilon < 2h,$ take $n$ large enough that $\left(\frac23\right)^n < \frac{\epsilon}{2h}.$
Arrange the open intervals in ascending order as $J_1, J_2, \ldots, J_m.$
Set $q_0 = 0, p_m = 1.$ In $J_i,$ for $i = 1, 2, \ldots, m,$ take a closed subinterval $[p_{i-1}, q_i],$ where:
$$
q_i - p_{i-1} \geqslant \left(1 - \frac{\epsilon}{2h}\right)|J_i|.
$$
Construct a polygonal chain $Q_0P_0Q_1P_1\cdots Q_mP_m$ of points on the graph of $\phi^*,$ where:
$$
P_i = (a + hp_i, b + k\phi(p_i)),\quad Q_i = (a + hq_i, b + k\phi(q_i)) \qquad (i = 0, 1, \ldots, m).
$$
Because $\phi$ is constant on each of $J_1, J_2, \ldots, J_m,$ and because in particular $\phi(p_{i-1}) = \phi(q_i)$ for $i = 1, 2, \ldots, m,$ the length of the chain is:
\begin{align*}
& \phantom{={}} \sum_{i=1}^m\|P_{i-1}Q_i\| + \sum_{i=0}^m\|Q_iP_i\| \\
& = \sum_{i=1}^mh(q_i-p_{i-1}) + \sum_{i=0}^m\sqrt{h^2(p_i-q_i)^2 + k^2(\phi(p_i)-\phi(q_i))^2} \\
& > h\sum_{i=1}^m(q_i-p_{i-1}) + |k|\sum_{i=0}^m (\phi(p_i)-\phi(q_i)) \\
& > h\left(1 - \frac{\epsilon}{2h}\right)\sum_{i=1}^m|J_i| + |k|(\phi(p_m) - \phi(q_0)) \\
& > h\left(1 - \frac{\epsilon}{2h}\right)^2 + |k|(\phi(1) - \phi(0)) \\
& > h + |k| - \epsilon.
\end{align*}
I hope it will be clear from the Triangle Inequality - without me labouring the details in the same way - that the length of any chain $Q_0R_1R_2\cdots R_lP_m$ of successive points on the graph of $\phi^*$ is at most $h + |k|.$
It follows that the arc length of the graph of $\phi^*,$ defined as the least upper bound of the lengths of all such chains, is well-defined and is equal to $h + |k|.$
Second update
I'll try to explain in gory detail what has been confusing me so much,
to reduce the risk of confusing others! The function
$g(x) = \sqrt{1 + f'(x)^2}$ is defined, and has the constant value
$1,$ on an open set $E \subset [0, 1],$ whose complement (a union of
countably many scaled, translated copies of the Cantor set) has
measure $0.$ Therefore, any extension of $g$ to the whole of
$[0, 1]$ is Riemann integrable, and the value of any such integral
is $1.$ It does not follow that $g$ itself is Riemann
integrable on $[0, 1]!$ There is simply no definition of the Riemann
integral that applies here.
As far as I can tell, the best that can be done using the Riemann
integral is to apply e.g. section 11.2 of Vladimir A. Zorich,
Mathematical Analysis II (first edition 2004), according to which
$E$ is an "admissible set", and:
$$
\int_E\sqrt{1 + f'(x)^2}\,dx= 1.
$$
This is a proper Riemann integral (Zorich also gives a definition of
an improper Riemann integral, which adds nothing here), but I find
this little consolation.
Best Answer
Using Taylor expansions, for small $x$: $$\cos x=1-\frac {x^2}2 +\mathcal O(x^4) \tag{1}$$ and $$\sqrt{1+x+\mathcal O(x^2)}=\sqrt{1+x}+\mathcal O(x^2)\tag{2}$$ With that: $$\begin{split} \frac{1}{n}\sum_{k=1}^{n}\sqrt{\left(2k^{2}+2k\right)\left(1-\cos\left(\frac{2\pi}{n}\right)\right)+1} &= \frac 1 n \sum_{k=1}^n \sqrt{\left(2k^{2}+2k\right)\left(\frac {2\pi^2}{n^2} +\mathcal O\left(\frac 1 {n^4}\right)\right)+1}\\ &= \frac 1 n \sum_{k=1}^n \sqrt{1 + \frac{4\pi^2k^2}{n^2} +\mathcal O\left(\frac 1 {n}\right)}\\ &= \frac 1 n \sum_{k=1}^n\left (\sqrt{1 + \frac{4\pi^2k^2}{n^2}} +\mathcal O\left(\frac 1 {n}\right)\right)\\ &= \left(\frac 1 n \sum_{k=1}^n\sqrt{1 + \frac{4\pi^2k^2}{n^2}}\right) +\mathcal O\left(\frac 1 {n}\right)\\ &\longrightarrow \int_0^1\sqrt{1+4\pi^2x^2}dx \simeq 3.38... \end{split}$$