Same differential expression but different values Sorry if this a basic question, but I found two different expressions for the same $\mathrm dV$.


*

*$\mathrm dV$ is the volume of a small cuboid, made of three different sides $\mathrm dx$, $\mathrm dy$, and $\mathrm dz$. Hence, $\mathrm dV=\mathrm dx\cdot\mathrm dy\cdot\mathrm dz$

*$V=xyz$ for a cuboid with sides $x$, $y$ and $z$; so $\mathrm dV=\mathrm d(xyz)$;
Now apply the product rule to get $\mathrm dV=xy\mathrm dz+yz\mathrm
    dx+xz\mathrm dy$.


In both the derivations the only formula we have used is $\text{Volume of a cuboid}=\text{Product of its three sides}$. Then, how can the same $\mathrm dV$ refer to two completely different expressions? 
I at least don't see any way how $\mathrm dx\cdot\mathrm dy\cdot\mathrm dz$ can equal $ xy\mathrm dz+yz\mathrm dx+xz\mathrm dy$. Is this even possible?
 A: They two different things. $V = x y z$ is the total volume enclosed from the origin to a point $\boldsymbol{r} = \pmatrix{x & y & z}$.
The physical interpretation of ${\rm d}V = {\rm d}(x y z)$ is finding how much the total volume changes as the location $\boldsymbol{r}$ changes.

The total volume of the three extensions above is
$$ {\rm d}V = y z {\rm d}x + x z {\rm d}y + x y {\rm d}z $$
But just because you named it ${\rm d}V$ doesn't mean it is the same as the infinitesimal volume element ${\rm d}V = {\rm d}x\, {\rm d}y\, {\rm d}z$


In general, to derive ${\rm d}V$ given a solid described by all points $\boldsymbol{p}(x_1,x_2,x_3)$ within a domain you perform the following calculation
$$ {\rm d}V = \frac{\partial \boldsymbol{p}}{\partial x_1} \cdot \left( \frac{\partial \boldsymbol{p}}{\partial x_2} \times \frac{\partial \boldsymbol{p}}{\partial x_3} \right) {\rm d}x_1\,{\rm d}x_2\,{\rm d} x_3 $$
Where $\cdot$ is the vector dot product, and $\times$ the vector cross product. $x_1$, $x_2$ and $x_3$ are any three independent quantities used to describe the location of all points in the volume.
The above can be derived from the geometry of the problem, and the volume of the tetrahedron volume described by three vectors.
For example, you are integrating over a cylinder with $\boldsymbol{p}(r,\theta,z) = \pmatrix{r \cos \theta \\ r \sin\theta \\ z}$ and $r=0 \ldots R$, $\theta = 0 \ldots 2 \pi$ and $z = 0 \ldots \ell$.
$${\rm d}V = \frac{ \partial \pmatrix{r \cos \theta \\ r \sin\theta \\ z}}{\partial r} \cdot \left( \frac{ \partial \pmatrix{r \cos \theta \\ r \sin\theta \\ z}}{\partial \theta} \times \frac{\partial \pmatrix{r \cos \theta \\ r \sin\theta \\ z}}{\partial z} \right) {\rm d}r\, {\rm d}\theta\, {\rm d}z $$
$$ {\rm d}V = \pmatrix{ \cos \theta \\ \sin\theta \\ 0} \cdot \left( \pmatrix{- \sin \theta \\ \cos\theta \\ 0} \times \pmatrix{0 \\ 0 \\ 1}  \right) {\rm d}r\, {\rm d}\theta\, {\rm d}z $$ $$\boxed{{\rm d}V =  r {\rm d}r\, {\rm d}\theta\, {\rm d}z}$$
A: In Euclidean space, the volume element is given by the product of the differentials of the Cartesian coordinates, viz $$dV=dx\,dy\,dz$$
However, what you have done in the second part is seemingly take a function $f(x, y, z)=xyz$ and composed its total differential 
\begin{align}
df &= f_x \,dx + f_y\, dy + f_z\, dz \\
&=yz\, dx + xz\, dy + xy\, dz
\end{align}
They are both very different objects.
