Why is the mass of small elements taken as $∆m$ in center of mass of a continuous body? A continuous body has continuous distribution of mass. Doesn't $\Delta m$ mean $m_f - m_i$? But, is the mass Changing? If yes, how is the mass varying? Why is the mass of the small elements in a body taken as $\Delta m$? Why isn't it taken just as $m$ (mass of the small element)?
 A: It's a good question.
Normally speaking, we consider a quantity written as $\Delta x$ to conceptually mean a "change in a quantity". For example a "tiny change in time" or a "tiny change in position".
However, the usage in case of center of mass calculation is in a different conceptual sense. We can think of chopping up the body into some chunks with each chunk being of mass:
$$ dm = \rho dV$$
Here I am considering that we have a 3-D body but it could be that we have 2d or 1d as well.
In a way, we can think of the two approaches as actually being the same. Suppose you had a rod of length L lying along the x-axis from $x=0$ to $x=L$. You can imagine a situation where tiny rods of mass $dm$ are placed consecutively with tail of one at head of previous as time passes to reconstruct the original rod.
A: You are right: it is a confusing notation. Usually it is used to "construct integrals", where it represents a finite-sized small chunk of the body, which is later assumed to go to zero size when a limit is taken.  I prefer to use something like $m_n$ for these "chunk masses", and reserve the $\Delta$-notation for coordinate variables in which your "final minus initial" idea makes sense.
For example, consider a bar of length $L$ which has a changing linear mass density such that if you place one end of the bar at the origin of the $x$ axis, the density is given by $\lambda(x) = \lambda_0 + a x \; (0 \le x \le L)$. What is the mass of the bar?
We start by dividing the bar into a finite number of equal-length chunks, $N$ (which you should think of as something like 8, so you can draw a picture and identify one of the chunks, $m_n$). Then we give an approximate expression for the total mass, $M$, which will become exact once we find a Riemann sum, and take the limit as $N \rightarrow \infty$:
$$
M \approx \sum_{n=1}^N m_n
$$
Notice that I have labeled the 8 chunks $m_1, m_2, \ldots, m_8$.  These are the $\Delta m$ chunks that confused you.  Now express the mass of an individual chunk using the density (at the position of the $n$th chunk, $x_n$) and the length of the $n$th chunk:
$$
M \approx \sum_{n=1}^N \lambda(x_n) L_n
$$
The length of each chunk is $L_n = L / N$, but on our chosen coordinate axis, it is more useful to express it as $L_n = \Delta x_n = x_{n+1} - x_n = \Delta x$.  This lets our approximate expression take the form of a Riemann sum, of which we can then take a limit:
\begin{align}
M &\approx \sum_{n=1}^N \lambda(x_n) \Delta x\\
M &= \lim_{N\rightarrow \infty} \sum_{n=1}^N \lambda(x_n) \Delta x\\
 &= \int_0^L \lambda(x) dx\\
 &=  \int_0^L \left( \lambda_0 + a x\right) dx\\
& = \lambda_0 L + \frac{a L^2}{2}
\end{align}
This idea can be generalized to higher dimensions where either
$$
m_n = \sigma(x,y) \Delta A_n = \sigma(x,y) \Delta x \Delta y
$$
or
$$
m_n = \rho(x,y,z) \Delta V_n = \rho(x,y, z) \Delta x \Delta y \Delta z
$$
and to other, e.g., spherical, coordinate systems where you will need to construct $\Delta V$ in terms of those coordinates.
