In the derivation of the formula for volume of a solid of revolution, how does $Δx$ "become" $\mathrm dx$? I am currently learning about the formula for the volume of a solid of revolution formed by the rotation about the $x$-axis through 2$\pi$ radians. I believe this is called the "disk" method. 
Referring to the section of this website "Volumes for Solid of Revolution", I am able to fully understand how one would eventually arrive at the formula:
$$
V \approx \sum_{i = 1}^n A(x_i^*)\Delta x.
$$
I also understand the next portion, which states that the exact volume is then:
$$
V = \lim_{n\to \infty}\sum_{i= 1}^n A(x_i^*)\Delta x.
$$
Where I have doubt is what it is next equated with, i.e. this statement of equation:
$$
V = \lim_{n\to \infty}\sum_{i= 1}^n A(x_i^*)\Delta x = \int_a^b{\rm d}x\, A(x).
$$
I understand fully how we can use integration here. What I don not understand is that how "$\Delta x$" is now replaced by $\mathrm dx$. Don't we include $\mathrm dx$ to show that we are "integrating with respect to $x$", not to represent any sort of length? And yet, $\Delta x$ was in fact supposed to represent an extremely small length. How am I supposed to understand "the transformation" of $\Delta x$ to $dx$?

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$The identity (Note that $Δx = \dfrac{b - a}{n}$)$$
\lim_{n → ∞} \sum_{k = 1}^n A(x_{n, k}) · \frac{b - a}{n} = \int_a^b A(x) \,\d x
$$
is the corollary of the definition of Riemann integral. For any $n \geqslant 1$, the notation $Δx$ means a small length, whereas the notation $\d x$ means an infinitesimal length.
In fact, the “derivation” given on that site is just intuition. To prove it rigorously, denote by $D$ the solid and define the section set$$
S(x) = \{(y, z) \in \mathbb{R}^2 \mid (x, y, z) \in D\}. \quad \forall a \leqslant x \leqslant b
$$
Note that for any $a \leqslant x \leqslant b$,$$
(x, y, z) \in D \Longleftrightarrow (y, z) \in S(x). \quad \forall (y, z) \in \mathbb{R}^2
$$
By the definition of area and volume,\begin{align*}
V &= \iiint\limits_D \d x\d y\d z = \iiint\limits_{\mathbb{R}^3} I_D(x, y, z) \,\d x\d y\d z\\
&= \int_a^b \d x \iint\limits_{\mathbb{R}^2} I_D(x, y, z) \,\d y\d z = \int_a^b \d x \iint\limits_{\mathbb{R}^2} I_{S(x)}(y, z) \,\d y\d z\\
&= \int_a^b \d x \iint\limits_{S(x)} \d y\d z = \int_a^b A(x) \,\d x.
\end{align*}
Here $I_B$ is the indicator function.
