Divide the interval into $ n $ equal parts, $ a = x_0 \lt x_1 \lt \cdots \lt x_n = b $, with $ x_{i+1} = x_i + \Delta x_i $.
Suppose you want to approximate the curve between $(x_i,f(x_i))$ and $(x_i+\Delta x,f(x_i+\Delta x))$. You could simply approximate it with the straight line between the two points, whose length is
$$\sqrt{\left( f(x_i+\Delta x) - f(x_i)\right)^2 + (\Delta x)^2}.$$
In the picture below, the black line is the graph $y=f(x)$, and the green line is the line that joints $(x_i,f(x_i))$ on the bottom left and $(x_1+\Delta x,f(x_1+\Delta x))$ on the top right.
Then you would have that the arc length is approximated by the sum of the lengths
$$\text{Arc Length} \approx \sum_{i=1}^n \sqrt{(f(x_i+\Delta x) - f(x_i)))^2 + (\Delta x)^2}$$
and take the limit as $n\to \infty$. Unfortunately, the expression in the sum is not of the form necessary to view it as a Riemann sum, so you cannot turn that limit into a limit of Riemann sums, and from there to an integral.
So we take a slightly different approach. Instead of approximating the length of the curve from $(x_i,f(x_i))$ to $(x_i+\Delta x, f(x_i+\Delta x))$ with the straight line between the two points, we will approximate it with the tangent line to the graph of $f$ at $x_i$, from $(x_i,f(x_i))$ to the point $x_i+\Delta x$. This is the blue line in the picture above.
If $\Delta x$ is small, then we know the tangent line is a very good approximation for the curve on $[x_i,x_i+\Delta x]$, so the line will be a good approximation to the length of the curve.
Now, the tangent line to $y=f(x)$ through the point $x_i$ is given by
$$y = f(x_i) + f'(x_i)(x-x_i).$$
At $x_i+\Delta x$, the line goes through $f(x_i) + f'(x_i)\Delta x$.
So this tangent line goes from $(x_i,f(x_i))$ to $(x_i+\Delta x ,f(x_i)+f'(x_i)\Delta x)$. The length of the line between those two points is
\begin{align*}
&\sqrt{\Bigl( (x_i+\Delta x) - x_i\Bigr)^2 + \Bigl((f(x_i)+f'(x_i)\Delta x) - f(x_i)\Bigr)^2}\\\
&\quad = \sqrt{ (\Delta x)^2 + \left(f'(x_i)\Delta x\right)^2} \\\
&\quad = \sqrt{\left(1 + \left(f'(x_i)\right)^2\right)\Delta x^2} = \left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x.
\end{align*}
Adding all of these, we get an approximation to the arc length:
$$\text{Arc Length} \approx \sum_{i=1}^n \left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x.$$
Now, these can be seen as Riemann sums. So if we take the limit as $n\to\infty$, the approximation gets better and better (because the tangent gets closer and closer to the curve, giving a better approximation). At the limit, we get the exact arc length, and the limit of the Riemann sums becomes the integral. So
\begin{align*}
\text{Arc Length} &= \lim_{n\to\infty}\sum_{i=1}^n\left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x\\\
&= \int_a^b \sqrt{1+(f'(x))^2}\,dx.{}{}{}
\end{align*}
By the arc length formula in the polar coordinates,
$s = \int_{0}^{\theta}\sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta$
Since $d\theta = \frac{d\theta}{dr}dr$,
$s = \int_{0}^{r}\sqrt{r^2 + (\frac{d\theta}{dr})^{-2}} \frac{d\theta}{dr}dr$
Hence $s = \int_{0}^{r}\sqrt{1 + r^2(\frac{d\theta}{dr})^2} dr$
Since $r = a\sqrt{2\cos 2\theta}$,
$\frac{dr}{d\theta} = -\frac{2a\sin 2\theta}{\sqrt{2\cos 2\theta}}$
Hence $\frac{d\theta}{dr} = -\frac{\sqrt{2\cos 2\theta}}{2a\sin 2\theta}$
$(\frac{d\theta}{dr})^2 = \frac{\cos 2\theta}{2a^2\sin^2 2\theta}$
$\cos 2\theta = \frac{r^2}{2a^2}$
$\sin^2 2\theta = 1 - \cos^2 2\theta = \frac{4a^4 - r^4}{4a^4}$
Hence $(\frac{d\theta}{dr})^2 = \frac{r^2}{4a^4 - r^4}$
Therefore $s = \int_{0}^{r}\sqrt{\frac{4a^4}{4a^4 - r^4}}dr = \int_{0}^{r}\frac{2a^2}{\sqrt{4a^4 - r^4}}dr$
Best Answer
In cylindrical coordinates, i.e. $(r,\theta,z)$ with $x=r\cos\theta,y=r\sin\theta$ we have $$ x'(t)=r'(t)\cos\theta(t)-r(t)\theta'(t)\sin\theta(t),\quad y'(t)=r'(t)\sin\theta(t)+r(t)\theta'(t)\cos\theta(t). $$ Therefore $$ (x'(t))^2+(y'(t))^2=[(r'(t))^2+r^2(t)(\theta'(t))^2]\cdot[\cos^2\theta(t)+\sin^2\theta(t)]=(r'(t))^2+r^2(t)(\theta'(t))^2, $$ and the formula to compute the arc length in cylindrical coordinates is: $$ \int_a^b\sqrt{(r'(t))^2+r^2(t)(\theta'(t))^2+(z'(t))^2}\,dt, $$ assuming that the curve is defined on $[a,b]$.