Same differential expression but different values

Question

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

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

Answer 1

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 ${\rm d}V = {\rm d}x\, {\rm d}y\, {\rm d}z$

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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 $z = 0 \ldots \ell$.

$${\rm d}V = \frac{ \partial \pmatrix{r \cos \theta \\ r \sin\theta \\ z}}{\partial r} \cdot \left( \frac{ \partial \pmatrix{r \cos \theta \\ r \sin\theta \\ z}}{\partial \theta} \times \frac{\partial \pmatrix{r \cos \theta \\ r \sin\theta \\ z}}{\partial z} \right) {\rm d}r\, {\rm d}\theta\, {\rm d}z $$

$$ {\rm d}V = \pmatrix{ \cos \theta \\ \sin\theta \\ 0} \cdot \left( \pmatrix{- \sin \theta \\ \cos\theta \\ 0} \times \pmatrix{0 \\ 0 \\ 1} \right) {\rm d}r\, {\rm d}\theta\, {\rm d}z $$ $$\boxed{{\rm d}V = r {\rm d}r\, {\rm d}\theta\, {\rm d}z}$$

Answer 2

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.