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 $dx$?
 A: No, the journey from $\Delta x$ to $dx$ in the notation is there to remind you of how approximating sums, in the end, turn into a definite integral. One could say that "$dx$" in the notation represents a "vanishingly small" length in the definite integral – even if that is lacking in logical rigor. But we don't need logical rigor in notation, as notation is just symbolism.
But let's start with what you've already know: the derivative. Recall that here one of our notational conventions is
$$\tag 1 \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \frac{d y}{d x}.$$
The quantity $\Delta x$ starts off as an ordinary real variable, to be thought of as going to $0.$ Meanwhile $\Delta y,$ which is a function of $\Delta x,$ is dragged along to $0$ as $\Delta x\to 0.$ If the limit exists, the right side of $(1)$ is notation for this limit. Think of $\Delta x, \Delta y$ as turning into the "infinitely small" $dx, dy.$ It's good notation, because it reminds the reader of where everything comes from (although the notation is often abused).
On to the integral: Recall the notational convention
$$\tag 2 \lim_{n\to \infty}\sum_{k=1}^{n}f(x_k)\Delta x = \int_a^b f(x)\,dx.$$
Here $n\to \infty,$ and correspondingly, $\Delta x \to 0.$ If the limit exists, the right side in $(2)$ is notation for this limit. The $\sum$ sign has turned into the tall snake of an $S$ to remind you of "Sum", and $\Delta x$ has turned into $dx.$ It's no ordinary sum: We're indulging in the fantasy of adding up the areas of infinitely many rectangles, of height $f(x)$ and "infinitely small" base length $dx.$ In light of $(1),$ this makes perfect intuitive sense.
The journey from $\Delta x \to dx$ thus has much more to do with the geometric/intuitive ideas underlying the derivative and integral than to a notion of "with respect to".
I have a question for you: Why is this question occurring to you now, in studying solids of revolution, instead of earlier, when the definite integral was defined?
A: $\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$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.
A: When $a<b$ the typographical picture $\int_a^b f(x)\>dx$
has two different meanings:
$${\bf 1}:\qquad \quad\int_a^b f(x)\>dx:=F(b)-F(a)\ ,$$
where $x\mapsto F(x)$ is a primitive of $x\mapsto f(x)$, i.e., $F'(x)=f(x)$ $\>(a\leq x\leq b)$. This is meant when you are talking about "integrating with respect to $x$".
$${\bf 2}:\qquad\int_a^b f(x)\>dx:=\lim_{\ldots}\sum_{k=1}^N f(\xi_k)(x_k-x_{k-1})\ .$$
The "limit of Riemann sums" indicated here is defined in an intricate way which I don't need to explain here.
That (under reasonable assumptions) the two descriptions ${\bf 1}$ and ${\bf 2}$  give the same value is the content of the Fundamental Theorem of Calculus (FTC).
Now in the case of your rotational body you begin with a Riemann sum of type ${\bf 2}$, whereby $x_k-x_{k-1}$ is abbreviated to $\Delta x_k$, resp., $\Delta x$, and $f(\xi_k)$ is the area of a "median slice" cut between $x_{k-1}$ and $x_k$. You have to be aware that the miracle does not consist in the typographical metamorphosis $\Delta x\rightsquigarrow dx$, but in the fact that we obtain a well defined "limit over all refinements", and that this limit can be computed using the FTC.
