Is the mass of a body the same in both material coordinate and spatial coordinate? Consider the body $\mathscr{B}$ occupies volume $\mathscr{V}$ in material coordinates (where points are denoted by $\mathbf{X}=(X,Y,Z)$), and occupies region $\mathscr{R}$ with volume $v$ in spatial coordinate ($\mathbf{x}=(x,y,z)$). Assume the density could be obtained by any of the two relations below:
$$
\rho=f(\mathbf{X},t)  
$$
or
$$
\rho = g(\mathbf{x},t)
$$
The question is does the equality below hold?
$$
\int_{\mathscr{V}}fd{V}=\int_vg\mathrm{dv}
$$
, where $dV=dXdYdZ$, and $\mathrm{dv}=dxdydz$.
 A: Yes, the equality holds, if you're using these two sets of coordinates to describe the same physical quantity, i.e. the mass of a physical body.
Physical quantities and equations are absolute, in the sense that they and the process they describe don't depend on the coordinates you use.
Your goal is to compute a physical quantity as the mass of the body $\mathscr B$, i.e.
$M = \displaystyle\int_{\mathscr B} \rho $
and you use 2 different sets of coordinates, X and x, as parameters to describe the physical volume $\mathscr B$, its density $\rho$ and its mass $M$.
(Notation: you should be able to avoid any differential in the physical volume integral).
To easily get a numerical value, then you introduce here two sets of coordinates to parameterize the physical space and the physical properties. If your parameterizations are consistent with the physics, the result you get is independent of the choice of the coordinates
$M = \displaystyle\int_{\mathscr B} \rho = \displaystyle\int_{\mathscr V} f(\mathbf{X}, t) dV = \displaystyle\int_{v} g(\mathbf{x}, t) dv$.
From the mathematical point of view, you're performing a change of variables, $\mathbf{x}(\mathbf{X})$, resulting in the transformation of the elementary volumes
$dv = J(\mathbf{X}) dV = \left| \dfrac{ \partial \mathbf{x}}{\partial \mathbf{X}} \right| dV$,
where $J(\mathbf{X})$ is the Jacobian of the transformation, i.e. the determinant of the gradient of the transformation $ \frac{ \partial \mathbf{x}}{\partial \mathbf{X}} (\mathbf{X})$.
Now, performing a change of variables in the last integral (using $dv = J dV$ and changing the domain of integration from $v$ to $\mathscr{V}$), and comparing with the other expression, you get the relationship between the two functions you used to parametrize the density,
$f(\mathbf{X}) = g(\mathbf{x}) J(\mathbf{X})$.
In this process we don't even need to assume that we're using Eulerian and Lageangian coordinates: it is just a parametrization and a change of variables between parametrizations to describe the same physical quantity, and thus holds for every set of arbitrary coordinates, not only for Lagrangian and Eulerian coordinates.
