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The differential mass element $dmass$ is derived by multiplying the local density $\rho(x,y,z)$ by the infinitesimal volume element $dV$. For the example density $\rho(x,y,z)=xyz$, this gives $dmass = xyz \, dV$.

Conditions: The object has non-uniform density.; The density is modeled as a function of position $(x,y,z)$.; $dV$ denotes an infinitesimal volume element.

The differential mass element $dmass$ is derived by multiplying the local density $\rho(x,y,z)$ by the infinitesimal volume element $dV$. For the example density $\rho(x,y,z)=xyz$, this gives $dmass = xyz \, dV$.

Conditions: The object has non-uniform density.; The density is modeled as a function of position $(x,y,z)$.; $dV$ denotes an infinitesimal volume element.

No, $dV$ is not always equal to $dx \, dy \, dz$. In Cartesian coordinates, $dV = dx \, dy \, dz$, but in other coordinate systems, such as cylindrical or spherical coordinates, the expression for $dV$ changes to account for the geometry of the coordinate grid.

Conditions: The current setup uses rectangular/Cartesian coordinates.; In other coordinate systems, such as polar-type coordinates mentioned verbally, $dV$ may take a different form.

No, $dV$ is not always equal to $dx \, dy \, dz$. In Cartesian coordinates, $dV = dx \, dy \, dz$, but in other coordinate systems, such as cylindrical or spherical coordinates, the expression for $dV$ changes to account for the geometry of the coordinate grid.

Conditions: The current setup uses rectangular/Cartesian coordinates.; In other coordinate systems, such as polar-type coordinates mentioned verbally, $dV$ may take a different form.

For the volume case, the integrand is constant (implicitly 1), so integrating with respect to one variable (like $z$) simply multiplies the area of the base by the height, reducing it to a double integral. For the mass problem with variable density $\rho(x,y,z)=xyz$, the density changes with $z$, so the integration must be performed over all three dimensions to accumulate the varying mass contributions.

Conditions: The goal is volume rather than mass with variable density.; The integrand over the volume is constant for the volume case.; The density varies with position for the mass case.

For the volume case, the integrand is constant (implicitly 1), so integrating with respect to one variable (like $z$) simply multiplies the area of the base by the height, reducing it to a double integral. For the mass problem with variable density $\rho(x,y,z)=xyz$, the density changes with $z$, so the integration must be performed over all three dimensions to accumulate the varying mass contributions.

Conditions: The goal is volume rather than mass with variable density.; The integrand over the volume is constant for the volume case.; The density varies with position for the mass case.

$\rho(x,y,z)=xyz$ represents a variable density function where the density at any point $(x,y,z)$ inside the solid depends on all three Cartesian coordinates. It is used to set up a mass calculation for a non-uniform object.

Conditions: The object has non-uniform density.; The density is modeled as a function of position $(x,y,z)$.; This specific formula is the speaker's simplified example.

$\rho(x,y,z)=xyz$ represents a variable density function where the density at any point $(x,y,z)$ inside the solid depends on all three Cartesian coordinates. It is used to set up a mass calculation for a non-uniform object.

Conditions: The object has non-uniform density.; The density is modeled as a function of position $(x,y,z)$.; This specific formula is the speaker's simplified example.